Transcript for:
Volume of Solids by Integration

so we're at the point of of our of our series here where we do get to talk about applications of what you can do with this today we're going to discuss how you find the volume of certain figures and how you do it with discs and washers now I know you're probably thinking discs and washers what in the world does that mean what do those little washer doughnut looking things and discs have to do with finding volume or even integrals but I'm going to show you that it does it actually does so here's the idea I'm going to try to build this up for you so you kind of understand where this stuff's coming from instead of just give you the formula just like I did for derivatives just like I did for integrals I'm going to do the same thing for volume you follow me now do you remember the raan sums where we took a whole bunch of rectangles and found arbitrary points and then added their areas together yeah okay we are going to do the same thing with volumes you ready for it so I'll try to break it down again it's not going to be too too bad but we're going to talk about first the the volumes by slicing and what that means is what if we did the same thing we did for areas areas we we had this figure the one I draw the time and we sliced it up right and we made lots of little rectangles we made lots of rectangles we added up the widths times the heights of all those rectangles there by getting the areas of the rectangles that's basically the idea that we're going to use only on a volume type application so let's talk about that so right now we'll discuss the volume by slicing is it different liid solids as as a solid figure okay solid figure uh you have we don't mean we don't mean a piece of cake is different than Jello that's I was thinking that's what I mean Jello goes like this putting like this cake sticks together no not that kind of solid I mean like you have this this threedimensional shape that doesn't have any holes through the middle of it okay that's a solid so it could be liquid it doesn't matter what it is it's contained and it does it's not changing shape you understand that's all no holes in the middle of it either so here's the deal let's take some random fig that I'm going to create right now that so that that's maybe the side view of this figure and let's make it three-dimensional is that threedimensional enough for you can you see it okay okay so that's my my figure here's the idea what if we could take and make slices of this this thing if we can find the volume of a slice then we should be able to add all the slices together and find the volume of the whole figure do you see how this kind of parallels the area under a curve idea so we're going to try to find s let's let's make a slice so some random slice of this thing I'll take it here you guys having a hard time doing the whole threedimensional thing M okay yeah it's kind of difficult you get used to it after a while do you see the the slice I'm trying to portray right there so basically this is a weird shaped cake you did a bad job on your cake but takes two planes right two pieces of paper or laser beams or something we have this Nice Slice of this this cake you're going to serf someone your crappy made cake you got it no icing even what kind of a host are you but anyhow that's that's what we're going to do so idea is let's cut this into thin slabs we'll call it a slicing or a slab and we're going to try to use a rem sum use our sums that we used before to set up some sort of integral so this is our idea I'm going to write it out I'm going to try to walk you through this so you understand where is coming from so idea cut into thin slabs then we'll use a remon sum or or our summations so we use our summations to set up an integral you make a little side note here this is much much like the area problem very much only it's in three dimensions so in order to do this though well how do we find I'll kind of break this down so you understand it do you let me start over then do you understand what's going to happen with this slab what's going to happen that slab if it's much like the area problem what's going to happen to that slab to get very very thin very very thin very thin now think of a volume a volume area for you guys typically is base times width time height right but that's only of an of a rectangle it's only of a rectangle and things don't have to be rectangular so I'll say it this way volume is typically the surface area times its length does that make sense so whatever the surface area is or the in our case the crosssectional area times its width so if I found the cross-sectional area so I cut it this like I cut it this way I take my slab out I turn it towards you that's the cross-section can you picture that three dimensionally you cut like here I'm cutting it I pick it up I turn it this way what you're looking at right now is the cross-section if I take the cross-sectional surface area times its width that's going to give me a very good approximation of its volume right if the width of that slab goes like this that gives me an exact interpretation of that volume because it has very little width does that make sense to you very much like finding the area of those rectangles say hey I can find the area now take it to zero that's now an exact area if I add them all together and that's what we're doing here except in terms of volume so in order to do this we need to find the area of the cross-section to do this I mean that we got to find the area of crosssection well in general we would do this I'll try to break down so you see where all this stuff's coming from if we had just a a basic like rectangular prism if we had just a basic do you see the rectangular prism if we had just a basic rectangular prism and it was I don't know going through through some axis like that like maybe the x- axis if we wanted to find the cross-section well what we would have to do is take a slab or take a slice to find the area and we' slice it like this oops that's way off there we go we do some sort of a slice now tell me something if this distance is X it's on the x- axis and this distance is let's say oh what did I call it I think I call it y for that and this distance is y and this is z could you find the surface area of the cross-section would it have anything to do with X what would it have to do with Y and Z just Y and Z for the crosssection so so check this out if I asked you to find the volume of this whole thing here's how you do it you'd probably say oh well I know the volume of a rectangular prism the volume of a rectangular prism is just the base time the height time the width so so basically it's the y * the Z times the X do you follow me on that now I'm going to kind of prove to you that this is the cross-sectional area times the length you do you agree that this is the volume of this figure no matter what what's y * z y * Z is the surface area of the crosssection that's that's what that is so basically this says oh that's the surface area of the crosssection time x the length so this is surface area of cross-section and that's a length well now now here's the idea if that's the case then what we're basically doing here is making a whole bunch of these things do you see it making a whole bunch of those things and then adding up all the volumes very similar to making rectangles for areas oring rectangular PRS for volumes and we're going to add them all up so what the idea I needed to get across to you was that a volume is basically a surface area time length how many people feel okay with that surface area time length gives you the volume good deal now we can solve any volume where the solid is bound by planes that are perpendicular to the x-axis here here let me let me explain it to you if this was like this and it was not straight up and down could you cut it into slabs it wouldn't work right it wouldn't be a slab to start with that would be a problem here slabs yes slabs that are perpendicular that's very nice and easy we can do that so now we we we've got it down to if our sides are perpendicular to the x-axis we can start cutting it into slabs and approximating approximating our rectangular prisms do you see that the point if they're not perpendicular x-axis then this method is not going to work for us we have to do something else but if the sides are perpendicular we're good to go so now we can find the volume of any solid that's Bound by planes that are perpendicular to the x-axis you know I probably should say at points A and B so at at where we're starting and where we're stopping so if we only wanted to find the volume between here and here it would just have to be perpendicular at those points the rest of it really wouldn't matter just like the area under a curve doesn't matter if it's undefined it's just between the two intervals that that we talk between the interval that we talked about are are you foll me on that one so be at points at points a [Applause] basically where you start where you stop would you like to see how to make this integral up it's kind of cool it's kind of fun I think you'll enjoy it there's going to be a lot of picture drawing well to start off with a lot of picture drawing you ready for it sure I'm going to go back to this because I really don't want to draw that anymore are you guys all right with that I don't want to draw that uh it's kind of annoying so let's just say that from this picture I'm going to go back and you already know the idea I'm going to be cutting a whole bunch of slabs out you know maybe a little redraw it darn it yeah I'll redraw it cuz I have to talk I talk about other things plus I'm going to reverse it because it's easier to draw the top of it as at an angle so let's say that I do same idea but I do this one that's a little off there and I'm going to try to give you every step along the way so my first goal I'm going to cut into slabs okay slabs okay there's one I'm going to cut another one there's two I'm going to cut another one you see my slabs yes no I'm gonna cut all of these so dot dot dot all the way down my figure I'm going to cut this into a whole bunch of slabs and I'm going to do it this way I'm going to cut the slabs so that they're the same width this is going to make it easier to add together just like we did with the area problem I hopefully you're seeing the similarities between this and the area problem so I'm going to cut into slabs the same width how did I represent the width of my areas and how am I going to represent the width of my slabs yeah do you remember that we used Delta X to represent the width of that that little sub interval so I'm going to cut this into slabs Delta X yde so all these little little intervals all those little Cuts those slabs are all Delta x what you still so far so good yes no okay now would you mind if I took one of those slabs and just kind of blew it up not like but like made it bigger guessing some of your wishing we could just blow it up huh yes no more math too bad for you this is two dimensional you cannot blow it up maybe you could I bet you want to anyway I'm take one of these guys one of these slabs and make it quite a bit larger oh ran out space let me RW pretty close do you see it but I'm trying to get across to you oh this is going to be difficult not too bad so I've taken one of these slabs right here and I've made it just bigger so you can see a better picture now can you remember what we did on the area problem I cut each one into intervals what did I do within each interval someone tell me what did I do wasn't necessarily in half but you're on the right track arbitrary Point arbitrary Point great you guys both had the same time so I'm going to pick some random point I could put it in the middle it's going to look like it's in the middle but it doesn't necessarily have to be in the middle explain to me why the arbitrary point doesn't have to be in any particular Point taking limit it's going to do that yeah so wherever it is it's going to be within there right but it's going to be so darn close to everything else it's not going to matter where it is so I'm going to pick an arbitrary Point what did I use for arbitrary point do you remember X XK dot or c subk one or the other I'm going to use XK do because that's what I grew up with so arbitrary Point XK dot or C subk doesn't really matter um on the interval okay okay so that means that down here I'm gon to go all right on my width this is my sub interval that's Delta X this width is Delta X do you guys understand that this Delta X on this let's see Delta X all the way to the bottom Delta X on this little sub interval I am going to pick some arbitrary point now Sarah said the middle I'm going to put in the middle it doesn't have to be there just so we can see it easier I'm going to pick some random arbitrary point XK dot now think back to the area on the area problem when I found my XK dot what did I do with that XK dot remember the area problem you had the width right the width was very easy what were you missing what did I do with the XK dot the function and that gave me my height height now what I'm looking for here please please watch this I hope you're you're seeing the parallels here what I'm looking for here is not the width the width is going to be easy the width is going to be from A to B because it's it's bound it's bound by planes like this it's going to be from A to B this is not what I'm worrying about what I'm worrying about is this part this part is my cross-sectional area so what I'm doing here the width is easy the W width is Delta X in general it's going to be from A to B that's easy part in an integral that's very easy A to B no problem what I'm caring about is how in the world do I find the cross-sectional area because when you think about it this is not a rectangular prism prism this is going up at an angle right so what I need to do is find some place to get the cross-sectional area the place I choose to find my cross-sectional area isn't this far right right one or this far left one necessarily it's some arbitary point XK dot we just talked about how XK dot really doesn't matter cuz I'm going to smash them together so XK dot simply allows me to find my cross-sectional area of each sub interval say that one more time so you get it arbitrary Point what we're doing with the arbitrary point is finding the cross-sectional area right there that's the way that we're going to find the well the cross-sectional area surface area that way I can multiply it by its width again the arbitrary point doesn't matter because when I take a limit of it it's all going to smash together anyway so this point says hey find cross-sectional area at XK dot so pick an arbit Point arbitrary Point XK dot on each sub interval and find cross-sectional area at that point cool that means to me that I'm going to go down here and I'm going to do this my purple line my purple line that goes around that that's my cut at XK dot do you see that what that's going to create is just some surface area for me yeah you know so we're going to find the cross-sectional surface that's cross-section cross-sectional surface area at that point again it doesn't matter where it is since Delta X is going to become really small doesn't matter so let's see oh we're about ready we're about ready can I work over here so I don't get all bound up you don't want to get bound up in calculus calculus constipation then you're bound to get a brain fart get it that was pretty funny come on brain fart lose it on I just made that up that deserves more than your courtesy that was good just think about that later watch the video again it'll be on I'll leave it on there for you so you can enjoy that over your two-day break what's uh V subk arbitr Point nope that's XK dot what's v subk what's v volume volume subk means of whatever little interval I'm talking about you get it so this is every little interval so the volume at the first interval or second interval is going to be equal to well the area at x k Dot times what do you think times what what am I doing up here how do you find volume again what how do you find volume what are you what are you doing for volume we're we're beyond that yes it's length time width time height for a rectangular prism we do not have one of those here surface area times width or length yes that's what we're doing so the surface area is the area at XK dot do you get it that's what I'm trying to represent here the area at XK do the surface area times the width how long is the width very good Delta X this is cross-sectional area times length just like we thought of it cross-section are times L now how many intervals does this stand for one arbitrary interval very good so the this is just one interval how do I find all of them so basically I'm just talking about one slab right now how do I say I want to do all the slabs a sum let's add them all together so if we say okay this is one SL then this if I drop the K I say oh if I want more than one slab I'm going to have to add them all up starting at the first one and going through the nth one that should look familiar it's just like ARA right now only inside of it says oh find the area at XK dot time Delta X hey that's just for each one what this says is take every slab that you have every slab find a random point for each slab find the area of the cross-section at each random Point multiply each one of them times its width and add them all up that should give you an approximation for the volume approximate volume okay now you use your Calculus how do you make it better take liit as very good did you hear him over there said take the limit as n approaches Infinity that's going to say instead of a finite number of slabs which would be a pretty good approximation making an infinite number of slabs if you pack an infinite number of things into a finite space it moves it really close to zero doesn't become zero because you have to have some sort of length don't you if Delta X were zero you get a volume of zero that wouldn't make any sense you have to have something there but it's so small that it's not going to be an approximation anymore it's so small that that's going to be exactly what the volume is so yeah exactly take a limit so this is an approximation how I make it exact is I go all right take a limit as n goes to Infinity of the same thing that we had and I'm hoping you can figure out what what comes of this I think I told it to you before what's a limit of a sum what's a limit of sum that's what it is it's the interal we'll be we'll be doing these through using interal through well I'll explain that in just a bit okay just understand that this is going to give you an integral from where you start to where you stop where A and B are the starting and ending points of your volume and they have to be perpendicular to the xaxis otherwise you can't make slabs follow me a of x k dot becomes a of X the area function of your of your function and then the Delta X becomes DX and a of X is the cross-sectional area over the interval from A to B would you wish I feel okay with what we talked about so far now are you going to have to do ran sums no I'm just proven it to you why would you do Ron sums that's way easier shoot don't go that whole process but I proved it to you didn't I showed you how this all works that way when I give you this you're like that doesn't make sense well yeah it does it actually does it says this is cross-sectional area with whatever the width is and you're adding them all up from A to B that's all it says it's just a different way to say this right here easier to calculate though because you know how to take an integral don't you yes awesome U by the way does it have to be along the x- AIS no oh yeah did you have any questions on this I don't do it on purpose I think IW it yeah no see me after we could probably do the same thing according to the y- axis except except instead of being perpendicular to the the X it just be perpendicular y turn your head it's got to be that way so we do the same thing with respect to y so in our case the volume would equal hey where would you start where would you stop C or D C D C small to big of whatever the H whatever the area of the crosssection was according to Y and then you'd have D yeah that would be it s DX just checking sure sure you're awake good yeah Dy it's such a habit to put the DX isn't it you always want to put that but watch watch your variables so basically what this says in English is the volume of a solid is found by integrating the cross-sectional area from A to B that's what that says in English the the volume is found by integrating the cross area area of that cross-section of a threedimensional space from well C to D in this case or from A to B would you like to see an example about how this is done it's kind of cool it's not not too bad let give you this one sorry for those of you who are not thre dimensionally inclined these are probably really hard to draw so my bad well actually your bad I'm not the one screwed up I really am kind of a nice guy out of class but you wouldn't know it would you all the jerk things I say uh this is a one here and a five one and five now would you like to see how this is done appropriately firstly we're going to have to find now this is kind of a these are simplistic examples to start with just so you know very simple why are they very simple might make it complicated I'm going to make it complicated yes why is this simple does the function ever change no it's a uniform function the whole way okay it's going to be uniform U aerial cross-section the crosssection cross-sectional area the whole way through so what we know is that the volume is going to be the integral of the cross-sectional area from where the function starts and from where the function stops now first thing does this qualify do I have planes that make 90° of the xaxis or the Y axis one or the other yeah this is this is 90° and that's 90° so that's that's great that's what we need can you please tell me what's the area of the cross-section how are you getting that well you got to Circle at one end and that's your sure so if we cut this thing right down if we cut this and then turn it and look at it that's going to be a circle do you see the circle so the circle is pi r s cool okay in our case what is it how much is the pi r s what's the r uh the radius sure one the radius is one radius go well can you see the radius is one diameter is two radius has to be one so for us it's Pi 1 2 would you all agree that the cross-sectional area of this figure is simply Pi yes and it doesn't change there's no X involvement because it doesn't change it says it's a constant it's a it's a circle stacked up right next to each other so what our our Theory says here is that the volume should be equal to an integral from where it starts where does it start along the x- axis where does it start I'm start okay one to where of the cross-sectional area DX because we're going along the x- AIS that's an X so P how many people feel okay with this so far cross-section area it's just the area of a circle that's Pi in this case radius is one that is crosssectional area that's a constant there are no X's because it does not change if it changed we would have X's for sure it' be a function in terms of X now we're going from 1 to five because it's a restart there a restop there where our planes are perpendicular no problem hey can you find integral of Pi DX not too hard what's the integral of Pi DX Pi with the little thing that's the evaluation symbol from where to where so this says Pi * 5 - < * 1 what's < * 5 - < * 1 4 Pi that's kind of neat we just used well it is neat Until you realize you can do this without calculus but it's kind of neat that you're able to find the volume using calculus now let's just be clear that this is true because this is simply a cylinder right what's the volume of a cylinder High squ height hey think about think about what you're doing okay think just think the volume of a cylinder is the base area times the height you basically Stack Up circles that is literally exactly what we're talking about here it's the area and you stack up all the circles that's what we're doing same thing so it should work out the same thing volume is 5 * 1^ 2 * height the height in this case is really the length was 5 - 1 that's four huh 1 s is 1 * 4 is 4 45 this might not seem so impressive to you but getting the same answer is kind of neat right now unfortunately for unfortunately these don't always happen right you're not always going to get that formula now he's going to get that did you understand that so far though now we can just take it a step further we're still kind of basic basic okay we're going are we going slow enough for you yeah you understand everything right this slow up then we're going to start talking about some more unique shapes not just cylinders but we're going to get some some other things so what we'll do now is say let's suppose you have some function and you start revolving it around an axis do you see that it's going to create a solid The Sweep out area is going to create some sort of a solid you with me let's let's get a picture of those before we do anything too drastic let's see what they're they're making and what we're going to call this subsection is a solid of Revolution a solid Revolution means the solid you get if you take some like a piece of paper or solid shape and revolve it around an axis you're going to get a solid of Revolution let me give you a for instance here if you took that shape and you what is that shape this you took that shape and you revolved it around the x-axis what's it going to make so basically you do this you'll sweep it out that's a horrible idea in see if I can find something better this is much more there now you see it take it revolve it that's just created a cylinder in space do you see the cylinder in space open right to the right page too awesome so this if I did yeah that's going to create exactly what we had right here this is basically a rectangle that's sweeping out an area an a solid Revolution sorry a volume out volume so this would create this shape basically I didn't draw it very nicely but it creates that cylinder that we were just talking about well I really jacked That Thing Up forget this hang on I got to make it nicer for you guys than that that's basically the shape that you get what about this so what if I took that yeah if I took a half circle and I revolved it I'm not going to get a half sphere am I full sphere which means that to get our our solids you can still deal with a function because if I took a full sphere and revolved it sorry a full circle and revolved it it still make the same exact shape actually some people would say Well it sweeps out twice the volume because have twice the sides but it makes the same exact overall shape it makes the sphere still so this if I did yeah sure I'm going to get GL not an egg if it's a half circle I bet hopefully it's a it's a sphere what if I took this figure what would I get if I swep that around [Music] mm sure it's going to give me a cone absolutely I about one more it's kind of a fun one yeah I'll [Music] how about that a donut what do you think another with the hole so a paper towel roll you basically get a paper towel roll that's what you get you're going to take this solid thing and sweep it around but there's a hole in the middle of it so dut I couldn't say yes because Donuts have it's actually called a Taurus they have curved sides you can't do that this will have the straight side still because you're sweeping that out it is rectangular but it's going to have this hole right in the middle of it so this would be basically a cyinder within a cylinder you sweep out this you leave this you see the figure I'm trying to draw for you it's that threedimensional the solid space is actually this and this and this empty the empty cylinder in the middle doesn't contain anything you have to swept anything out for that now the reason that we're going to go a little further is these shapes are pretty easy these These are easy these we have standard formulas for this this would be hey cylinder this is sphere we have formulas for the volume of the sphere we have formulas for the volume of the cone we have formulas for this it's a cylinder minus cylinder that that's all that is what we don't have formulas for as a Where I Leave You is this what if I said now what I'd like you to do is take my favorite function from A to B and sweep that thing out now that's not a regular shape that actually has a curved Edge so I can't use any of those geometrical principles but what if I did that well do you see the picture I'm going to get kind of weird looking that's kind of like that isn't it basically a base basically this and it's also a basee from A to B now two questions before I let you go number one thing is it still a solid yes cool number two thing are the sides still perpendicular to the x-axis which is cool which means that we're going to be able to use this stuff that we talked about to do something like this and that's what we'll start next time all right well we're going to wrap up talking about volume of these threedimensional objects which is a really cool idea what it says is that if f is firstly continuous and if it's bounded between a and b bounded between a and b and we rotate it around the x-axis well if you think about that picture as bounded by by A and B another way to say this by the way if you get this on your homework notice that you could get instead of bounded by A and B it could say say bounded by y = sorry x = a and x = b are you okay with that x equals A and B are vertical lines are they not that would be bound between those two vertical lines vertical lines are perpendicular to the x-axis so if we have a continuous bounded function between a and b and we rotate about the x-axis it will have sides that are perpend to the X and this implies that we have a solid that's been rotated around the x-axis using that we can talk about our solves of Revolution we just talked about those last time we talked about cylinders we talked about the uh rectangle rotated around the xaxis Mak cylinder we had that paper towel looking thing right toilet paper looking thing where we rotated a rectangle that was hovering above the x-axis we had a cone we had a sphere we had those shapes now it says well if you have just some random function f ofx as long as it's continuous as long as it revolved around the x-axis in such a way that sides are perpendicular to the x-axis we should be able to use some method in order to calculate that volume would you like to find out what it is sure it's kind of cool and I don't think it's going to take too much for you guys really to get did it um what I'll do first is draw you a picture of the Revolution so I'm going to draw this in three dimensions or try to at least that's close enough feel look with the shape see where it's coming from so rotating that solid sweeping out the whole volume question so the sides that are think there the the front the back that's what you mean I mean the front and the back that's exactly right so right here this is perpendicular it looks like it's a little curve but that's just because I want to throw the show the three dimensions but if we had this thing here it'd be it'd be completely fat flat just like a vase sitting on a table right it's it's flat and it's flat it's the plane is perpendicular to the x-axis right now are you all right with that now questions I have for you uh first question put something right in the middle what's that point say it again yeah exactly if we have something starting a ending in b the threedimensional space should also start at a and end at B that's sure for sure next question I have for you oh this is a little off isn't it that's a little better if I were to take this figure and I were to slice it because that's what we're about to do is find the volume by slicing if I were to sliced this just right right in the middle like this slice right there what is the cross-sectional shape going to be Circle why a revolution and both sides are perpendicular okay very good so we have a circle you understand that right now let me ask a question is this some random function or is this a specific function have I used like x s have I used any specific function here I just took a general curve so my question is if I take any continuous curve between a and b and I revolve it what's the cross-sectional shape always going to be Circle always going to be a circle you understand that if I take a a shape and I revolve it it's going to give me some Circ circular or C yeah circular slicing that's kind of interesting right so every one of these if I look at the slice it's going to have a distance what do we call that distance if this is my my shape what do we call that say what radius that's right now what we're going to do is we're going to find the volume by slicing here is the idea that we came up with last time what we came with last time is that the volume should be some integral of the surface area from A to B across my interval do you remember doing that from last time refresh your memory on your notes or or the video that's what we came up with from our Sol Revolution we said hey length time width time height is great but we don't always have the same length time width time height so what we're going to do is say hey let's just find the the surface area of a cross-section and we'll integrate it from A to B what that says is that the surface area can change shapes and sizes because we have a variable in there we just need to find a function that represents that Sur liate again we need to find a function that represents that surface area and then we can integrate it integrate it is just adding up all those little slices from here to here do you follow me on that so we'll slice them all up real fine like real nice and then add all those things up that's adding all the surface areas of each little slice from A to B that's what that means tell me will feel okay with that so far all right that's the idea now we're going to get a little bit more specific I'll walk you through every step firstly note the a ofx represents the surface area of the cross-section you okay with that area equals the surface area of the cross-section that's what we mean there now we just talked about this this is the cool this is the cool part oh my gosh this is a really cool part actually the surface area of the cross-section what is the cross-section what was that shape that we just talked about ccle ccle very good that's PK r s that's a circle note that anytime you revolve something around the x-axis the cross-section will have have to be a certain you're not going to do this you're not going to go uh revolve Square no that doesn't work right when you revolve something you're you're evenly re rotating it around so it's going to sweep out a circular pattern that's what it's going to do whether it's really big or really smaller changes the cross-section at one cutting is going to be a perfect circle you follow me and the area the perfect circle is going to be P Pi R square whatever the r happens to be now let's talk a little bit more about R from here how in the world are we going to find R because we would definitely need R up here wouldn't we definitely need R what is our R don't don't think overthink the arbitrary Point thing we're a little bit past that that's how we came up with this idea was arbitrary point at each sub interval and then taking it to Infinity so we have very little width between there it's the the r is okay think about r as being perpendicular to the xaxis then R is the same thing is the height of your function at any given point x right so that you know that your height here is just F ofx if this were some point and now maybe I see what you're saying if you pick some point x sub one then the height of that one point is f of x sub1 or the same thing at the as the radius at that point so here's what I'm going to say if this is a circle then the radius of that Circle has to equal F ofx at any point x yes no the is is the is the radius the height of your circle at this point the radius is the height of your circle right and changes depending on where your X is the radius would be here the radius would be here that also is exactly the same thing as the function the height of the function so what I'm trying to get at is just just look at the board here real quick the height here which is the function's height is the same as the height here which is the radius is that true therefore the function equals the radius that's another way to look at it so radius equals the function well that's kind of cool that let us make a substitution if we have a subx which doesn't make sense right now because that's not in terms of X is it and we needed it in terms of X so a ofx = PK r² a ofx = < not r because R is not in terms of X right now we need to make it in terms of X what's going to do wi in there and then what very good how many able to follow that hey guess what you're done you're done that is the method cuz look what we were looking for is a of X the area in relation to some variable X and now we have it this is in terms of x what it says is in order to find the volume by slicing you go from where the interval starts to where interval stops of Pi FX s DX and it says you know what it's not even that hard it says you take your function you plug it in you square it and you multiply by pi now someone tell me something about the pi what can you do with the pi you can always bring it out because that's a constant that will always be there I show it here so you don't forget about it but you can easily pull that out of your integral this right here this is called the method of diss why are they called the meth method of diss do you think making SES in a circular object yeah which is a disc you're basically adding up adding up discs circular you're adding up discs that's why they call the method discs now would you like to see an example yes okay let's do a couple of them first one's going to be real basic basic here's how some of these things are worded find the volume of the solid of revolution where I'm going to give you some function let me give you some interval is revolved around the x-axis so we're going to find the volume of the solid Revolution that means we're going to be revolving something of Y = 3kx on the interval 1 to 4 and that's being revolved about the xaxis so we're going to be making some circular shape it's roled around the x that means it's can have perpendicular sides which means we can do the slicing thing if we had this we couldn't do the slicing thing because we won't be able to make circles we have to be able to make circles for this to work you got it that's what where the perpendicular comes in now also this could be this could be worded a little bit differently I need to be prepared for that I'm not going to write out but you can you can hear the difference this could be worded in a manner such as this find the volume of the solid of revolution of the area Bound by y = 3 < tkx x = 1 x = 4 and the X AIS does that make sense to you it says take it between this function whatever this function happens to look like it looks like that about the x = 1 x = 4 and the xais that comes up with the same exact area that you're going to be sweeping out you see the point so it can be worded differently don't let the wording Jack you up it's going to give you balance it's going to give you a function and we're revolving that so again this could be written just like this how I'm going to hopefully word it for you or it could be written find the the volume of the solid Revolution that is bound between this y x = 1 x = 4 those are your vertical lines and the x- axis that's your horizont or yal Z could be the x-axis as well you okay with that so far so that's the picture we're going to have we're going to take this thing and spin it around now we go real slow and see if we can do this together so we know that volume volume is some integral from A to B of Pi fx^ 2 DX let's see if we can fill that out appropriate thing where is my integral going to start do you think very good where's you going to stop what's going to be inside of my integral first thing oh yeah you're certainly going to have a pi for sure that doesn't change and then you're going to have the function f ofx now in our case what is f ofx 3 rootx very good so we'll have 3 rootx and what do I do with that 3 rootx do I know mhm notice how the three is with that rootx in this case because that is part of your function right you're squaring the whole function so we have our PI we got our PI we got our F ofx that's 3 rootx got our Square don't forget the square last people forget the square can't forget the square don't forget the square you feel okay with this so far notice how the integrals are going to be not so bad well sometimes they are a be I'll give the honesty sometimes they're nasty because you're squaring function I mean if I had given you like x^2 - 3x + 4 we had to square that that's not that fun uh but this one's going to be at least a little bit nicer you can see it right so don't forget to square it what we'll get is 1 to 4 Pi tell me about this what happens there say what now both parts not just the square root but also the three so what are we going to get that was pretty nice see how nice of a guy I am today told you I was in a good mood yeah we get 9x see where 9x is coming from tell me something else I can do with this integral good I'm going to pull up both the nine and the five from 1 to four of X DX no substitution necessary nothing hard don't make these harder than they have to be if you can do an integral without a substitution and it's fairly easy do it okay do that don't trick yourself on these problems you're going to get 9 pi times what so we'll be evaluating from one to four you know I'm going to show you a slightly different way you can consider this this is another way you can do this problem instead of having the the two here with the four and the one you know that's a constant right you can pull out those constants before evaluation so here I could write this as 9 piun / 2 * x^2 from 1 to 4 that's also an appropriate way to write it so you could just look at something a little bit smaller there you just have to not forget about the 9 Pi / 2 but when you do you're going to get 4 2 - 1 2 that's going to be 15 hey it works if you do it that way of course it works if you do it that way it has to be the same yeah it's going to work either way don't forget the pi too I don't want 135 over two don't forget that Pi that changes things what did we just find by the way volume that space I drew before that's being rotated around the x-axis it was it'd be easy to find the area but now we can find the volume how we're doing is we're saying okay all we're basically doing is this finding the height of the function at each point revolving that height around it's making a dis and adding up the surface area of those disc that's what that's doing how many people understand this example now would you like to do something a little bit cooler this is pretty cool already but what I want to do now I want to see if we can come up with the formula for a sphere by doing some of this calculus stuff you ever found out you ever know I mean I'm sure there's many ways to come up with a formula for the volume of the sphere but have you ever thought about actually how they did it how would you do it take take this sphere and say hm well that's got to be uh 4i R Cub over three obviously I mean clearly if you have a sphere that's going to be the volume right which is the volume of I think it is 4 pi r Cub over three should give you a volume of a sphere so how do they find that how would you find it would you actually cut up a sphere and try to organize it into rectangles and add them probably not water displacement you could do that you could definitely do water displacement as long as you know the exact value of Pi which no one does but then come with a rectangle that has that exact thing it's going be hard to do probably estimate it yeah probably estimate it but to get an exact wait why don't we use some calculus and figure that thing out let's see if we can find derive the volume of a sphere you ready for this sure this is good stuff to me the first time I saw this I was like dang that's pretty cool I thought that was really interesting now let's think about how you would get a sphere what shape would you have to revolve around the x-axis in order get a sphere half circle why not a full circle because be redundant it would be redundant it would say well I I don't need that extra part just a half circle would do it wouldn't it take a half circle revolve it all the way around it's going to create a sphere so let's start with that all right try to do it twice and look about the same must be the way I'm going to do it okay so we don't need this bottom part we don't we don't need this but that's what it's going to create is this bottom section when I sweep it around it's going to create some sort of sphere how far is this and how far is this from each other no I mean how how far what's the coordinate of that x coordinate Z z r would be I'll give you that z r is this one that's R how far is this one how far is this one Whatever the radius is That's How what it happens to be so notice that the the only the only two things you know about circle is where the or where the center is and how much the radius is and we know the center is at 0 0 these that's what we're implying and the radius is R whatever R happens to be I can't say one I can't say two cuz it's would only be the volume for that specific sphere I want to find the volume of any Sphere for any radius and that's what we're doing right now so we know this is netive R to R and this is R High whatever that happens to be at its highest point oh question how do we find a function see we need to find a function don't we to be able to integrate this thing we got to find a function what's the function for half of a circle what's the function for half a circle do you know the equation of a circle what's the equation of a circle then you mumbling you're guessing it you're guessing think I know say what now X Y squ yeah you kind of need to know that one how you get circles X2 + y^2 equals radius squ very good now what variable denotes our function in terms of X here what what variable what variable would you need to solve solve for to say hey here's a function you solve for y yeah very good to say here's a function of X you solve for y right R is going to be a constant for whatever type of sphere we're going to have okay so we're not going to be dealing with r as a variable R is just going to be whatever radius is so in our case we'll solve for y hopefully you can see this we're going to subtract x s and then what are we going to do now if you square root it you would get that yes no tell me something about this which one don't we need yeah actually you could take either one of them but you don't have to have both if you had both it would have the whole Circle so what I'm saying is that the solid one that we have is the plus version the dotted one we have would be the minus version do we need the full circle to make the sphere no we don't no we just need one of them so take either one the negative one or the positive one I prefer the positive one because well that's easier to deal with so we're going to go okay even though it has this plus minus we'll consider this to be the upper half circle you know what that's about all the all the setup we're going to need to do do we have a function that has x's in it do we have a starting point and a stopping point yeah let's set up our our integral we know the volume equals let's do this where's my integral start ladies and gentlemen where does it end R okay what needs to go on the inside of our integral remember that well now here's an interesting interesting concept do we have planes that are perpendicular to binding my yes or no it actually stops doesn't it and at that point so this point consider the tent you could consider a tangent at that point yeah but it actually stops it it stops right there it doesn't go it doesn't lean in do anything like that but it tangentially yes we have those those ending points as well so it touches that one point is perpendicular so it does fit this it means we can slice it into little bitty circles is the limit of it no it it could it could be angled downward as well as long as it doesn't do this here's the difference it can do this could do that what it couldn't do would be this it could do that it can't come back on itself but it can't as long as it's going outward yes because you'd be missing this this volume okay that's what you'd be missing do you see the point there literally do you see the point she be missing you'd be missing that piece the little would get locked off because it takes it to the perpendicular so this one when you went down like this it's fine you're not missing any everything's encompassed it's all in there yeah it's cool uh you can create little little slices but this one you'd need a separate uh way to do that and that would be with washers as we'll talk about a little bit that would be two different integrals you see the point okay so yeah this one works because yes technically it's you don't have planes which incorporate your function and a perpendicular but if you have perpendicular uh boundaries of your function which is great that's what we need now okay what's on the inside let's go quickly Pi I'll give you the pi what else do we have in there parth bracket okay bracket I like the bracket why do we need the bracket we're going to square something we're not going to forget about it so we're going to square it right now and have a DX on outside of it saying whatever's in there whatever want to put my function it's got to be squared what's my function told you I'm a nice guy today nice guy good numbers so this is an integral from r to R of Pi great what happens here do I sell any parentheses I'm still multiplying by pi so yes I do still so far so good okay what now you could pull the pie out sure if you'd like to do there's a couple options that you have right here on doing this intergral one of them will be to distribute that and separate your integrals into two integrals and then pull the pi r s out and pull the pi out of the the second integral with the x s or you could treat it this way doesn't really matter show hands how people can make it down that far right with it now here's where you got to be careful what is the variable that we are integrating with respect to so that means we treat r as a constant value that means you do not do R Cub over 3 that's not the case here x is your only variable are you are you with me on that that's the variable so what you do you you don't pull that out NE you can't pull that out that would mean that you you'd be multiplying right you're not multiplying by that X you're not so that's why I said you have options here you either separate this into two integrals you could do and put a big old bracket around it because you're multiplying by pi that's one way you could do it or you just consider it this way and this is the way you guys L me down so I'll do it this way you have the pi you're multiplying by this integral what's the integral of R 2 well think of it like a constant if that was a three what you'd get over here would be 3x wouldn't you I hope so integral of three would be 3x if that was a one you would get X true this is not a three it's not a one but it's a r 2 we treat that like a constant so you're going to get R 2 x it's a constant integral of a constant you just add an x to the back end of it think about it what's the derivative with respect to x r s so that works that works remember we're just anti- derivating here differentiating whatever that is minus now this one that is a function of X it's tied to the DX so go ahead and do that you're going to get X Cub 3 and we're going to evaluate from R to R now I'm going to move over here for this one but are you okay with that so far yeah yes no people can't hear you yeah okay are there any questions on that one so far so you can take an integral of this R 2 x integral of that super easy now let's just plug it in where do I plug in my RS to my RS or to my x's clearly my x's that's my variable I'm dealing with so quickly we're going to have Pi we're going to have R 2 * r - RB over 3 minus don't forget with the minus you got to plug in the negative R r^ 2 * r - R Cub over 3 are you okay on where all those things are coming from we got our R 2 R here and then R Cub over 3 no problem then subtract the Whole Thing r^ 2 Rus R Cub over 3 now be very careful with the signs don't lose anything don't mess anything up that's the real key with these definite integrals I got negative R cubed I got positive RB 3 because that's still going to be negative negative negative is plus don't forget any brackets yet not done RB - RB 3 + RB - R Cub over 3 2 R Cub - 2 R Cub over 3 yes no common denominator would make this well if you multiply by 3 over 3 you're going to get 6 rb/ 3 - 2 RB over 3 now they have a common denominator hey check it out what's 6 R Cub over 3us 2 R cub over three or what we know and love ha falling with the spirit Isn't that cool I love that one that's kind of fun so nice and precise and interesting that we were able to find it took us this long this long in calculus to be able to find the volume of the sphere isn't that neat but by slicing up and using little bitty circles spere how many people understood that feel okay with that good all right so we've already accomplished how to find the volumes of these solids which have been rotated around an axis we we said basically that if we rotate a solid around an axis the cross-section is going to be a circle if we add up all those circles in other words diss then we're going to get the volume of that figure does that make sense to you so to add up the surface area of all these discs this infinite number of discs you're going to get a volume out of it the idea of vol my washer stems from this problem what would happen say if I had my famous function called f ofx from A to B now here's where started okay this is what we were just talking about if you sweep this around the x-axis it will create a volume create a solid we'll be able to find the volume by taking slices of with our our discs that that's true however what happens if there's another function between F and the x-axis namely let's call it g of X what if I swept that thing around rotated about the x-axis it's also going to create some sort of a solid do you see it only the solid is going to have a hole in the middle this is actually the idea of a vase the other one that looked like a vase but it was a solid vase it' be like a vase out of concrete you couldn't put flowers in it this one would really have be a vase if I sweep this around it's going to be like taking this piece and Rota out there's going to be a hole in the middle kind of like a toilet paper roll almost like that but it's going to be a fancy shaped toilet paper roll I mean that'd be kind of cool but that's what it would similar that's what it would be like let's see if I can draw it for you so you get a threedimensional picture this might be hard for some of you to draw do you get the picture at least the idea of what I'm trying to portray here does it look can you see that that's the rotation if I rotate that out what it makes is some outer surface yes and then some inner surface where the the inside of that inner surface is actually Hollow let's think about what's going on here first thing that I want to talk about is the crosssection still a circle circular in nature sure still circular in nature if I take a cross-section bam but let's think about this for a second if I think about it well if it's circular in nature then then this should work for us the volume should still be some integral from A to B should be some integral from A to B how we did the volume of any solid is we integrated the surface area of our crosssection remember that we took the surface area of cross-section we integrated the surface area of our cross-section from A to B that was basically it so we still have this idea integrate from A to B the function a of X DX where a ofx is the area of the crosssection that seems pretty easy now all we got to do is find out how much is the area of the crosssection well let's think about it for a second a the cross-section well that should be think about think about the crosssection here what's the bigger disc you see the two discs that we're going to get out of this we're getting a disc made by F ofx and inside we're getting a dis made by G ofx what's the Bigg one so what if we did this what if we took the area of the cross-section of F ofx and subtracted the area of the cross-section of GX would you agree that that's going to give us the surface area the cross-section of the the piece that's actually filled in I I'll try to explain it with the picture again the area of f ofx remember that this is F ofx this outside one and this is G of X this inside one so if I take the area of the cross-section of FX and I take the area of the cross-section at G of X and I subtract them basically what I'm taking is a big disc from F ofx minus a small disc from G ofx and it's going to give me the region that the difference between those those circles basically and that's going to be the area between f ofx and g of X we if you understand that you okay with it okay cool well let's think what that means then oh how do you find area of a circle Pi r s p r 2 so if area is PK r s for a circle can you tell me what our R is here the r is for my f ofx my outside one okay I'll give you it's going to be y it's going to be Y at the value X wherever X happens to be what is y wherever X happens to be FX remember FX is the function right so how do you find out the height right at this point ah well you just take your function the function's value at that point remember F ofx says the function's height at the point x you with me on that so wherever I'm at I'm given the height by just plugging in X to F ofx that will be the function f ofx so the height here is f ofx that means the radius is f ofx what's the height here so here's basically all we said I'm going to try to recap this for you so you understand where the form is coming from we take this figure that's created by F ofx minus G of X that that that's basically the area we're going to sweep that area around the X AIS it's going to create some sort of a toilet paper rule uh that that's the idea a fancy one because it's not uniform in shape it's not rectangular but that's the idea here so we we sweep that out we have this volume it's solid but it has a hole in the middle of it what we're thinking is if we can find the cross-sectional area then we can integrate if we can find the cross-sectional area we can integrate across our notice how we still have perpendicular sides perpendicular to the x-axis we needed that remember for the discs you need to be able to get actual discs so we had to have those that work still so this is still true we're integrating all those discs adding them all up adding them up from A to B of the function a a ofx that's all the surface areas of those discs no problem whatsoever you just got to find the surface area of the disc the surface area of the disc of the I should say washer in this case because we're going to make that transition of the washer is the F ofx Circle minus the G ofx Circle that's that's what we're talking about here the area of the F ofx minus the area of the G ofx well we know areas pi r squ our R in any case of these cross-sections is f ofx or G ofx respectively so radius of our circle of our cross-section that's just the height of the function let write F ofx it's just the height of the function it just says G of X so if we put put in these areas we should get p r 2us pi r squ hopefully you're seeing where these things are coming from I'd love for you to understand why these formulas actually work what are we what are we really doing what's this stand for no this whole thing oh the area for the first function good whatever the outside one is the top we'll say the top one in this case the top one that's the radius the height of the function is the radius square it times it by pi you get an area this is the the inside second inside whatever you want to call it the G of X in this case radius square * pi that's the area area of the large disc minus area of the small disc gives us the area of a washer the surface area of a washer uh if we want to be a little bit fancier about this we'll probably just factor out the pie just back to the pie out and what that says is we actually have the expression for a ofx that we're looking for we're going to do the volume is an integral from A to B Sure Pi is going to be there we're just going to integrate the surface area of a washer from A to B that way right there is called the volume by washers that's what we just found would you like to do an exam okay I was hoping so otherwise the lesson to stall out be like okay so what do we do now party time not so much should e okay that's very wordy but I wanted to make sure you understand exactly what we're doing this is what we're doing we're trying to find the volume of the Sol that's created when the area between two functions f ofx and g ofx in this case between the interval on the interval 0 to two is rotated around the x-axis that's the whole idea now we should be able to set this thing up and find the volume of that um you're subtracting a few functions it Poss compose them can you like of mean likeus plug G of X it's not a composition no you can't do that sure you might be able to but it's not going to give us the volume that we're looking for okay now oh what's important for us to know before we start this problem top say it again which one's on top why would that be important if you subtract them incorrectly you are not going to get the right answer you're going to get some sort of negative can things have a negative volume no that's crazy to think about but no you can't have negative volume well maybe the opposite of complete vacuum I suppose that's not even black hole still has something in fact it has a lot of mass actually the center of black holes so no we can't have negative volume uh what we could have is positive volume but we need to make sure we set it uppr right in order to get positive volume so it says you need to find the one on the top you need to find the one on the bottom that way when you do subtract them the S said correctly when you subtract them correctly you get the right answer we need to know which one of these is our F ofx and which one is our G of X that we said a problem right so how would you do that not quite because I'm already giving you the you make it harder than it is if you set them equal to each other you're if you set them equal to each other you're going to find the places where they intersect if they do intersect these don't intersect I don't think they intersect someone do the math real quick in your head I believe they don't uh but X2 + 1/2 that goes like this x goes like this they're not going to intersect okay so even if you set them equal to each other you get x2 how they do- X don't know I don't really care but no that's not the way do they M CU you have one F right there X is let's see I got to see I got to see no they do not intersect discriminant tells us that you get b^2 - 4 a c that's negative true so b^ s - 4 a c is negative that means you do not have a solution for a quadratic formula which means these things can cannot except intersect in real numbers therefore no these don't intersect shame on you people for saying yes not a shame it doesn't really matter uh but no they don't your se but that's what that would be T doing for you so if you set them equal to each other you're going about it the wrong way that's too much work for this problem again you'd be making it harder than it needs to be this is not hard to set this thing up all you need to do is figure out what's on top for the given interval how do you figure what's on top for the ah I gave you the interval if I didn't give you the interval that's when you would set them equal that's when you solve to see where they intersect you get me on that so what do you do here plug the number yeah one's always going to be on the top of each another one that you might check by by seeing if they do intersect send them equal to each other do they cross over that's what that would do for that interval here I'm just telling you I told you already this one is going to be one function this one's going to be another one one's already going to be on top of the other for the entire segment so between zero and two One's On Top otherwise I would have worded it a little bit differently so if one's always on the top plug in a number what number you gonna plug in one I picked something in the middle of it one I don't doesn't really matter but plug in one if you plug in one here you're going to get one and a half you plug in one here you're going to get which one's higher so this is going to be higher than this one right it's always on the top I drew the picture for you but that allows you to set up even without a picture can you tell me where we start on this where whatever we going to have in here good Pi then a big old bracket because we're going to have a couple functions being squared in there and that that Pi we can even pull that out of our integral if you'd like to it's a constant it can do that what goes in the inside first tell me tell me first okay good what are you going to do with that function then what are you going to do uhhuh subtract what now now I'm going to put it in a bracket just so you see that this is the second function so notation wise I want to want to be consistent let's see if we can get it we've got our our formula here which says you have the pot no problem that's going to give us the well that's basically our formula from area to Circle we have the outer disc minus the inner disc but we got to square them because they're the radius squared how many people feel okay sitting that problem did you feel com this side do you guys feel all right with that you feel comfortable getting that this one was on the top of our G ofx this one's first this one's second now what probably go for it see if you can pH that out so fing all that out if you doing your head great if you want to write off the side great just get the right get the right answer make sure you fold it correctly uh this is going to be 1/4 plus X2 + x 4 and then you have this - x^2 I would like to know if you got the same thing on your paper that I got my paper did you get the same thing yes no good okay good what happens here cancel each other out which is kind of nice see the thing about it you could do this to two different ways if you really thought about it you could find out the volume of F ofx and find the volume of G of X and just subtract them couldn't you which that's another way to derive this formula by the way it's the same exact thing you'd have the same balance right put them together it'd be subtracting integrals between the same bounds with the common Pi you'd Factor the pi same thing you could do it however you're creating more work because when you do an integral of f ofx or the area of F ofx and do the integral of the area of G ofx you're doing two different integrals and doing all the bounds when sometimes you get this situation where you can combine like terms sometimes things go away so it's nice to do this rather than two separate integrals do one larger integral you seeing the point on that that's why I give it to you this way so X s's gone love that what do you say we pull that Pi out we got 0 to 2 1/4 + x 4th DX the pi is going to stay there can you take your integral of 1/4 + x 4th what are we going to get say what now X okay X over 4 that's great 1/4 x doesn't really matter it does need to be in parenthesis though has to have a bracket plus what and we'll be evaluating that from 0 to two so I'll have my Pi out there I'm going to do 2 over 4 plus 2 5 over 5 and then I'll subtract but when you plug in zero you're going to get zero so I'm just going to show it this way just to make sure I don't have anything wacky going on with that zero make sure I have it up there that's going to be our our volume when we multiply by five so 12 plus 32 over 5 69 how much 69 okay I think you catch do that for you but 69 Pi don't forget the pi 69 Pi 10 what did we just find it's kind of cool we actually find the volume of Some solid whatever it was as long as you know the function on the top and the bottom Some solid revolve around the x-axis that we found out how much if you were going to pour this out of concrete how much concrete you'd need 69 pi over 10 yards or feet or whatever cubic feet cont that order right yes hello I need a 69 pi over 10 yards of concrete H unless you know calculat like you got it son I got that you got see see contractors might know how to do that be the greatest concrete contractor ever how many people understood this example feel okay about it now can you do this when we revolve around the Y we've been going around the X the whole time can you do it this way probably I hope so the ideas aren't going to change just says what happens if you have a figure going up and down rather than left and right and the formulas will be very similar the only difference will be well of course the variables will be slightly different yeah so let's talk about volumes that are perpendicular where the I'm sorry where the planes of the interval are perpendicular to the Y axis so volumes where the crosssection is perpendicular to the Y AIS before we had the cross-section perpendicular to the x-axis that's why we're able to take and make all those little discs out of it here we got this way well if we if we talk about it the diss method then would end up looking like this instead of from A to B we usually use from C to D for the y- AIS so we go from C to D we'd have some area function but the area function is still going to be a circle Pi r^ squ Dy where U of Y would be your function in terms of Y not in terms of X where U of Y would be your function in terms of Y hence the Y instead of X that's the idea for the discs for the washers very very similar still from C to D still a Pi still the radius squar some outside function minus an inside function squ hopefully it looks familiar to you it's basically the same exact idea only you're going to the Y AIS to the X I'd like to give you an example if you don't mind on this would you like to see one sure now the process won't be any different it really won't just have different variables but this stems from from this idea what if you were to take by the way I only know how to draw one function like this so that's why I do the same one over and over I took it and went that way basically this is U of Y then the solid would look like this well it's not in proportion but hopefully you the idea we're just going this way with it taking this solid Rota around the y- AIS we'll get that sort of shape notice how we're perpendicular to the Y all the cross-sections will be perpendicular to Y that means we can integrate from instead of a to B we usually call it again C to D we can still integrate from C to d add up all those cross-sections basically let's try it let's do I'd like to give you several examples on this so I don't know if we'll finish this today but I do want you to see a lot of them uh so you get kind of the handle on it so we'll start kind of nice and easy work a way up from there so let's let's say this y = x is revolved around the Y AIS and it's bound by Y = 2 and Y = 0 what I'd like you to do is find the volume now you can draw a picture of this but you don't need to I want to draw the picture so you see what's going on on but you don't need to draw the picture you just need to understand what all these pieces mean okay here's the idea there's x s you with me now bound between 0 sorry bound between y = 0 and y = 2 what is yal 0 and what is yal 2 hor they are yeah in this context they're horizontal lines y remember y equal a constant equal a horizontal line yeah so what this says is when you go up to two it's bound by this and it's bound by this this is Y = 2 and Y = 0 so essentially I'm not worrying about this piece of the graph I'm just worrying about this piece now here's the problem am I going this way or this way it says around the Y so I'm I'm basically taking this shape and making this out of it making that shape out of it do you think the volume will be different if I go this way versus this way that's an interesting question why don't you figure that out later see if it is right now that's not our question the question is can you take this go this way and uh between those basically lines and find the volume of it question when you um we're taking it with respect to Y we want to Fig out volumes the same and we turn it with respect to X our BS change as well if you turn it this way yeah I don't even have any bounds you said Bound by two y y right you have to respect this way okay I see what you're saying so like that right there yeah okay yeah you could do that but I'm not asking with respect to x i I'll give you the the long story short no the volumes not going to be the same one's going to be completely different than one if you take this and roll out this way or this way you're going to have different volumes significantly different volumes okay it's not going to work the exact same way could you do it revolve around X yes but that's not what I'm asking for I'm asking for can you revolve it around the Y it'll give you a completely different looking shape even this if you revolve it this way it's going give you kind of like a bullet point right this gives you some sort of a weird looking cone thing that's the technical term for it at least weird looking cone thing so you have you have this shape almost like a flower coming out at you that's what that's going to be now here's the issue I'll give you a little note as well when you're revolving things around the Y AIS you need to have them in terms of Y which means you need to solve for x don't let that confuse you I'll say it one more time when you're revolving around the Y AIS it's got to be x = y y y y y y y does that make sense when you revolved around the x- axis you had y equal something in terms of X that's what we did I eras the problem but that's what we did over here you had all X's didn't you f of x g of X there're in terms of X that's because we're going around the x- axis and our bounds are in terms of X what are our bounds in terms of here why that you better have y to work with then so the note for you when revolving around the x axis functions must be in terms of X that means solved for y when revolving around the x- axis put functions in terms of X cuz your bounds are in terms of X you have planes that are perpendicular to the X it's all about the X solve for what when revolving around the y axis it's backwards put everything in terms of Y your functions needs to be in terms of Y your bounds are in terms of Y your planes are perpendicular to the Y in terms of Y solve for x that's the idea notice that we're revolving around the Y you're using this thing you really are you're going from C to d u of Y that's why I put U of Y over here guys so that You' see it has to be in terms of Y not X you're going around the y axis got in terms of y d y so for us let's see about our function if y = < TK X is this in the right in the right form to go around the Y AIS this would be around the xaxis perfect oh my gosh it' be awesome for the x-axis in fact if you want to right now you can probably figure well you have to do a little bit work for bounds because your bounds would change I do a little bit of work for that you'd have the square root of two kind of awful no you have to you have four that wouldn't be so bad so you could do that and figure out that the volumes are not the same uh but that's not good enough for going around the Y how do you make it go around the Y get xal y get x equals very good and you get that by square on both sides sure we're going to get if you square both sides X = y^2 what's the function that we're talking about X or y^2 M look what this does this says no no no I don't want the sare RO of X anymore we'll consider it y^2 from here to here just that little piece though just that little piece so we'll have let's see if we can do our our integral where are we starting ladies and gentlemen zero somebody else where are we ending what function goes on the inside oh you know what I forgot a couple key parts of course we want to fill that out right I get so excited just doing this got this we do mean this of course we got to have a pi what else are we going to have in there what do I put in for my my function what's my function and then what squ mm so these are in terms of y yal 0 Y 2 that's not a problem we put this one in terms of y That's our y s we fill out our our formula that we got now why do we have discs and not a washroom yeah we only have one function there's nothing between that so we can use the dis method just fine there's not another function to take away volume from it can you do that integral that's actually pretty basic let's go ahead and do that we'll get PI from 0 to 2 y 4 Dy that's Pi y the 5th over 5 because we know how to do integrals real well from 0 to two Pi will do 2 5th over 5- 0 what we're going to end up getting is 32 pi over five if I show a hands how many of you will feel okay with that so far would you like to see if the volume is the same go another way if you want to it's not hard I mean why don't you try on your own what that'd be a good practice for you right now set up if I switch this to going around the X and leaving the bals the same nope got to figure out the balance do that way don't just give up on it come on try something try something this part should be pretty easy right if we're going around the x axis we got the pi we got the r that's our function squ X and we got the square orig have you set up just like that that way yes guys over here too okay the problem is our balance how do we find our bounds up if we have y if we have y's right now how do you figure out plug plug them into x what are you plugging them into if y equals the sare root of x and you know Y is 2 and Y is z can you figure out X's that's doesn't this look similar to a u substitution when you're changing balance yeah very similar idea so this says plug in 2 for Y 2 = x 0 = x square both sides x = 4 and x = 0 you follow me on that so now we'll have an integral from 0 to 4 this is going to be a pretty easy integral it's going to be Pi 0 to 4 of X DX s < TK of X2 is X that's kind of awesome that's great then we'll have Pi x^2 over 2 from 0 to 4 Pi 4^ 2 2 - 0 equals same way different yeah quite quite a bit different actually not quite a bit different I guess it's off by just 1 30 second U but anyway one one that's huge P 32 it's off by P 32 huge difference huge but it well is different not the same so now that you see that I'm satisfied hope you understood both these examples is the calculus hard no really you know integrals right you know them and they have to give you nice ones to be able to do them so they're they're nice uh just remembering how to do it how to set it up and being prepared don't worry we have time still uh being prepared that you need to know what variable you have to have the function in terms of all right I I get to forewarn you here in the next section I'm going to give you a different way to find volumes it's going to be volumes by cylindrical shell and in that case your variables can of switch if you're going around the Y you want in terms of X and if you're going around the X you want in terms of Y that switches so be careful on on these two me on these methods if you're doing discs and washers your axes match up around the x- axis in terms of X around the y axis in terms of Y for the next method it will switch around the Y in terms of X around the X in terms of Y so get really comfortable with this that way you don't forget about it you got me now we'll stop there today we'll talk talk about a few more examples next time all right so we're in the practice stage of whatever section Ron 5.2 I believe we're in the practice stage and just doing some examples so the questions are normally worded something like this find the volume of the solid that's contained between a couple lines they'll give you boundaries or not contained between these two functions and revolved around they got to tell you which axis we're going about so basically this says find the area that's between there and revolve it around the x-axis that's going to make some sort of a solid that's going to have some sort of a volume do you find you with me on that now just as refresher if we are revolving around the x axis note that your equation should be so solved for y they should be X in terms of X so y equals and then a whole bunch of x's that's what solve for y or in terms of x mean so around the X must be dealing with x's you got it around the Y must be dealing with y's so my question is if we're going around the X are these okay absolutely okay yes we're going around the X they are in terms of x in terms of X means solve for y and these are solved for y so this is in terms of X and in terms of X which says it's all the terms have x's in terms of X the terms are based on X you get it so those are perfect for us now the problem is do I give you any bounds right here do you know which one's on the top right now do you know which one's on the bottom right now so those are things we're probably going to need to figure out in order for us to set up our integral remember that the volume the volume is I think I gave this to you last time a to B Pi fx^ 2us g ofx 2 DX did I not give that to you but this implied that F ofx was the top function G ofx was the bottom function so we need to find those things out we also need to know well it says the the volume uh in which the area contained by contained by we need know where to start and stop so where the area starts and stops according to the Y or the X in this case because with x's that's right according to the x axis now your headphon you're okay with the the idea of this so far so how do we find out this goes back to our area problem very similar idea how do we find out where these things are contained where the area is contained say it again that's exactly what we would do to find the area between two curves right well hey all we're doing now is finding the area between look at that the area between two curves and then taking the circles when we rotate them around that's all we're doing here so this really says take the area between the two curves and revolve it around that's what we're doing so what we need to do first is exactly what you said let's set these things equal to each other figure out which one's on the top so what we want to find is the intercepts where they intersect where they intersect by they I mean the functions and also which one is on the top okay so let's find out where they do intersect if we set them equal to each other this is kind of nice because it's already Sol for y basically we have y = x^2 and Y = X Cub set of them equal to each other gives us X Cub = x^2 still okay with that so far solving that well we get what what would you do to solve that probably because you're going to be using some sort of factor X Cub - X2 = 0 now what if x^2 * X - 1 = 0 can you solve for my x's please Z perfect not so bad in fact it's very similar to what you did for finding an area because that's exact the same idea that we're doing only revolving around some sort of an axis now do you know where your integral is going to start and stop that's pretty great so we can already start filling this thing out the volume is going to be from0 to one not a problem that's where these functions intersect therefore the area must be contained between those two things okay they got to be between there somewhere now what we need to do is find out which of these functions is on the top for the interval and which of these functions on the bottom for that interval how do you do that remember doing this remember doing that I gave you the graphic organizer I said plug a number in there plug it into X cub and x2 in this case figure out which one's on the top and which one's on the bottom what number are you going to plug in between zero and one 1/2 be great .5 whatever something between there it doesn't really matter as long as it's not zero or one so plug in 1/2 to both of them this gives you 1/8 this gives you one4 which one's on top x s is a bigger number than I'm sorry 1/4 is a bigger number than 18 therefore X2 is on the top X Cub is on the bottom is this coming back you at all do you feel okay with our setup thus far well we already know where we're starting where we're stopping let's go ahead and figure out the rest of this thing the rest of our integral we already know what our f ofx is going to be in this case the top function and the bottom function is going to be in this case so let's continue what's the next thing I should write here ladies and gentlemen yeah they're all going to have a pi that's from the circular nature of revolving around an axis then we have a big parentheses what goes next just X squ now you don't have to use brackets you can use parentheses as well just make sure you have the right idea we take the top function we Square that's what it says take the top function Square it we have that take the bottom function Square it we have that top bottom now we're set we're ready to go you feel okay with doing all this so far not bad do you see how similar it is to find an area very very very similar can you do the integral sure what are you going to do first probably pull the pi out square this you're going to get X to 4th Square this you're going to get x to the perfect perfect now when we do the integral remember to actually do the integral what this means notice how I'm writing these these things I've seen some notation problems in some of your work uh notice that when I do the integral firstly I actually do it I do x to the 4th and I don't plug in numbers to X 4th I first need to go X the 5th over 5 of course also notice that when I do the integral I've stopped writing the integration I have the bracket sure but I've stopped writing the integral that's not appropriate if you take an integral you stop writing the integral - x to 7th 7 and this is how you show you're going to evaluate off to the right hand side between 0 and 1 also there's a few people who have forgotten that sometimes zero is a number you have to plug in is not going to give you zero in this case if you plug in zero yes you're getting zero no problem but if I had like an xus one and you plug in zero that's not going to give you zero so be very careful on that so let's go ahead and do the rest of this if we do hi plug in one here and one there you're going to get 1 15us 17th and just to make sure I'm not forgetting anything I'm going to check to plug in zero 0 over 5 I know that's zero 0 over 7 I know that's zero so I'm going to write the zero just to show that I've done it that I haven't forgotten about it you don't want to forget about that thing so basically we have 1/5 minus 17th do 1/ 15 - 17th for me please 2 35ths or in other words 12 35ths wait yeah should be 735th - 535th should give you 2 35ths it's all good you guys able to get 2 35ths 2 pi over 35 don't forget the pi and that's important right I want to leave that hanging that's going to really change the answer any questions on that one before we go to the the next one remember this is practice you you you know how to do all the stuff at this point we're just kind of practicing get get a real good feel for it are you ready for another one or do you have any questions okay let's switch something up a bit and talk about a different type of problem uh maybe not around the x- axis maybe we go around the Y AIS so the same idea I'm not going to write the whole thing out area contained by y = x^2 x = y sared that should be fun around y AIS now could you please tell me something would you please tell me that if I'm revolving around the Y AIS what's what must my functions be in terms of so solved for X solve for x means in terms of y means your terms are y's that would be solve for x so in our case which one is okay for revolving around the Y AIS is this one okay for revolving around the Y is this one that's perfect this is what we want for revolving around the Y for the method of discs and washers that's what we want this one not so much if we're revolving around the Y change it how are you going to change it y very good now I think we're neglecting something here really when you take a square root you should have a plus and minus right however this function is only going to intersect the positive version of that if you if you think about that you have well I could draw it I suppose y = x^2 you have X = y^2 that's the only intersection that's happening so we're going to neglect the negative version of that xal y^2 just because it wouldn't intersect anyway do you follow me on that and you could show it by actually setting it equal but you're not going to get a real solution out of it it's not going to happen or you get some false solution so are you okay solving for x in this case take a square root plus and minus yes but the the positive the negative really isn't going to affect us in any way before so we're going to ignore it for us now what do we need to know after that to set remember our our whole goal is to try to set this thing up what's on top we need to balance remember terms of Y so everything's got me terms of Y nice and neat someone said bounds how do you find bounds now check it out do you see why you need this in terms of Y if I set the the functions in terms of y equal to each other I will be solving for y's right and this is along around the Y AIS it would be useless to solve for X's if I'm going around the Y AIS that wouldn't make any sense that's way I have in terms of Y all right well set them equal to each other I know X = Y I know X = y^2 therefore if x = aunk of Y and X = y^2 that means < TK of Y = y^2 so far so good solve for y how would you do that H not yet without just by looking at it how both sides Square both sides might be the best way to go how do you get rid of a square root square not square root a square root but Square it if I Square both sides it's legal to do you can do pretty much theary anything you want to to an equation almost anything you want to to an equation as long as you do to both sides which means that even if even Joe if you had change it to a 1/2 power to get rid of the 1/2 you'd still multiply by two correct that would be essentially the same exact thing as squaring it so Square both sides notice you're going to get y = now say what you said Alan what do you do next bring there you go 0al y 4us y now what would you do sure we we can Factor as much as needed here y = ybus 1 now when you do this problem you're going to get y = 0 and Y Cub - 1 equal 0 Can You Solve for y yes yes you can add one and take a cube root and you're going to get still one the only way this is working is if yal 1 that's the only way that happens with Cube not negative 1 because it's not a square it's a cube so the only way you're getting one from cubing a number is with positive one negative 1 cubed is negative 1 that's the only thing that that would work if you want to show it explicitly you'd have y = the cube root of one and that is one but I show of hands how many people feel okay getting that far yes yes it is just some algebra but it's kind of nice to review that so you don't get stuck on the algebra when you're trying to do calculus that's the biggest problem of calculus hey do we know where our integral starts and stops where along the y or the X very good because this is in terms of Y we solve for y so already we know that V equals 0 to 1 in terms of Y now H the idea which one's on top changes when I'm talking about the Y on top doesn't mean this way anymore on top means this way right now do you get that so really we're asking which one is on the right and which one is on the left you still do the same way though you still I mean basically you don't even have to draw a picture or think of it you just have to plug in a number and figure out which one's bigger that goes first which one's smaller that goes second that still works only bigger this time means right smaller means left and then you'll be revolving that now your head you're okay with that I know we're talking lots of theory here not lots of theory but it's it's abstract because I can't draw a lot of the stuff very well so how do you do it same thing we just did here same thing 0 and one we'll take a number and we're going to plug it in to right here what number are we are we thinking about you could do 0.5 I might do uh 1/4 because I can take a square root of it very easily and I'm going to have to take a square root do you understand maybe take 1/4 so if I plug in4 just do it to both this gives you a half this gives you 116th and we can clearly see which one is bigger and which one is smaller by using some numbers that we actually understand so which one is a bigger one so thenare root of Y goes on the top nice graphic orer huh tells you which one exactly goes where which one is on the bottom now of course bottom means left top means right but you set it up the same the same way same way so which one in other words is going to come first the square Ro of Y or Y squ okay fill out the inter for me so what am I going to have Okay C get the pi square root of Y I heard squared very good and then yeah don't make the mistake putting a plus right we're actually subtracting volumes here subtracting uh I'm sorry surface area and then revolving that so I'll make that bracket what's the next one I heard you I'm going to write it this way feel comfortable sitting up the integral notice really the setup and the plugin in numbers is the hard part here hard part the integrals most of them are going to be very doable for you because otherwise we get stuck you know maybe I'll pull that Pi out of here we got a constant anyway we may as well pull that out square root of y^ s that's kind of nice I just get y minus this is going to give you y 4 Dy not such a bad integral to do after all we get a pi we're going to get y^2 / 2us y 5 over 5 and we're evaluating from 0 let's plug in the one first that's what it says to do so we're going to do 1^ 2 over 2 that's just 12 - 1 5 over 5 that's just 1/5 I'll check the zero but I know I'm getting 0 - 0 that's still Zero 12 - 1/5 should be three bam not too bad not too bad as long as you understand the idea here that find out what they they actually intersect set up the integral appropriately make sure you find out which one comes first which one comes second and then it's all about the formula just plugging it in by a show of hands how many people feel okay with what we've talked about so far now I'm going to give you one more uh I would consider this to be kind of a tougher problem but it's going to take us a little while for me to explain it to you but I want you to see it I don't think I've given you any homework on it I might I don't think that I have I don't think I changed the assignment but I will certainly give you something to do with this so see if you can comprehend the idea here what if I ask you to find the volume of the solid contained by this and that is revolved around y equals 3es that's interesting question it says right now we're not going around an axis right now it says imagine this you have some sort of an area right and instead of revolving around a y AIS or an x- axis you're saying pick some line revolve it around that can you still do it answer is yeah we can as long as you understand what the concept is that we're doing here now I'm going to draw the picture of what this is what this is you don't necessarily need the picture to do it uh you can do it the way I showed you as long as you understand what's going on so I'm going to illustrate it so you do understand the way it's going on first thing I want you to get picture of is what taking place now that's a pretty easy function that just says y = x yeah and this one says y = x Cub which looks essentially like this for this interval and if you really want to find out where they intersect where where could you do it or how could you do it okay so if I set them equal I would get x = x Cub I would get 0 = x Cub - x I would get 0 = x^2 - 1 for the sake of sorry for the sake of keeping it easy I'm just just going to focus on the positive here we're going to get 0 1 and 1 I'm going to just so you make a note of this I'm going to ignore this one to keep it easy for us also you can think back to odd and even functions if I were to Revol well no that's not true because we revolving around something that's not XA disregard that statement to keep it easy I'm going to keep between zero and one just so we make it easy on ourselves you can do another you you could also do this with the negative portion of this but it would be a different integral you'd have to do two different ones no that won't any of the volume area of course it will if I remember that this is going to this would also do this right I'm saying let's just to find out what this is going to be let's forget about this for a second because you'd have to do a different integral anyway you have to set a completely different integral anyway cuz they cross over one's not onp top of the other one the whole way so you have to do it a different one anyway let's just focus on this one for right now and see how to do the problem in the first place do you understand what I'm trying to do here so all I'm really saying is this is zero and this is one that's what I'm saying now so we see the area what are we going around this no this are we going around this no what is yal 3 house that is it a oh come on you should know is it this way or this way it's this way y = 3 y a constant just gives you a horizontal line Y = 5 would be at 5 Y = 2 would be at 2 y 3 over2 would be at one and a half so if this is one which it is one one what we're talking about is can you revolve this figure around that treating that like the axis now that's a really interesting question isn't it it's kind of kind of crazy can you can you do it well we're going to think through it as to how in the world we might do this and it's going to stem from this formula but we're going to change it just a little bit by the way questions on that I asked you that though already I think oh yeah this way so we're ring around that one now let's think of what what actually is going on here um what does this represent in our formula the top sure but what is this well let me let me I'll kind of split it apart for you again A to B Pi FX 2 yeah minus G of X2 DX you know before I go further I better make this clear also what should our variables be in terms of x's or y's xes because we're really going this way around that it's almost like an x- axis that's been moved up a little bit do you get why we're in terms of X and not y if it was this way yes why basically goes uh where is the where are the planes perpendicular to that you're going to find the area of and then consequently the volume they're perpendicular to the xaxis still so they're they have to be this way in terms of X well what does it mean when I do this what what are those things what do those look like C they're circles so what is f ofx say it again it's a radius what is g of x a radius now that worked very well because F ofx can be considered a radius if you're going from the X axxis can't it this would be the radius of that figure being R and this would be the radius however I'm not going around this now I'm going around this so if we can find out the area of those cross-sections remember doing the cross-sections by washers if we can find out the area of those cross sections incorporating this y = 3 we should be able to do very similar thing let's think about it if I'm to revolve so this is FX is basically your radius if I'm going to revolve this thing around the three halves can you tell me what the r it's going to be so hard for me to draw I don't know if I can do it um I'll try for it's hurting my head too many circular shapes there's so many there's so many going on but it almost looks like an eyeball eyeball but that's flat this is flat the end this is flat the end uh and there's this little space that goes around you see the space that goes around that oh you know what this would make it a little clearer um this right here there we have that space that's going around that what we want to figure out is like a sideway B kind of like that yeah very much with the open bottom and open top if this is y = 3 here was basically the idea for going around the axis look find the radius square it times it by pi find the other radius inside radius square it times it by pi and then subtract them that's the surface area of the outside one minus the surface area of the inside one and then we integrate the surface area that was the idea do you get the idea well now this is very similar let's just think the outside one is or what the inside one is when when you go back to here what function is this steming from look at the function execute what function is this one stemming from okay would you agree that the outside function is X cubed yes and the inside function is X yes let's find a radius that's based on three halves how would I find out the distance oh this thing for a second at any random point x okay how would I find the between a constant line 3 let's say right here 3 and this is what I'm looking for how far is that say it again it'd have to start with three halves wouldn't it because three halves a little bit higher have to start with the three Hales CU that's here three halves and then you said minus minus what ah would you agree that at any point the height of this function is X cu therefore at any point the distance between here and here is 3es are constant minus X cubed and that's exactly right can you do the same thing for a random point from here to here how much is that 3 Hal it's where we start that's our constant minus X that the height of the function at X so right here says this whole thing from here to the xais is 3es the thing from here to here is X so if I take 3 minus X I will find the radius now that's kind of hard for some people to grasp do you grasp it how people do it feel okay with that now is that for the entire thing like if you were to do above radius is the radius right so we use the fact that this is apart from the xaxis to figure out the radius whether or not this is above or below that's the same distance it's still 3- did you get that you sure so radius is radius doesn't matter how you get it that's the radius now we're going to go ahead and fill this out this out actually using this idea this was the radius that's why I had I walked you through this this was the radius this is the radius correct so for us when we have a y equals a constant line and we're going around that that actual axis it's going to augment this just a little bit it says well really what we'd have is the y equals a constant minus the in this case it's the outside function not the top function the outside function that's weird to think about outside function inside function outside inside now I don't know if they give you that form in the book or not I don't think that they actually do uh but they expect you to understand the idea that all you're doing is taking a sliver of it a slice a washer finding out the radius look at that the radius here and the radius here squaring them times them by pi and then subtracting them you find the surface area of the outside Circle minus the surface area of the inside Circle in this specific case why moving its point of Revolution Chang the area of this of the actual area wouldn't the area still be the same it's on x-axis it's just location space no think about how much further away you are this is actually touching the x-axis right okay and you're just going right around it this is sweeping out think about if you're this far away you're this far away out here too right but the inside part is not being swept out cuz the shape of the object hasn't changed this whole thing is being swept out though so if I do this it cover I go I'm going to take this little pan right here and I'm going to go like that how much space is it covered I don't know this much whatever that is uh 37 pi over4 no I'm just kidding awesome if I could do that then how much is this with the sound effect does it sweep out more volume or not because it covers more space of course it covers a lot more space it'd be easy if we could if we had could just do the same thing but it's not the same thing this goes all the way around something that's further away than the x- axxis right now yes that formula is it FX or oh you know but uh I just forgot this no no no no no what did I do to you guys goodness me no I was right I was right but and C get sare radius that was where my mistake radius squared and radius I just had too many little brackets sorry I got confused radius hopefully you understand the idea I just Jack it up for you good catch Scott radius squar and radius squared you got it this is the radius of the inside this is the radius of the outside you're squaring it subtracting them and Times by pi that is the surface area of the cross-section of the big one minus little one that's what we're talking about and then we'll integrate that let's try to fill this thing out where do you start take your integration where do you no where do you start sure it's still zero right still one zero two one what do you have [Music] inside now inside the pi you can put P outside that's fine it doesn't really matter inside inside there what's next what's our C in this case yeah so when you think about it's just 3 - XB squar sure square of course but that's saying the outside function that's the larger of the two radi the outside function minus inside function give me that M keep going yeah perfect I would love it if you would understand where this is coming from do you understand where it's coming from radius of the outside big one radius of the little one no problem this is the surface area after you subtracted those two things can you do the integral sure yeah sure we can what's the first step in doing the integral what would you do deal Roots firstc sare yeah probably Square them keeping our signs correct is crucial for you also I I'd like to encourage you to combine start combining like terms if you have them I don't know if you're going to have them here or not but if you do have them my goodness combine them U with before you start doing integral some of you guys are are doing integral you're setting up a problem and you have like five terms three of which can be combined you're doing the integral of all five terms why why I'm why why not combine your like terms before you do it saves you twice the work it saves you so much work okay I'm off my little so box now let's go ahead and figure the rest of this uh if I do this this is 94s minus top minus the bottom say what oh you know what it really is but you can't see it here it is the top minus the bottom all right but it's just a different axis you just have to set you have to set it this way yeah kind of reverse Think Outside Inside okay uh good question 9us 3 X Cub I believe plus x 6 very good that's the first one hope I don't run out of room here next one gives you 94s notice how it's still in parenthesis I'm not dealing with that yet I'm not doing in my head I'm making sure I have the signs right even I do this take your time and do the signs right- 3x plus X2 and I need a DX don't forget about your DX it says what you're integrating okay get double check on my work make sure that I've done that correctly yes yes yes can see that far 9/4 - 3xb + x 6 then minus in parentheses 9 4 - 3x + x^2 dist yeah you know what we are going to do that we're going to distribute negative we're going to look to see if we can combine anything after that we have 94s we have - 3x Cub + x 6 - 94 + 3x - x^2 DX I'm going to move over here do you have any questions on the picture that I drew firstly okay wow okay so Pi 0 to one oh that's nice I like zeros to on those are those are real cool what happens here simpy this is why you'd simplify before you do integral because otherwise you'd integrate 94s and you have 9/4 X twice and you're going to eventually get rid those numbers anyway why not get them now and and don't waste time integrating the other things you can't combine any of that I'm going to write in the appropriate order I'll write x to the 6 I see a - 3x Cub I see a - x^2 and a was it a plus 3x can you do the integral of course yeah it's actually not bad right it's the setup that's hard to understanding [Applause] that let's check your ending roll it with my integral to make sure we both did them right did you guys get it right as well same thing same same now we're going to plug in one first we'll plug in zero next if I plug in the one if I if I evaluate I'm going to get 17 I love 1's two man they're nice 34s 1/3 and 3 Hales I'm going to check the zero but notice how every term is going to go to zero so I'm going to write at the back end of this yeah I did check zero it's just going to be zero so far so good notice please stop for a second notice how if the zero had come first it would be zero minus all this stuff you follow making it all negative that would be the case for the other part of this the between negative 1 and zero you follow now if that went around three halves notice that that volume is going to be huge which is why I almost made the mistake of remember the odd function how you just double it that's not the case for volumes because you're taking something way lower way above that three Hales it's going to be much larger volume you'd have to do a completely different integral for that did you get it so we set up the same but it' be a different volume Al together so combine all those has anyone done 1 7us 34s 9 84 how much 89 17 - 34 - 13 okay Pi * 47 84s is 47 pi over 84 I got to be honest gotta be honest your Calculus guys know how to set this stuff up you're plugging numbers in that's so great make sure you plug the numbers in right that that's the biggest mistake I've seen on tests all the time