hello everyone in this video we will discuss section 16.3 of our textbook which is all about triple integrals what have we done so far in this chapter well we've reminded ourselves what it meant to integrate a function of one variable and then we stepped this up to integrating functions of two variables which required a double integral which you can do in either the order dxdy or d y d x as long as you update your bounds as appropriate see the previous section there's no reason that we can't go up a dimension and integrate functions of three variables if we were to do that we would need a triple integral instead in some order on dx dy dz and you can put these in any order as long as your bounds are accurate for the order you've put them in now in every single case here you're integrating over some kind of region for a function of one variable you're integrating over an interval in the x y plane and that interval is one-dimensional because it lies along the x-axis here you are when you do a function of two variables and you integrate you integrate over a region in the x-y plane that's two dimensional and so you get an area times a z value ends up giving you a volume as an output for this whereas this has a length dx is a length along the x-axis times a y value height um so dx uh is a particular x value um times yeah a length along the x value times length along the x-axis times a y-value that's this is going to end up giving you an area so the result of this is an area that's all i'm trying to say very poorly um the result of this is a volume and the result of this is well since we have three the region we're integrating over is a region associated with a volume so it's a three-dimensional region as opposed to a two-dimensional or a one-dimensional region and therefore the result you're going to get is whatever is the equivalent of thinking of a quote-unquote volume or a size of some sort um in four space because f of x y z as a function of three variables lives in four space because it equals a fourth variable which we will call w that gives you your fourth dimension um the problem with that is we cannot visualize things in four space we can't draw it though we can certainly still draw the region which lives in three space and the result of this we are going to call a hyper volume because it's not really a volume anymore if it's in four space but whatever the equivalent of size however you want to think about it in four space is so if you get higher than three space where you consider volume you usually call it a hyper volume okay so today we want to talk about these triple integrals integrating over regions in three space functions of three variables let's tackle this we're going to define it in a similar way as we did for a double integral for a function of two variables and the idea behind that was you started with a rectangle and you split it into a bunch of other rectangles of equal area and formed rectangular prisms we're going to do basically the same thing except because our region lives in three space instead of it just being a rectangle it has to be three-dimensional so let's call it b for box and if you're going to form a box which can also be thought of as a rectangular prism in three space your x values live between some particular constants as do your y values so i'm going to put x between a and b y between c and d and then your z's need to live between some constants as well now the natural thing to do would be e to f next but of course e is associated with 2.71 dot dot euler's number and so i cannot use e here and so we're going to go crazy and we're going to use p and q instead as our constants for z okay so idea for a double integral over for a function of two variables was take a rectangle split it into a bunch of smaller rectangles of equal area i'm going to take for this my box in three space and split it into a bunch of smaller boxes all of equal volume instead and so let's talk about how this would be done well if we just want to talk about the x's if we want to split this up into a bunch of boxes of equal volume and so that all have the same side length x the same other side length y and the same other height z the width in the x direction will be given by delta x equals b minus a over n where n is the number of pieces or the number of new boxes that you're using so we're going to say n pieces of equal width delta x you can do something similar for your y's except you can potentially split it into a different number of pieces so let's use that delta y equals d minus c over some other number m so this is m pieces of equal width delta y and then once more we do this for our z's delta z is going to equal q minus p your larger z value minus your smaller v z value divided by the number of pieces you want to use in the z direction so let me use another letter l for that and this is l pieces of equal width that we'll call delta z okay delta x delta y and delta z are all going to be particular numbers and essentially what we're doing is we're splitting the original box into a bunch of smaller boxes that have side lengths as determined by delta x delta y and delta z which will also determine the volume of your little boxes within the full box in summary then we can then say that the integral over this box in three space for our function of three variables x y z d v because the region we're integrating over is three dimensional and hence its size is associated with a volume v for volume this is going to be given by the limit as all of delta x delta y and delta z go to zero of the sum of the volumes of all um [Music] my little boxes times you're going to use a candidate ordered triplet in each one of these boxes and plug it into your function to get the quote-unquote height at that point whatever that means in four space so we're basically going to add the hyper volumes together is the idea so we're going to take f of x i j k y i j k z i j k and that's going to be multiplied by a delta v and again what's the idea here this is going to give us a hyper volume one dimension up from a volume whatever size is there we're calling a hyper volume um you're integrating over something that's three dimensional again that's the delta v which is the same as delta x times delta y times delta z and if you multiply that by basically your fourth dimension w um this is what you would get so you have your volume of each mini box times the w value the output of your function of three variables at some candidate ordered pair in your mini box and then you add together all the hyper volumes that you get formed like this using each of your mini boxes in a candidate point from inside each one all right what does this mean well you could also think of this first part as saying that delta v is going to zero and what would that mean well this is just saying that you're splitting your original box into a bunch of mini boxes of volume delta v if your volume is going to zero that means you are getting more and more boxes um the closer this volume gets to zero so when it actually gets to zero instead of a bunch of boxes you basically have a bunch of points but that's the idea closer this gets to zero the more mini boxes you're using and the better your approximation begins is and then it's exact when you take the actual limit delta v again is the volume of each mini box so you can think of it if your volume is going to zero you can also send each of your side lengths to zero and accomplish the same thing and so we would still be having the sum over i j k of f of x i j k y of i j k z of i j k but i'm going to write delta v as delta x times delta y times delta z instead because this will allow me to think of this as a triple integral based upon how i've set this up the order of my triple integral is going to be dx dy dz i'm integrating the function of three variables f of x y z and then with the order i've chosen here the bounds on z are p to q the bounds on y are c to d and the bounds on x are a to b so this is one way i can set this up and of course it's not the only way there's nothing special about the order i put here as long as you have a box when you're integrating over a function of three variables you can integrate in any order just like when you have a rectangle for a double integral you can integrate in any order and just swap the order without having to do anything but swap you know your two integrals with their bounds kept the same okay so other orders are equivalent since b is a box if all your bounds are constant you can integrate in any order you like and just move the integrals around in your order keeping your bounce just like we could extend the idea of integrating over a rectangular region when you're doing a double integral um by essentially taking a non-rectangular two-dimensional region and closing it in a rectangle splitting that into a bunch of mini rectangles and only taking candidate points within the mini rectangles that are actually hitting a piece of the region boom that's how you do that you can do the same thing um with a box and the idea would be you take your three-dimensional non-box region and you enclose it in a box break it into a bunch of mini boxes and then take a candidate ordered triplet from each side of one of each of those mini boxes that actually touched the region and plug that into your function to get your quote unquote height hyper height whatever that means in four space with all of this in mind let's do an example we have been asked to integrate this function of three variables given by x y z squared over the region in three space and again the dv is telling me that this box is in three space v for volume volume is three dimensional this box has x's between zero and one y is between negative one and two and z's between zero and three step one choose an order to integrate in any order it does not matter simply because all of your bounds are constants so pick any order you like let's say i do dz dy dx because that's the first thing that comes to my mind i'm integrating the function of three zero variables x y z squared and this is a triple integral the outermost part is x which lives between zero and one the middle part is y which leaves between negative one and two and the innermost part is z which lives between zero and three there's a lot of variables going on as usual if you need to label your bounds so you plug the correct thing in my innermost integral is with respect to z so my bounds are z values if i go to integrate the innermost integral with respect to z x and y are treated as constant multiples and they come along for the ride the integral of z squared is z cubed over 3 and then i'm going to evaluate this between the z values of 0 and 3. if i plug in those bounds for z i'm going to get x y times 3 cubed over 3 and then minus x y times 0 cubed over 3 but when i plug in 0 it wipes this whole thing out and it contributes nothing so i don't need to write that down if i don't want to simplifying a little bit before i do anything else this is 27 over 3 which is 9 so this is 9 9xy simplified now the innermost integral involves gy so we want to integrate with respect to y and this means that our bounds here are y values so i'm going to label them as y values to remember to plug in for y when i'm done doing the inner integral with respect to y 9x is a constant multiple that comes along for the ride and y integrates to y squared over 2. we will evaluate this from negative 1 to 2. i am plugging in negative 1 and 2 for y and so this is going to be 9x times 2 squared over 2 minus 9x times negative 1 squared over 2 or 4 over 2 which is 2 times 9 18x in the first term and then negative 1 squared becomes plus 1 so i'll have 9 halves x being subtracted or 9 x over two nine halves x however you wanna write it our final step is then to integrate with respect to x and then plug in our bounds the integral of eighteen x is eighteen x squared over two the integral of nine halves x is nine halves x squared over two which we will evaluate from zero to one this simplifies to nine x squared minus nine quarters um x squared and i could get a common denominator there now um 9 times 4 is 36 so 36 quarters x squared 36 uh quarters x squared minus nine quarters x squared from zero to one 36 over no 36 minus 9 is 27 so i have 27 quarters x squared or you could wait and simplify this later that's also fine plugging in my bounds i have 27 quarters times 1 squared minus 27 quarters times 0 squared and that tells me that the hyper volume trapped between this box in 3 space and this function in 4 space is 27 quarters if you want to try to integrate this in a different order and you should get the same result so there's three orders you could do i did dz dydx you could also do dx dy dz or dy dz dx lots of options on the second page here we have a more involved example we have a function of three variables given by x plus y so this is living in four space we would like to find the integral over the region w in three space we know w is in three space because of the dv for volume um this w is the solid trapped by the x y plane the x z plane uh in the plane defined by this equation how i'm going to attempt to set this up is the same as i did for double integrals and step one there was sketching the region so i'm going to try to do the same thing here if i want to begin by sketching the region it would be useful for me to have some idea of what this plane looks like and the easiest way to get an idea of what this plane looks like is to figure out what all of my intercepts are so i can find my x-intercept my y-intercept and my z-intercept in order to plot this your x-intercept occurs when your other two variables are zero so set y and z equal to zero and you're going to get x over three equals one and so your x-intercept is 3. your y-intercept comes from setting x and z equal to 0 and then solving for y in this case you'll get y over two equals one so your y intercept is two and then finally for your z intercept set both x and y equal to zero in your formula for your plane you're going to end up getting that z over 6 is 1 which means that your z intercept is 6. this is helpful to us because let's do some visualization here our x-intercept is 3 our y-intercept is 2 our z intercept is 6. we're bound below by the x y plane we're bound behind by the x z plane so here's x y z and i should have also said in order to make this a finite thing um i also need the yz plane so the idea here is this thing is blocked behind by the xc plane and by the yz plane and it is cut off below by the x y plane and so essentially what we're getting here is the plane itself goes through all of these points so it's kind of slanting towards the z-axis and what you end up getting is essentially going to be a tetrahedron so you have something with a triangular base and one two three triangular faces this big triangle is this triangular face in the front and then we have one triangular piece or face in the xz plane um in another in the yz plane okay so this is what it looks like just fyi and you don't necessarily need to draw the picture i'm just doing it so that we can start to talk bounce you have to decide in order to integrate in and you can choose any order that you like you just would have to set up your bounds correctly for the order you choose i typically like to do dz dy dx and the reason i prefer this order is because it goes how i would expect it to usually you choose an x value and you plug it into y and then once you know your x value and your y value you plug those both in in order to get your z um so that's why i go this way so for instance here if i want to talk about the bounds for x this is the outermost integral they have to be constants what is the lowest x value and the highest x value that i hit in this tetrahedron the lowest is going to be zero and the highest is going to be three so my outermost bounds go from zero to three however given a particular x value so choose a particular x value what are the y values that would go with it well at a particular x value you would draw a strip in the y direction and you would be trying to figure out the lowest y value and the highest y value along that strip the lowest y value is going to be zero based upon how this looks and the highest x value y value rather is going to lie along this line in the x y plane let's figure out what the equation of that line is so redrawing in the x y plane in the x y plane we have an x-intercept of 3 and a y-intercept of 2. so we're trying to find the equation of this line which reoriented looks like what we see here now this particular line is going to have an equation given by y equals mx plus b where b is your y-intercept that we can see is 2 and m is your slope for our slope to go from our y-intercept to our x-intercept we go down two in the y direction and over three in the x direction so our rise over runner our slope is negative two thirds and hence the equation of this line is negative two thirds x plus two and this is going to be the upper bound on y so this is negative two thirds x plus two for that all right finally let's talk about our z's we can see that our lowest z value because the bottom of this tetrahedron is in the xy plane the lowest z value is always going to be 0 so there's my bottom bound and then my upper z value is going to come from figuring out if i have a particular x value i've chosen and i figured out the y value that goes with that x value then associated with that ordered pair is some z value up here along our tetrahedron we're trying to figure out what that upper z value would be and since it lies on this face the forward face so to speak of the plane of the tetrahedron the plane um you can get it by solving the equation for the plane for z so i'll make a note of that for us we are going to solve equation of plane for z in order to get upper z bound the equation for the plane way up there is x over 3 plus y over 2 plus z over 6 equals 1. and if i go to solve this for z i will have that z over 6 equals 1 minus x over 3 minus y over 2. and then i can multiply both sides by six to finish solving this one times six is six x over three times six is two x and six y over two is three y so this becomes six minus two x minus three y equals z and that becomes my upper bound for z so 6 minus 2x minus 3y up there and our integrand what we're actually integrating is x plus y all right here is our setup we now have something to integrate so we're going to have to integrate it the innermost integral is with respect to z so we're going to treat x plus y as a single constant which means that the single constant should get multiplied by the variable when we integrate so this is going to become x plus y times z which we are then going to evaluate we just integrated with respect to z so from z equals 0 to z equals 6 minus 2x minus 3y our two outer integrals are going to stay the same i'm just copying that down and then i need to plug in my bounds for z if i plug in for z i have x plus y times 6 minus 2x minus 3y minus what i get when i plug in z equals zero but when i plug in z equals zero it gets multiplied by this and wipes it out so i'm only getting this and then the d y d x on the end indicating my order of operations or my order of integration what variable to do in what order i am next going to have to integrate this thing with respect to y in order to accomplish that it is probably most useful for me to expand this the distribution so we do some distributing i'm going to multiply every term in here by the first x that we see so i'm going to have 6x minus 2x squared minus 3xy and then i'm going to add to that what i get when i multiply y times everything in here so this is going to be plus 6y minus 2xy minus 3y squared and then the dydx on the end of this we look and the only like terms i'm seeing are the negative 3xy and the negative 2xy so if you would like to you can combine those before you do your next integration step you don't have to but it might make your arithmetic easier later or your algebra um so i'm going to have 6x minus 2x squared minus 5xy by combining these two like terms and then we have the plus 6y and the minus 3y squared as well which i am writing in blue for no reason whatsoever we want to do this in order dydx next step is to integrate everything you see here with respect to y so i'm going to have 0 to 3 on the outside if i integrate everything in here with respect to y 6x is a constant integrated it becomes 6x times your variable y negative 2x squared is also a constant it's going to end up getting multiplied by y when you integrate it negative 5xy becomes negative 5y squared over 2xy squared over 2 so negative 5xy squared over 2 plus 6 times y squared over 2 and then minus 3y cubed over 3. which must then be evaluated from zero to negative two thirds x plus two where you are plugging in these bounds for y because we just integrated with respect to y okay there is um stuff happening here notice these threes cancel and that six over three can be reduced to three just to clean that up a smidgen and we want to plug in y for y and no matter what i do this is not going to be particularly pretty um so i'm just going to go for it you have options to make it try to look a little bit cleaner like you could factor a y out of every single term maybe that would make your life better you could combine these two sort of by factoring out a y squared there are a couple different things you could do but i'm just going to leave it and plug in as is if i plug in negative two thirds x plus two for y i get six x times negative two thirds x plus two minus two x squared times negative two thirds x plus two minus five x times negative two thirds x plus two quantity squared whole thing over 2. i have made it this far i still have these two terms to go and i'm going to have to carry on down here because i'm out of space and i write too large um so in addition to this we are going to have plus three y squared becomes plus three times two thirds x plus two quantity squared and minus y cubed would be minus negative two thirds x plus two quantity cubed and then minus what you get when you plug in zero for y but since there's a y in every term everything gets zeroed out and that doesn't contribute anything else so we're just going to take this long string of stuff and integrate it with respect to x if you wanted to do this you absolutely can you have to distribute 6x through here negative 2x squared through here you need to distribute out negative 2 3x plus 2 quantity squared and then multiply it by minus 5x square this then multiply by 3 cube this then multiply by the negative sign out front that is a lot of algebra it's totally doable just extremely time consuming and a little bit tedious if you do it all you will get 15 halves and i'm just skipping those details for the sake of time and a little bit of suffering let's be honest here now this is a perfectly reasonable thing to ask for a person to set up say on an exam and it's certainly a reasonable thing to ask for on homework where you have basically as much time as you need up until the due date in order to accomplish things but like on an exam there are so many places to go wrong in the algebra just multiplying all this stuff out uh chances are i would not give you something that long for simply that reason because this isn't necessarily this is seeing if you can do algebra multiplying things out combine like terms more than it's seeing if you can do calculus 3. so i care most about in this situation the setup and understanding how essentially you get your bounds depending upon the order that you're integrating in that is the most important part to me talking about this and also it's good to know how to integrate with respect to each variable as well but again on an exam or something i would not give you something this long-winded example three brings us to something a little bit different it asks us only to sketch the solid of integration w corresponding to the integrated integral that we see here so here is um the idea we have all of these different bounds determined by our order of integration our outer is dz so this is z equals 1 and z equals 0. my middle is d y so i have y equals root 1 minus c squared and y equals negative root 1 minus c squared and then my innermost is x and so i have x equals root 1 minus y squared minus c squared and then x equals the negative of the exact same thing i would like to take all of this in sketch my region of integration and so essentially my region of integration w you don't necessarily need to write this down but your z's live between zero and one your y's live between negative one negative square root of one minus z squared and positive the same thing while your z's live between the positive and the negative of the square root of one minus y squared minus c squared all right we're gonna try to sketch this as a triple integral this region lives in three space here's x here's y here's z the easiest part of this is probably zero to z or zero to one for z so we know our z values only go between zero and one boom so we're bound below by the x y plane and above by the plane z equals 1. that we know for sure from there here are two bounds on y that depend on z so we have for instance y equaling root 1 minus z squared if you square both sides you get y squared equals 1 minus z squared or y squared plus z squared equals one which in the y z plane would give you a circle of radius one but since this is only the positive square root that we started out with that means that what we are in fact dealing with is a semi-circle of radius one and i'll talk about which semi-circle it is in a second um and this guy is also a semi-circle of radius one if you do the same thing this is a semi-circle of radius one in the y z plane that only has positive y values so if you're in the yz plane um that would mean that your semicircle um is going to be out this way for only positive y values and you would not have your semicircle back there coming from this so in the y z plane we're going to end up with a semicircle let's do it in red so semi circle of radius one let me put a one here and also put a one there for giggles um so the semi-circle of radius one with all your y values positive however we have an extra restriction and our extra restriction is that our z's cannot be negative so with my y's positive the semicircle would actually extend all the way but due to this earlier restriction i'm going to stop it and i'm really only going to get a quarter circle rather than a half circle how about the other side of this this is the semi-circle of radius 1 again in the yz-plane that's nice to mention and it's the semi-circle of radius 1 in the y z plane that has negative y values because y equals the negative square root um and so that's going to be the back semicircle and so your y axis continues back this way somewhere there's negative one um and you're essentially getting like the other part of this semicircle and again this would normally stretch all the way down to z is negative 1 but because of this earlier restriction that z cannot be negative we don't have any of this we don't just extend below the x y plane so so far we have step one our z's lived between zero and one step two our bounds on our y give us a semicircle in the y z plane because we have y equals each of these equations and then finally our last bit here was dx and i put a z but i meant to put an x our x's are living between these two things well let's talk about these things if for instance i look at this guy this is saying that we have the equation x equals 1 minus y squared minus z squared we have the negative that on the other side if you square both sides you get x squared equals 1 minus y squared minus z squared and then rearranging you're going to have x squared plus y squared plus z squared equals 1 which is a sphere centered at the origin of radius 1. now we need to put some restrictions on this sphere in order to make this part make sense so this equation on its own involving the square root is not a full sphere it's a hemisphere and it's the hemisphere that appears with all positive x values so the hemisphere of all positive x values would have to live with a base out this way in the x y plane so let me see if i can draw that in i'll draw it in purple um the hemisphere with all positive x values is going to be here and it would extend with this face and it would also go down um this way as well and so it's sort of a hemisphere with its flat face it's flat side in the yz plane is the way that you can look at it however we can't have a full hemisphere because our z values are not allowed to be negative based upon our earlier restriction so we're getting the for this one we're getting the upper half hemisphere um of radius one that i have plotted where your x's are positive the piece of that hemisphere where your x's are positive and your z's are also non-negative all right so that is this piece the other side of this is the exact same idea and the only difference behind it is this is the one with the negative x values instead so this guy is x equals negative the square root of 1 minus y squared minus z squared and so this is giving you the quote unquote back hemisphere of radius one which would be behind the y z plane with its flat side on the y z plane and so that would um extend you know down into this territory down to z equals negative one but again we have the six restrictions so we're going to ignore that and we are going to pick up um this is a bad drawing i'm so sorry um we're going to pick up the piece of this hemisphere where you have positive z values so you're going to really get a quarter sphere the quarter sphere with flat side on the yz plane um with z values between 0 and 1. so essentially our figure our region w is the region given by the upper half hemisphere of radius one or i guess i could just say upper hemisphere it's not a half hemisphere it's a half sphere so the upper hemisphere of radius one centered at zero zero zero in the three space you could alternatively looked at this in a different order you could have started with your innermost integral and hence with your x's and you could have begun with essentially this would give you a full sphere of radius one and then you could go backwards and then restrict your y values if you take your hemisphere of radius or sorry your full sphere of radius one that's what these two things do together um your full sphere of radius one and you restrict it so your y's live between these two semi-circles what pieces of that does that cut off well if your y's live between these two semi-circles these two semi-circles together are forming the full bottom of this sphere in the x um y plane is what they would be doing um so we'll be giving you that full thing we would need oh no in the y z plane y z because those are two variables so this would be giving us a full circle if you projected down to the y z plane so that will not have changed anything yet you would still be with just these two looking at a full sphere of radius one centered at the origin the thing that keeps it from being a full sphere then is restricting from zero to one again going backwards once you add on that restriction you take the full sphere of radius one and you chop off any part of it that has negative z values which gives you your fat face on the xy plane and then your upper half atmosphere exactly like we see here so you don't necessarily need to think of it in this order you could also do the reverse order and figure out what's going on there's one illustration of how you can identify what your region looks like from your bounds that was so much fun we're going to do it more times so we're going to describe the region of integration for each triple integral that we see and so if we start looking at this my outer bounds for x and y are constants so we have that our y's live between zero and one and we have that our x's live between negative one and one and that's going to put a restriction on whatever is happening with these bounds which apply to z so the lower bound being z equals zero tells me that we don't have any negative z values so whatever i'm going to plot is on or above the x y plane we're going to have nothing below it in addition to this i know that my upper piece of this is z equals root 1 minus x squared and this is in three space because it's all the way on the interior of a triple integral so this is going to be in three space this is the same as z squared equals one minus x squared when we square both sides or x squared plus z squared equals one and you may recall if you're in three space this equation gives you a cylinder of radius one centered along the axis associated with the missing variable so it's a centered along the y-axis so essentially what we want here is the cylinder of radius 1 centered along the y-axis with these restrictions on x and y respectively so we have that our y's are between zero and one and we have that our x's are between one and negative one and we also have that our z's um cannot be negative and so instead of a full cylinder of radius 1 centered along the y axis we are going to chop the cylinder so that it does not go below the x y plane so this would essentially give me a half cylinder instead and in addition to that we are only going to have that cylinder stretching from zero to one along the y axis i am going to attempt to draw this our y's are between zero and one our axes are between negative one and one so you can think of this as if you had a cylinder like this on its side and you cut it straight through the middle horrible drawing but this rectangular thing is going to be what i have there and then the piece of the cylinder i'm getting is stretching along the y-axis between y equals zero and y equals one and only has positive z values and so it's going to be like coming up out of here like so only in this piece so it's a piece of i guess we'll call it piece of the upper half cylinder of radius one extending from y equals 0 to y equals 1 along the y-axis and i guess i should even say centered along the y-axis to be specific okay so that's that description for b similar idea to start with we begin with the bounds on y and x being constants and so we're going to have some kind of rectangle that we start with in the x y plane our y's are between 0 and 1 and our x's are between negative 1 and 1 again so so far this sounds familiar this is exactly what we had seen at the beginning of the previous example the only difference here between this in the previous example is that we have that this bottom bound is not zero instead it's negative the same square root we had up here since this was the top half cylinder of radius one this would be the piece of the cylinder where your z values are zero with a cylinder of radius one extending along the y-axis um if you put a negative on it that is the other half cylinder of radius one um which is going to be with the negative z values instead so i will mention this is the upper half cylinder of radius one extending along the axis of the missing variable and so we have an x here and since this is associated with the bound this is about associated with the variable z you have z equals this the missing variable is this y um extending uh along y axis and then this guy is the lower half meaning negative z values lower half a cylinder of radius 1 extending along the y axis so the difference between this and the previous example is we get not just the upper half cylinder but also the lower half cylinder because our z's are allowed to be negative so we get the exact same thing we have here but also the mirror image of it appearing below the x y plane so this is going to be the piece of the full cylinder of radius one where your y values live between zero and one so piece of the cylinder of radius one centered along the y-axis from y equals 0 to y equals 1. so instead of an infinite cylinder it's only a piece of the cylinder that is our region w we have one last one to go here um and there's a lot going on if we look at this we can see that our x's live between negative 1 and one this is good to know these two equations are associated with the middle integral which is with respect to z so i can think of them as z equals positive and negative root 1 minus x squared and this right here is a semi-circle of radius one and it's in the x z plane because those are the two variables that appear and because it's the positive one you're going to have that z is greater than zero for this one the bottom guy is also a semi-circle of radius one and it's in the xz plane because you have z equals a negative square root it's going to be the semicircle that has z less than zero so stretching between both of these with x being allowed to go from negative one to one gives you a full circle of radius one in the x z plane so our x's went from negative one to one and our z's correspond to a full circle of radius one in the x z plane the xc plane is like our left wall so to speak so there would be a full circle of radius one associated with that i'm dotting it in for now because whether or not it's actually going to be full um for our final figure depends upon what our y's are doing so let's get to that we are restricting things so that our y's are negat never negative and so that means that we cannot have anything that extends behind essentially the xz plane if we did then that would be negative y values so so far we're okay because the circle i drew was in the xc plane itself not behind it and so that's that restriction anything we have is going to be coming frontwards out with flat side on the xz plane next up we have this upper bound and this upper bound is y equals the square root of one minus x squared minus z squared since this inner integral is with respect to y and we've seen before if you square both sides of this and move things around this came from the equation of a full sphere of radius 1 centered at the origin however since this is y equals the positive square root this is the upper um i guess upper is bad ways to say that it's going to be the hemisphere of radius 1 where your y values are always positive and if your y values are always positive this is going to force your flat side is on the xz plane so this is going to be once again very challenging for me to attempt to draw but the idea is this hemisphere is kind of coming out forward towards us and does not extend backward through the xc plane so flat side on the xc plane um and then the rest of this coming out and since the only restrictions are x between negative one and one that doesn't change me having the full circle corresponding to our bounds on z here in the xz plane and the only thing that really gives us a restriction on this sphere then is making it so we can't have any negative y values so that's why we get a half sphere instead of a full sphere and this is the half sphere that we are looking at um all right so you could restate what i just said we have the hemisphere of radius one with flat side on the x z plane and y being positive so hopefully this gives you a little bit of an idea of how you can look at bounds and figure out what region goes along with them if you have any questions about this or anything else from section 16.3 be sure to get in touch