Hey, so in this section, we're going to do exactly what we've been doing with volume. But instead of taking a, like in previous examples, a washer, right, these washers, or these discs, right, we're going to use volume by the shell method. So the difference between this one and the previous sections is 2.2 calculated volume in which the washer or disc was perpendicular to the axis of rotation.
So if we rotated around the x-axis, that means our disc would have to be horizontal because the axis of rotation is vertical. Therefore, our disc, or in this case, our washer, has to be horizontal. Same here. If our discs were vertical, this means that our axis of rotation had to be horizontal. And sure enough, it was around the x-axis here.
And so just taking a look at one more, notice that here our washer was vertical. That means our axis of rotation had to be horizontal. So essentially, for 2.2, All our method had the disc or washer that was perpendicular to the axis of rotation. But in 2.3, the only difference is now we're going to just take the same exact type of stuff, but our axis of rotation will be parallel to our shells.
So if I want to go ahead and just take a simple graph here, let me draw it here. Okay, in an interval from A to B, A to B, here's A and then B. Let me go ahead and draw those lines. And so let me go ahead and now shade in blue.
this region R that is above by the graph y equals f of x, which I just drew, and below by the x-axis. So they're just bounding it by this graph and the x-axis. Okay, and then I'm going to, there we go. And then let me mark, this is going to be y equals f of x. Okay, so here it says a region R, so here's our region R, that is bounded by, above, by the graph F of X.
It could keep going forever, but it says below by the X axis, so we have all this. And left and right on, from A to B, right, that's the closed interval. and then it revolves around the y-axis. So now our axis of rotation will be y. So as we do that, I just want to remind us how we ended up getting those disks or cross-sections in the first place.
Recall that we took a, this is like x sub 0, this is like x sub n, and then in the middle somewhere, we took some random x sub i, right? And because we can take, it's the midpoint, uh, or it could be the midpoint or left endpoint or right endpoint right we just called it x sub i star or x sub i sometimes just anything between x sub o and x sub n right and then we drew this rectangle which rectangle all right and then we Made it, shaded it, and made it look like a rectangle. Okay, and then what did we call this, right?
And all of you know, you're like, yeah, so we called this thickness delta x because it was the base of our rectangle times high base times height, right? Well, now because we're going to be spinning this across the y-axis, and you can only see what it's going to look like here, right? Like, I'm so sorry, that's not right.
There we go. and we're going to spin it. This is going to be empty, but this is going to be the solid, right?
And because it's going to spin around the y, that means our axis of rotation is vertical, and therefore our Rectangle also has to be vertical and we're going to rotate this rectangle across the y-axis. And when we do, we get one shell of the entire solid. And we'll draw the solid in the area below.
But right now, I just want to focus on that here, the axis of rotation. And I want to put a note, make sure you note this. So note that.
The axis of rotation, AOR, is parallel to the rectangle. It's going to become a shell in a few seconds, but for now we'll go ahead and call it that. And then, because it's the axis of rotation is parallel to the rectangle, this is going to end up, when we rotate it, after rotating, we get a shell.
Now, because the shell itself will have this thickness, if you can only imagine this shell here. like this and then a thickness right it's gonna have a shell this little thing and it's gonna look like oh is it looks like just like the washer it is but it's more of like the thickness of the shell like turning around so um like a pot you know like at the top you can see a thickness on the pot on top of a pot right so that's what that is so we're what we're gonna end up having to do is Because we're going to be taking an, you know, we're going to take one volume of one of the shells, then take n many volumes of the shells, and then infinitely many, and then we can get the definite integral, right? Just the same process as before, right?
Process never changes, problems do. So because we're going to be taking an infinitely amount of these shells, I mean, we're going to have this thickness to worry about, like we did with the washers, like those radii, right? So here, but you know, we can always think about, well, if this is I, X sub I sub star, right, really what we can say is this is going to be this piece here, right, that middle part.
And the endpoints, well, this will be on the left, X sub I minus one, and X sub I in the next one, just like with X sub n. right? The next one would be X sub n minus 1 and then X sub n minus 2, right? But because X sub i is in the middle somewhere and who knows, right? Left hand or whatever.
We put the X sub i star to indicate that it would be a midpoint, just like we have been since the beginning of chapter 1, right? When we did the Riemann sum there. And so really what we're doing is this delta x ends up really just being this.
thickness right but because we're doing volume it's going to make it a little bit different so let's go ahead and note up here that we still get that delta x we're really used to and if i really wanted to calculate exactly that delta x i would just take the right value minus the left value so really let's call that x sub i minus x sub i minus one so that's really the thickness and it's really just delta x right and we're just finding that distance Okay, so that's pretty much what we have so far, right? And we know we're going to spin it to make a volume, but that's the idea right now is to understand that the rectangle we're going to be taking that will create a shell for us when we do the volume is parallel to the axis of rotation. And that thickness of that shell that we're about to rotate is going to be x sub i minus x sub i minus 1. Okay, so now...
and I wrote this in a note right that is different and that's because that is always going to be para the pit word parallel right that's in it okay so now that we have this interval for Delta X right and let me go ahead and write Delta X right up there so you I don't want you know it's not it's not supposed to be sneaky it's just supposed to be Delta X represented in this tiny little interval that's all so the love the height X sub I star so you'll notice here That x sub i star is that same exact x sub i star that we had up there. And f of x sub i star is just the y value there. So because we called x sub i star this, sorry, that is not right. There we go.
Because we called X sub I star the midpoint of this interval, right? Just like we have before in the previous chapters and sections. This is the Y value is just F of X sub I star, right? We could call it Y X, you know, but the Y value that coordinates is just the F of X sub I star. Okay, so Here when that rectangle is revolved around the y-axis we get a cylindrical shell and I just did that to save some time a little bit.
So in doing that again we're going to take this piece here and then rotate it around the y. So and we're only we're gonna take this one here and so that way maybe you can kind of see it a little bit. So if I take this, let me draw the graph, there we go.
Doesn't look very good, but you get the idea. And then we draw the barrier. Okay, so now I'm just leaving it as a solid region.
And of course, I'll just paint it. And please do this. I know it seems like, oh, she's drawing it so many times.
But you know, the more you draw it, the more you understand what's happening geometrically. So at one point when you're doing homework or something, you actually don't end up having to draw it, but you could actually visualize it. So the more you draw it, the more your visualization skills get better.
And it's a skill you will need, especially if you're going into STEM and math and taking some physics, you really have to vision some of the stuff, visualize some of the stuff. So you make those connections. So I always end up drawing, I draw tons of pictures as you'll see. Okay, so now let's go ahead and take this rectangle. And I'm going to draw just a little bit bigger so you can see it like thicker, just so you see the shell part.
Okay. And then we called this, again, this is X sub I minus 1, and this is X sub I. And we said this thickness was delta X, right?
Okay, but don't forget that when we called this midpoint of it X sub I star, that way it doesn't really matter if we're taking left and right endpoints. That's really what it's about is just taking any endpoint, really. And then this would be the F of X sub I star.
So again, because this is the midpoint, right, we know how to calculate midpoints, right? Midpoints, the midpoint x sub i star is really just going to be the average of the two endpoints. That's how we found it in geometry or trig, right? Any sort of midpoint, that's what it is. So let's take x minus i.
Minus 1 plus X minus I all over 2. So we're just really just taking that average. Okay, so now I'm going to go ahead and try to do my best in drawing the shell. And again, here, I'm going to do this smaller one first. And it's not going to come out perfect the first time.
But... At least we get some idea. So you can see this thickness here and let me go ahead and shade that of the shell. And if you want to go to your book, the textbook has actually some pretty nice pictures of the shells. And then let me go ahead and...
so you can kind of see now the shell that's happening right like and it's going to be this a flourishing like this and you'll have some shading here right and then really what you have is a um get that looks like this the whole solid you know looks like that And this is all solid here. Right. And then it's like an empty spout going in.
So you're really just filling up that little area of water. And then all that blue is solid. So when you spin that rectangle, that delta X that we just thought was a base times height before an area and 2.1, we just like said, oh, yeah, it's just this.
But once we got into volume, it became a thickness for us. and we rotated it. So because we're rotating it now around the y-axis here, we make a rectangle parallel to the axis symmetry and we're going to spin it and it makes this shell. So notice it's this thickness of a shell in which we have to find the volume for.
And so one rectangle, I'm sorry, one shell gives us a certain volume and then I can take n of those And how many n? Infinite many. So then I shoot the limit to n approaches and then I get the definite integral for volume. So this is the work that we're going to have to do. It really is about involving the building the formula that we have.
And it's very important we build the formula because if I, your whole life you've been given a formula and you believe in it. But you never did it for yourself, right? We just told you what Pythagorean theorem was. We never showed you how the mathematician got that, right? But in here, because you're in second semester calculus, we will be showing how we get formula.
We're no longer just believers, right? We're going to own it and we're going to prove it. Not only we're going to be believers and have faith in those formulas, it is only because we have proven them.
And so we really have to believe in our work at this point. So it's up to you when you start to really care about that. but you should really begin that process now of like let me get let me see let me let me question this a little bit right so i can tell you the formula of course but it doesn't mean much to um calc to math students if we don't show it and it's there's such a it's such a simple way of um obtaining the formula that i think it's just a great way of like showing students and Again, it's that same process of the Riemann sum thing where you take one rectangle and then in this case we're going to take one shell, n shells, infinite shells.
So again, we'll just build it just like that. Okay, so here let's start. So the first thing we're going to do here is take the volume of one shell.
Let's just do one and then obviously all the other stuff is just pretty simple, right? So the volume of one shell is always going to be the area of the base times the height. Again, those formulas never change since you were in arithmetic, right, when you were doing those a long time ago.
So we're going to just go ahead and use that same formula. Okay, and then, so what is the area of the base? Well, we know that here, that area of the base of the shell has to be this radius, bigger radius minus the smaller radius pi r squared right so again it's going to be this bigger one here we'll call it r sub i so let me do it in purple here so you can see so it's going to be this bigger one here we'll call r sub i and then the inside one which I'm going to do right here That's going to have to be r sub i minus 1. So in our case, it was called x sub i and x sub i minus 1, right?
But I just want to make it clear that really that shell, that area of that base, is going to be that outer circle minus that smaller circle, right? So it's r sub i, right? But in our picture, we call that radius x sub i and x sub i minus 1. So let's write that out.
So the area of the base here would be the area of the outer circle, which is pi x sub i squared minus, so that's, I put a parentheses around there, minus the area of the inner circle, which is pi x sub i minus 1 squared, right, times, oops, not a star, times the height. Okay, and then if I went ahead and just played with this a little bit, I do see I can factor out a pie, right? I do see I have a difference of two squares. And then I have x sub i star, which is just the midpoint height, which we can just leave because that was the only height we really talked about. So let's go ahead and try to do this.
Let's factor out a pi. It's all multiplication. So I think we can go ahead and just factor out the pi and then put f of x sub i star outside and leave the x sub i minus, sorry, squared x sub i. I minus 1 squared.
And then I realize that that delta x is the difference, right, that distance, x sub i minus x sub i minus 1. So right here, let me write it here, and then don't forget that delta x is x sub i minus x sub i minus 1. So I'm going to go ahead and notice that This is going to be a difference of two squares, a dos. So I'm going to go ahead and factor that out. So I get pi times f of x sub i star times x sub i minus x sub i minus 1 times x sub i plus x sub i minus 1. And again, that's just the difference of two squares, right?
If you have a minus b, a squared minus b squared. That is just the difference of two squares, right? A minus B times A plus B.
We learned that way back in algebra and polynomials. So again, we're grabbing all these tools from all of our life and using it. So we're using a lot of pieces here. Okay, and so then I want to go ahead now and take that, what I reminded myself of, that delta X was just this. So let me rewrite that now.
So then I get equal to pi f of x sub i star delta x. And let me highlight that yellow. And then we have x sub i plus x sub i minus 1. Now I do recognize this piece here. Now I see the relationship between x sub i star and this numerator and I see that the numerators are the same except I do have that midpoint where I have to divide by two but you know I know some algebra tricks where I know I can go ahead and and deal with that and that way I can just replace this last factor here with x sub i star.
Okay so let's go ahead and try to do that. If I want this to be x sub i plus one plus x sub i minus one over 2, I need a divided by 2 there, right? In order to replace this entirely with x sub i star, I have to have a divide by 2. Well, I can do that.
That's just arithmetic. I can divide by 2 and multiply by 2, and then I don't change anything. It's like multiplying by 1, right?
It's like I manipulate that number. So let me go ahead and do that. Let me go ahead and divide this by 2. but then multiply by 2 next to it and that way you can see it's just like having the original i'm not changing the problem i'm manipulating 2 over 2 which is 1 right 1 to be 2 over 2. okay now what i can do is this is equal to now i'm going to throw the 2 out in front and get 2 pi f of x sub i star delta x and then I get this piece here to be x sub i star.
So see how fun that is? Like we can manipulate This is to make a much cleaner formula. And this is only one shell. We're only doing the one, right?
We still have to take n and infinite, right? Okay, so now, rewriting this nicer, we're going to get 2 pi, the x sub i star, f of x sub i star, delta x. So the volume of that shell is just 2 pi times x sub i star times f of x sub i star delta x, right? So really what it is, is the x value times the y value and then the width of the shell. And so this is only one shell, right?
One. Right, so now let's go ahead and take, well, we'll take n many and then infinite shells. So let's go ahead and take the volume of the infinite number of shells. Okay, and what is that? Well, this is now the volume is going to be equal to.
The limit as n approaches infinity of the summation from i equal 1 to n of 2 pi x sub i star f of x sub i star delta x. Okay, and then that is just by definition the definite integral from a to b and then 2 pi and then times x times f of x delta x. I'm sorry, dx.
Then that's the exact formula and you just prove that so no you don't have to have a question anything anymore I think the tough part was just getting that shell Situated you know and then the only trick I found You know bringing in is this piece here. That's that is something we really had to bring in right our own technique So as long as we can do that and bring divide by 2, multiply by 2, I get x sub i star here. And then I do get a 2 pi on the output. That's just a coefficient.
So it doesn't really, you know, that's just, that's okay, you know. As long as we can replace that, you know, anything to make our math more efficient, I think is really great. So as you can see, since this is a coefficient, we do put it in front of the integral here in the formula on the next page. And then you see the x, f of x, dx. So don't forget this x because it does change your problem that we'll see in the next example.
That you'll see that because you're multiplying by x, right? So it's just going to multiply how many x's. So the power rule, things like that are going to be a little different.
So don't forget that x. Okay, so that is it. That's the volume.
What are the key things about shell method? When you see the word shell, using cylindrical shells. Right, you know that you're going to take a rectangle that's parallel to the axis of rotation, and then you're going to rotate that that way. Okay, so let's try an example.
So here you have one over the square root of x, and it's going to be bounded by the x axis. So you shade up and the vertical lines x equal one, and then x equal 4. Okay and then I'm going to move this line over a little bit. Perfect.
Find the volume of revolution by revolving R around the y-axis. So let's go ahead and shade in our region R. We're doing the same process as before. So region R, there you go. And we're going to rotate it around the Y axis.
So if we do that, and then there's that volume, right? So you see like the hole in the middle that you can fill up, right? And then...
And so you can fill that up and it's like a hole in the middle and then you have the solid around it. And then you would be rotating that bigger solid around the y-axis. So really the width would be, so you can kind of see that it's going to be pretty wide.
And now when we want to integrate, we see that we can take one rectangle. So if I took one rectangle, I don't know, some random area here, right? Because we don't really care.
and then we want to rotate that and actually let me make that in red So we get an idea of the shell that's happening. So this is one of the many shells we're going to take, right? So let's go ahead. Now that we have, we see it and we're letting visualized it.
Let's go ahead now and make that formation. Now we are rotating it across the, um, Y axis. Our rectangle was parallel to it as it should.
And we rotated it exactly the way it should. And we do know that we don't need to worry about f of x sub i star midpoints anymore because we dealt with that. We have the formula now.
We know that all we have to do is just take the volume, which is 2 pi from a to b of x times f of x dx. Like we already did all that hard work just so all of this can be real nice and quick and smooth. So because we're only taking, we're not taking 0. This is where it might get a little bit confusing, but recall, because some of you, I know what you're thinking.
You're saying, oh no, you got to take this big radius minus the small radius, like the washer method, right? We don't. Recall that our interval here was only from one to four.
So it was going to be, so it's four and then that radius here. So it's only going to be this shaded region, even if it's rotated around the Y. If the integral, if this was 0 to 4, then yeah, we would have to do that shift, right? We would have to like integrate that empty cylinder in the middle. But because our r's is 1 to 4, it's kind of nice.
We only are taking that solid part, which is really nice. So let's go ahead and build it. We get 2 pi from 1 to 4 of x times 1 over square root x dx. Okay, and then we get 2 pi.
The integral from 1 to 4 of x times x to the 1 half, negative 1 half, dx. So I'm rewriting the square root of x as x to the 1 half, and because I want to put it on top of, you know, as a numerator, it's a negative, right? And that way we see that we have same base, product in between, we add exponents. So let's go ahead and add those. We don't always write it, but we know it's there, it's x to the first.
So then we get... 2 pi definite integral from 1 to 4 of x to the positive 1 half dx. And let's go ahead and just integrate that real quickly. So then we get times x to the 3 halves over 3 halves evaluated from 1 to 4. So here I'm going to get 2 pi times 2 thirds, which gives me 4 pi thirds. And then I'll get times and then I'll evaluate x to the 3 halves at 4 and then 1 minus 1 to the 3 halves.
Okay, so then we get 4 pi thirds times and then 4 to the 3 halves, that's 2 cubed which is 8 and then minus 1. So this is just going to be 7, right? And then 7 times... the 4 pi is 28 so then we get the volume is equal to 28 pi over 3 cubic units so it's kind of nice even though some of you i'm sure were like uh i'm i'm gonna forward through all this hard work up No way I'm going to do all of that. But honestly, it's worth it because if you sat through this with me, you really have the deep, you fully have a deeper understanding of what we're doing. That, okay, we took one shell, we're going to take infinite shells, we're taking that midpoint of f of x sub i star.
So, you know, not only are you capable of using the formula, but now you really truly understand how and why you're using that formula. Okay. And so just like before, if, you know, we can do it with respect to x, we can also do it with respect to y.
All right, so in this example, we're going to take the function y equals square root of x and bound it with the x-axis 0 and 4 and revolve it around the x-axis. So when we do this, and Because let me, here we go. I'm going to do it like this. And then I'm going to go ahead and make vertical lines at x equals 0 and x equals 4. Okay, and then we can see that we're going to shade the area here. Do it a little bit quicker with a thicker marker.
Okay, and then a little, here we go. And so it looks like that and then when we rotate it, it's going to be and then you can see you know how it's going to look and rotate as you turn it. So the one thing is when we want to do the method of cylindrical shells but about the x-axis.
So there's our little, let me highlight that yellow, there's that little pattern for you red flag. So when they want to revolve, they want you to do it by shells around the x-axis. That's your red flag that you're going to have these, you're going to be using in terms of y.
And so if I go ahead and draw a rectangle that represents one of the shells. parallel again parallel right to the axis of revolution it would be horizontal so i'll just take it at a random spot so let's go just make it pretty thick like that and so you can see that that little rectangle there will end up spinning itself so let me try to draw that duplicate do like a little bit bigger so you can kind of see the shell so that radius i'm sorry that other circle did it came out a little slanted but you know you kind of get the idea what you're seeing here and so again remember that um this is just going to be the um the base times height right like so we know with cylindrical shells that we have circles as our base. So the area of the base would be pi r squared, right?
And then we need, we have two of them. So the shell that it makes is going to be this radius here, but then the height, notice it's a little bit different. If we just look here, we know that it's going to be this formula, y, right? It will be, I mean, dy will be this thickness.
The base times height we get from the formula and so we can see that now this height if I like turn it right this width of this rectangle is going to be along this curve right here to the line y I'm sorry x equals 4. So we can see that this here is this rectangle's width will change as it goes along the graph of the square root of x and then it and bounded between on that x equal 4 line so the the distance or the height here is going to be this you know we always take right to left right you know when we want to find distance we take the bigger one subtract from the smaller one so we see here that we're going to have to take this line x equal 4 minus whatever y value is here or x value is here the graph so if the graph if it's here it's going to be 4 minus whatever that value is so we can see that we're going to have to use that line x equal 4 as our right bound for our rectangle and then that height of that rectangle however you want to look at it sideways or not I always look at it sideways so I look at its height and then the dy is the as the base but sideways So looking at sideways, we could see that we're going to have the line x equal 4 minus whatever x value is going to be on that function. It's going to just get like go like get smaller and then a little bit wider until it reaches that x axis. So in this case, when we do this, we're going to go ahead. And the first thing we need to do is we're going to know that we have shells around the x axis.
We know we have to make everything's in terms of y. Meaning that we're going to go ahead and have to take the first thing we're going to have to do is take Y and rewrite it in terms of Y. So we're going to rewrite.
y. So if y is equal to the square root of x, this would imply that x squared is equal, oh sorry, y squared is equal to x. Or we could just say x is equal to y squared, just kind of make the x on the left. So that will be our integrand function here, okay, and then we'll have times y.
But remember that Our limits of integration here again will be from here we see zero to two let me put zero there right so here's a and I'm so sorry that a in the formula it is c so here is c and then here is d. So we can see that if I look sideways I can see the interval goes from zero to two and of the function y squared, right, if it's in terms of y, and then the height on that is going to have to be 4 minus that function. So now let's go ahead, we'll do the lois, we'll just kind of be more organized about it. So c is equal to 0, and then d is equal to 2. And then the last part would be to go ahead and let x equal to y squared Then we can go ahead and say the volume is equal to 2 pi definite integral from 0 to 2, right?
There we go. Y times, and now we can go ahead and get our height of our function. So the height of our function in terms of y. is going to be 4 minus y squared dy.
So let me write a little note here. So the height is 4 minus the function g of y. And really what I'm doing is I'm looking at it sideways and I see like the top is 4 and then the bottom is everything below it is the function. Or you could do right minus left, however you want to see that.
So then it just becomes a normal integration problem, but it's just in terms of y. So the first thing I'm going to go ahead and do is distribute that y in there. And then we get 2 pi definite integral 0 to 2 of 4y minus y cubed dy. And then we can go ahead and put 2 pi times, and then we can have 4 times y squared over 2 minus y to the fourth over 4 evaluated from 0 to 2. And so this would be equal to, just kind of simplifying here, right?
We get 2 pi times 2 times, oops, let me go do that in black. two times so I'm just taking this coefficient out and then I'll just square the y so then I'll have two squared minus two to the fourth over four and then minus and then put in zero so you'll have two zero squared minus zero to the fourth over four you So again, we always see that zero in there. We see kind of the pattern with that. It's going to be zero. And then if I just simplify what's there, I'll get 2 pi times.
And then 2 times 2 squared is 2 cubed, which is 8, minus. And then 2 to the 4th is 16. And 16 divided by 4 is 4. And therefore, we get 8 minus 4 is 4. four and four times two is eight pi and then cubic units.