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Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the y-axis.

$ y = e^{-x^2} $ , $ y = 0 $ , $ x = 0 $ , $ x = 1 $

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01:53

Wen Zheng

01:48

Amrita Bhasin

Calculus 2 / BC

Chapter 6

Applications of Integration

Section 3

Volumes by Cylindrical Shells

Campbell University

Harvey Mudd College

University of Nottingham

Idaho State University

Lectures

0:00

Use the method of cylindri…

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02:16

03:40

01:18

03:07

01:16

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00:59

we want to find a volume that is generated by rotating the region bounded by the curve. Why equals uh E to the negative X squared. And uh the lines Y equals zero, X equals zero And Ux x equals one. Uh So the region that is bounded by this curb. And these lines, that region is going to be rotated around the y axis. And we want to use the method of cylindrical shells to find the volume generated to get a better idea idea of what this region looks like. I have a uh this curve and these three lines graft on the dez most graphing calculator. All right, so here's the portion of e. Actually it's craft incorrectly. Let me fix this. We raised to the negative X squared. There we go. Okay. Uh So here's the portion of the curve Y equals E to the negative X square between ex uh equals zero and X equals one. And of course this is the line Y equals zero. So we're looking at this little region in here, let's blow it up a little bit. We're looking at this region in here uh is going to be rotated around this y axis. So we're going to take this region and spin it around the y axis. And we want to use the method of cylindrical shells to find that volume. Well volume. Using a cylindrical shells is going to equal to times pi time she entered role. Looking at the uh region. Again, we're going to integrate from X zero up to x is one uh X times our function. Why? Which is E. To the negative X squared D. X. So once we calculate this definite integral uh That will be the volume when this region is rotated around the y axis. And this is using the method of cylindrical shells. Well to evaluate this integral. Uh First um let's go back to let's just go off to the side here. What is the integral of X. He to the negative X squared D. X. If we use U substitution let U equal negative X squared. Then D. U. D. X Would be -2 X. Uh so d'You would be -2 X. D. X. Let's get a space here. So I wanna rewrite this integral in terms of use now this will be each of you. So in a girl. Alright each negative X squared. Um Each a negative X squared will be E. To do you. No X. D. X. Yeah is equal to -1/2. You. Uh So if we want to put A D. U. In here we need negative two X. Dx. So if we want a negative two X. D. X. So that we can write negative two X. D. X. As D. U. That means I want to put a negative too negative two. Let me write this negative two has two times this. But if I'm time seeing this by negative two we have to balance it by multiplying out here by negative one half. Negative one half times negative two is positive one. So I'm not changing anything by multiplying by one. But the E. To the negative X squared is E. To the U. Uh Since negative X squared is my U. E. To the negative X squared is E. To the U. The negative two X times dx That's my D. U. And then of course I have the ties by negative 1/2. So this integral is really uh the integral of uh X. E. To the negative X squared dx. This is the hardest part of the problem. Not that it's hard but when I find this integral I'm basically finding the integral of X. Each negative X squared dx. So I'll know what the anti derivative of X eaten negative X squared is. Once I know the anti derivative, then I just had to evaluate it between the limits of X X one and x zero. So once I calculate this I'm going to get an expression uh in terms of you and then I'll substitute back in what U. Equals. So then I'll have uh an expression in terms of X. Which will basically be the anti derivative of X. Each negative x squared dx. So Uh integral v. To you. That's an easy one. The enemy derivative of EUU. Is each to you. So this is equal to negative 1/2 E. To do you Which is really equal to negative 1/2. Uh Since you is negative X squared negative one half. Each of the eu is negative one half E. To the negative X squared. So not our final answer. But this is the anti derivative of X. Each negative X squared. If you took the derivative of this, if you want to check your work, if you take the derivative of negative 1/2 times E. To the negative X squared, you will get X E to the negative X square. So negative one half E. To the negative X square is the anti derivative of this. So running out of space, I'm going to rewrite this uh volume formula. Uh Using cylindrical shells rotating around the y axis. I'm going to rewrite this formula down here and we're actually almost done. Okay. The volume in that region is rotated uh around the y axis. Using the method of cylindrical shells is equal to two pi Times the integral from 0 to 1. Uh E it was X. E. To the negative X squared D X. So this will be the volume of the region. Once it's rotated around the y axis. Now, the anti derivative of X. E. To the negative x square is uh this expression right here in the red box. So this volume is simply going to be two pi times The anti derivative. The negative 1/2 E. To the negative X squared uh evaluated between X is one And X0. So basically we have to uh calculate the value of this definite integral. So we take it the anti derivative and we evaluated at one and then we subtract the same anti derivative evaluated at zero. Now just to make life a little easier, we could do the two times in one half and we can do that and you'll still have this negative. So let's let's take this one negative one half. That's times and all this. And let's move it out here just to make things a little bit easier. So two pi times negative one half will be negative pi. So we really just have to take negative by in times of by E to the negative X squared Evaluated between one and 0. Well that's going to equal negative pi times while each of the negative X squared when X is one is E. To the negative one. Then we have to subtract e two negative x squared evaluated at zero, which would be E. To the negative zero square T. 20 which is one. So uh multiplying negative pi by each of these uh negative pi times negative one is pie negative pi times E. To the negative one is a negative pi. Each a negative one but each a negative one is 1/8. So we can write this as minus pi Over each of the one or 8. So there is our volume when we rotated that region around the y axis. Using the method of cylindrical shells

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