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Use the method of cylindrical shells to find the …

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Problem 63 Hard Difficulty

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

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

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

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Video Transcript

The problem is use the method of cylindrical shelves to find the volume generated by rating the region bounded by the curves of the given axis here y is equal to 2 negative x y. Equal to 0 x is negative. 1 x equals to 0 at x, equals to 1, so we can do. The aphis is x y and it's a graph of y equal to e to negative, x and x, equal to negative 1 and about its axis. This is x. If 1 is the region. Is this part and rotate this region about axis x, equal to 1 by using the miser riddle of celindrical shells we can see the volume is equal to 2 pi integral from negative 1 to 0. Here x is a distant distance from the point to the x x to the axis of the line. So here it is, is 1 minus x times function. F x is 2 negative x, so notice that this is the distance was a point to the axis between the point and the axis distance. So this is 1 minus x. Now we can use integration by parts to solve this integral. The formula is integral from a to b: u times, vix is equal to u times v from a to b minus integral from integral u prime v d x from a to b now for our problem, we can write. U is equal to 1 minus x and prime is equal to 2 negative x, then prime is equal to negative 1 and v is equal to negative to negative x. Then this integral is equal to 2 pi. By using this formula we can say this is 2 pi times u times v, so this is negative 1 minus x to negative x from negative 1 to 0 and minus integral from negative 1 to 0 prime times, so this is to negative x dx. So this is equal to 2 pi, plain 0 and negative 1 to these functions. We have. This is negative: 1 minus negative, 1 minus minus 1 is 2 and e to 1 is negative and minus integral of e to negative x is negative to negative x. This is past negative x from negative 1 to 0 point, so this is equal to 2 pi 2 e minus 1 plus this is 1 minus e. So i answer is 2 pi ti.

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Calculus: Early Transcendentals

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