💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here! # Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer.$y = e^x$ , $y = 0$ , $x = -1$ , $x = 1$ ; about the x-axis

## $V=\pi\left[\frac{e^{2}}{2}-\frac{e^{-2}}{2}\right] \approx 11.39$

#### Topics

Applications of Integration

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##### Top Calculus 2 / BC Educators  ##### Heather Z.

Oregon State University ##### Samuel H.

University of Nottingham Lectures

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### Video Transcript

we are given curve in your region and the line and rest defying the volume of the solid obtained by rotating this region bounded by these curves about the specified line. And were asked to sketch the region. The solid and a typical disk or washer. The curves are Y equals E. To the X. Y equals zero X equals negative one. Guys can't see what's on and X equals positive ones. And the line of rotating is about the X axis. It's like there's really no danger in. Yeah. 1st I'll sketch this region house. Yeah. Mhm. Yeah. Yeah that's the thing. So we have our curve Y equals E. To the X. Has a Y intercept about 01 It looks something like this. Now we also have the horizontal line. Y equals zero like this and the vertical lines I think he's going to do just kept me off stop before. Yeah, she's finishing. And so this region and green is our region are now the line that we're rotating around. This is our X axis which is this line and red don't bust. Okay. And so we can draw the solid by reflecting across the axis like just don't bust. We can talk. So the solid is this blue shape just really like Children just like girls gymnast. Yeah. Mm I'm just I know and like all those shots have never landed. It's like how did you know what in a typical disk? She is this red disk here when you're going if not the last. Now the volume of the solid by the disk method, this is going to be the integral from X equals negative 12 X equals positive one of the area of the disk, which is two pi times the radius of the disk, which is our function E to the X squared dX. He was like, this is equal to two pi times the integral from negative 1 to 1 of E to the two X. Dx. Kids jokes like taking the anti derivative. Well, this is pi times E to the two X from negative 1 to 1, which is pi times, sorry, not to pi. This should be pi times R squared. And so this is going to be hi over two times E to the two X, which is pi over two times E squared minus E to the negative second. And this is approximately 1139. Either way, I'll accept either answer. Ohio State University

#### Topics

Applications of Integration

##### Top Calculus 2 / BC Educators  ##### Heather Z.

Oregon State University ##### Samuel H.

University of Nottingham Lectures

Join Bootcamp