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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.

$$y=e^{-x}, y=1, x=2 ; \quad \text { about } y=2$$

$$

V=\pi\left[\frac{5}{2}-\frac{1}{2} e^{-4}+4 e^{-2}\right]

$$

Applications of Integration

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Missouri State University

Campbell University

Idaho State University

So I sketched out to set up here for finding volume, and we're going to rotate this volume around. Why Ankles, too? So the equations were given. Why equals one X equals two and then why equals e to the negative X And that's just a decay function there. So first I want to label the area that were rotating the area is whatever is bounded by all three of those equations. So here, that's going to be the participated in red. We want to rotate around the line. Why equals to? So let's say that's here. So we're taking this whole part that's in red and rotating it up and around. Why equals two. So it should be pretty clear that there's all of this missing space As I'm creating that volume. It's not gonna be totally solid. There's gonna be like a doughnut. There's gonna be that empty part in the middle. So because there's that missing space, then we know this set up here is a washer set up. So your formula for the washer set up is high times the integral R squared minus R squared. You pick your very well. Based on what access you're working with and here, like was too is like working with the x axis. See how those are parallel. So I picked D X for my variable, which means for the limits I need to use excess as well X one and X two. So now I need to go back and figure out Well, what is that? X one and x two? What is the starting x value in the final X value for the area? Well, if I'm looking at this area, the tribute just something I can inspect and see those X values as zero. And to write the given vertical line and the access says you were into then the next thing to figure out is what is capital R and lower case R Capital R A's What if this volume wasn't missing anything at all? So if I'm going from, why equals two to the shaded area and I wanted to say, Hey, I'm not missing anything, I'd have to color it in like I just did in blue, right? I'd have to fill in all that missing stuff, which means it would go from that horizontal line of two down to the curved line which is e to the negative x so thinking about it. But we do area upper minus lower here, I think, to minus either the native X for a lower case. R it's what am I taking away? Well, the stuff I'm taking away is this missing block between our area and the line of rotation. So that comes from that upper bound of to minus the horizontal line one. So two minus one. Or you could just say one is lower case are so filling it in capital R squared to minus E to the negative X squared and then lower case R is wine. So one squared, which is still just one. And then at this point, we could expand that binomial. Maybe you have software you can solve this integral. So solving through. We get pi times for e to the negative too, minus e to the negative for over two. Plus I've over two and that's an acceptable final answer for this volume.

Oklahoma Baptist University

Applications of Integration