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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 = x + 1 $ , $ y = 0 $ , $ x = 0 $ , $ x = 2 $ ; about the x-axis

$V=\frac{26 \pi}{3}$

Applications of Integration

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Numerade Educator

Oregon State University

University of Michigan - Ann Arbor

Idaho State University

this problem here. What we're gonna want to do is we're going to want to be looking at, um, determining the volume of a solid, um, by rotating it around an axis. So the first thing you want to do is sketch what this solves gonna look like who always see that the line we're focusing on is X y equals X plus one. So I have a point here, and then what we'll have is a ruler. Um, look, something like this. Perhaps that would be our graph. And then we want Thio create a line like this now that we have this in place, um, we want to also sketch other lines that we have. We know that why goes from we have the line y equals zero right here. We also have X equals two. So 12 And that's really what we're gonna have for our solid this region right here is what we're gonna have. And then what we're gonna do is we're going to wrap it around the x axis. So when we wrap it around the X axis right here, we have our solid. We'll do some bloom. So this right here, we're gonna wrap it around the X axis. So it's gonna go like this and what we're gonna end up getting if you can picture this is a solid almost like a a cone shape or like a horn shape. It's gonna look something like that. And what we're gonna have is this is not gonna be a washer. It is going to be a disk. So now we want to do is find the volume of it. So the volume is going to be equal to pi times the integral from 0 to 2 of the radius squared DX. We know the radius is going to be determined by X plus one, because if we're right here, it's X plus one away from the X axis. If we're here, it's X plus one away from the X axis. So with that in mind, we now want to replace the word radius with X plus one squared. Now that we have that, we see that V. The volume is equal to pi times the integral of, um, I find it easier to foil this out so we'll just foil it out to be plus one DX. Then we're going to differentiate this. What will end up getting is pie times. You'll get X cubed over three plus X squared, plus X evaluated at zero into. When we do this, we're gonna end up getting that. The volume is equal to pi times, um, 27 3rd minus one third. That's gonna be 26 3rd. So our volume is 26 pi over three, and that will be the final answer.

California Baptist University

Applications of Integration