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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, y=0, x=2, x=4 ; \quad \text { about } x=1

$$

$\frac{76}{3} \pi$

Applications of Integration

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University of Michigan - Ann Arbor

University of Nottingham

Boston College

Okay, let's draw this first. So we can see what area they're talking about that we're going to rotate. So Y equals X. That this green line? Because you're 001122. That's why we'll six Michael zero. That would be the X. Axis X equals two. Maybe this vertical line X equals four. That vertical line. So you can clearly see the area that we're talking about, the area and here. Okay, now we're gonna send it around the line. X equals one. That's kind of in the wrong place. Just a little bit. When I'm moving over, just a little bit, X equals one right here. So, we gotta cut a slice because we're going around a vertical line, we cut horizontally. Okay, when that sliver goes around, it's gonna leave a hole. Okay, get out some terrible picture. Okay? But when this sliver goes around, it's gonna leave a hole. So we have a washer. Okay, it looks like this. There's the line white goes one right through there. Where's the whole coming from? Well, it's from this blue line right here to when it rotates around and it ends up over here. Okay, So in between those two holes there is nothing I mean in between those two lines. Okay, what's the thing look like? Well, the bottom is flat and then the inside is kind of lower than the outside kind of like that. Okay. That's pretty weird. All right, so, to find the volume, it's uh pie big R squared minus little R squared H. Where h is the thickness of the disc or shell or washer which is deep y. Big R is the distance from the axis of rotation out to the big piece. All right so watch me draw it's from here to here. Okay well this much is one and from here to here is some X. Value. So it's X -1. Which ex well it's x equals 4 -1 which is three and that's constant throughout this whole shape. Okay. But then little R is a problem and here's the problem. Oops little or changes depending on if you're down here or if you're up here so down here little R. Is four minus two. We'll see at the bottom and then at the top little or is from here to there. So just some Why value Which Y? Well it's seeing the line Y equals X. I'm sorry some X I'm sorry it's some X value. Okay from here. From the Y axis out that's X. But you're going to integrate dy so you have to change it to Y but conveniently X is equal to Y. So I would have got that one. Right anyway. All right so here's what I got volume is somewhere to somewhere and we'll worry about that in a minute. Pie big R squared minus little R squared. Dy plus pie big R squared minus little R squared. Do you want? Okay so my picture is getting kind of messier. Let's see if I can get some of this out of here. It well pretty much takes the whole thing away erasers. Huh? I would have known. All right so yeah down here you can see that this is the distance between the two sides. Is too is too it's too it's too it's too it's too and then right here right here it stops being too. Well that's when the line x equals two hits the line, y equals x. So that's at y equals two. So it's like well 0 to y equals two And then y equals to two. Well it stops when it gets here which is when y equals X. Hits X equals four. So that's 4-4. All right so we got 0-2 pi 16 -4. Which is 12 do you I that's 12 pi. Why? From 0 to 2? That's 24 pi plus This one is 2-4 pi That's a 4 9 and nine. Almost made a big mistake there. 16 minus y squared. Dy Okay I forgot to integrate. Oh no I got to be in a great part here. It comes 16 y minus y cubed over three From 2- four. So 64 64 3rd minus 30 to minus eight thirds 64 -32. That's 32 minus 64 plus eight. That's 56 3rd and that's negative 56 3rd 32 times three. That's 96 96 -56. That's 43rd pie. Okay so 24 pi plus 43rd pie, 24 times three, plus 40. That's 72, 112 for three Pi. That's the volume.

Oklahoma State University

Applications of Integration