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Let $ T $ be the triangular region with vertices $ (0, 0) $, $ (1, 0) $, and $ (1, 2) $, and let $ V $ be the volume of the solid generated when $ T $ is rotated about the line $ x = a $, where $ a > 1 $. Express $ a $ in terms of $ V $.

$a=\frac{V}{2 \pi}+\frac{2}{3}$

Applications of Integration

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for this problem. We have a triangular section of the grid that we're going to look at the, um, vortices of my triangle R 00 10 and 12 So we have a little triangle here. I'm gonna use color in blue, and we want to generate a solid by rotating this about the line X equals a So we're going to rotate around here? We don't know what a is. We just know that is larger than one. So it's gonna be rotating somewhere around that line. And our question is we want to find the volume and we want to put we actually wanna write a in terms of the volume so we'll find the volume of that solid, and then we'll solve her A. Now, when we rotate this around that, um, around that line, X equals A. This is a really good problem to use, um, shells for because you can imagine we're just stacking these shells one after another. I'm just going to draw a little red line here. This will be a representative shells. You can imagine taking that red sliver. They're rotating it all around the X axis and the volume for that representative piece you can imagine that's going to be the circumference of the circle, that it makes times the height, times that little tiny D X. Because it's just a little it's got a little bit of a thickness to it, and we're going to stack all those shells up So the shelves are going to stack from X equals 02 X equals one because I can take all these vertical slices and wrap them around that line X equals a So I'm stacking from X equals 02 X equals one. I know I'm gonna need a DX here. That's the infant s Immel thickness of each one of these pieces. And the formula we're looking for is to Hi are times the height. Okay, so I have the formula. I want to get rid of those arts and H as though I want everything in terms of X. And that's my volume. So what do we know? Well, first of all, I'm going to actually going to erase this. I just can move it a little bit easier. I'm going to move my two pi. Since they're both Constance, I'm just gonna move them through my integral. Simplify things up a little bit. Now, what's the radius from this line? X equaling a out to that red representative slice there. Well, that's going to be the whole distance. A minus. This little piece. Well, that little piece right there is X. So my radius is a minus x. Okay. What's the height? The height gets me to that line that connects 00 to 12 Well, we can write the equation for that line by inspection. It has a slope of zero, so it doesn't have to mean it has an intercept of zero. So I don't have to add that plus B on the end and my slope. I'm going up to over one. So my line is y equals two x. So that is my height. And this will probably be easier to take, um, to integrate. If I took everything without parentheses so I can multiply that and say that that's to a X minus two x squared. Mm. So let's take our integral integral of to a ex. Well, it's become to come X squared, divided by two. So that becomes a X and two X square. That becomes two X cubed over three. Okay. Oops. I forgot to put the squared in there. It's a X squared. Apologize. I think I said it, but I didn't write. It is a X squared minus two X cubed over three. And I'm evaluating that from 0 to 1. So what is my volume? My volume is two pi. If X is one, this becomes a minus two thirds if x zero everything zero. So I'm just going to be subtracting zero from this number here. So that's my volume. Now I want to solve this for a so let's divide both sides by two pi. That gives me a minus two thirds. So a is the volume divided by two pi plus two thirds.

Rochester Institute of Technology

Applications of Integration