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# Refer to the figure and find the volume generated by rotating the given region about the specified line.$\Re_1$ about $BC$

## $\frac{2 \pi}{3}$

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Applications of Integration

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for this problem, you need to be looking at the book. It's referencing a certain region and a certain line that we're going to be looking at as we generate a volume. So for this problem, we are looking at Region One. So that is the area. I'm going to try to keep the same color scheme as in the book. That's the area under this line. Okay. And what am I going to do with that? I'm going to rotate it around the line B C. Well, BC is Y equals one. Okay, because this is B. This is C So we're rotating it around. Why equals one? So let me just put this on here. There's my rotation. So as I rotate this and I generate a solid let's see what we have. If I just take a general piece, just a random piece right here, Representative slice, This is going to be a washer because I'm going to have an empty space where that white is. And then once I hit the blue, I'll have that as it goes around. This is a washer. What is the volume of a washer? Well, a washer is going to be two circles. First I have the outer circles. I'll call that my big our pi r squared minus the inside circle, which is pi little r squared times the height. If I want, I can also pull the pie out. I could say this is pi times big R squared, minus small r squared times the height. That's how I find the washer method. Or find the volume using a washer method that's each individual washer. I could use that formula to find the volume, and then I stack all the washers up. So let's see what that looks like when we use calculus to solve an infinite sum of an infinite number of washers, all stacked up. As I stack these washers, I'm going to be going from X equals zero two x equals one, which means I can pull the pie through. I'm going to have the ours, which I'm going to look at in just a minute times the height and the height is that little infanticidal change in X or D x? That's that becomes my height. So what are my two radius is? Well, the largest radius goes all the way down to the X axis. It's one, and it's uniformly one all the way across this distance. So my capital are the big radius squared is just one. So what is our small radius? Well, the small radius and I'm going to put this in green here. Is that line going from the origin up to Point C, which is 11? Well, if I think about what the equation of that line is, this intercept is zero. The slope goes from 00 to 11 The slope is one or the equation is just y equals X. So as I go on that, um, my representative disk here and I go down, I hit that green line. That's the line Y equals X. So that radius is X and I'm squaring it. So this is based on our washer definition. This is the setup for the integral that we're going to take to find the volume of that blue section rotated around the line Y equals one. So let's actually do the calculus are the integral of one is just x the integral of X squared. What we add one. And we divide by that exponents and we evaluate from X equals 02 X equals one starting at the top limit of integration. I get one minus one third and I would subtract the lower limit of integration. But that's just zero. So I don't have to worry about that in this problem. So this is just two thirds times pi or to pi over three.

Rochester Institute of Technology

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Applications of Integration

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