# Use a computer algebra system to find the exact volume of the solid obtained by rotating the region bounded by the given curves about the specified line.$$y=x, y=x e^{1-x / 2} ; \quad \text { about } y=3$$

## $$V=\pi \int_{0}^{2}(3-x)^{2}-\left(3-x e^{1-x / 2}\right)^{2} d x=24 e-\frac{142}{3}-2 e^{2}$$

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Applications of Integration

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{'transcript': "were given a set of curves in the line and were asked to find the exact volume of the solid obtained by rotating the region. Bounded by these curves about this line using a computer algebra system, the curves were given R y equals X. Why equals X times either the one minus sex over to and the line is, Why equals three. First, let's just sketch graph of the region. So this is going to be in the primarily 1st and 4th quadrants. I'm sorry, just really in the first quadrant. So we have the curve y equals X, which is simply a straight line through the origin in the 0.11 We also have the curve X times E to one minus X over to this has a zero at X equals zero, and to find where these two curves intersect, you want to set X equal to X E to the one minus X over to, in other words, either one minus x over two SD equal one, which implies that one minus X two equals zero. So the X is equal to two, so they intersect somewhere at the 0.22 and otherwise this function is going to look something like this. So the area that we're interested in is this region that I've drawn in red here, and we're going to rotate it about the line Y equals three, which is up here somewhere. Now we have that the volume of the solid formed a rotating about this line. Well, if we rotate about this line of solid we get has cross sections parallel to the Y axis, which are washers. The volume is going to be integral from X equals zero to x equals two of the area of these washers, which is pi times The large radius. This is our top function. Y equals three minus. The bottom function y equals X squared minus smaller radius, which is our top function y equals three minus the bottom function, which is X times E to the one minus X over to square the radius T X and using a computer algebra system like Wolfram Alfa, this is exactly equal to high times negative two e squared plus 24 e minus 142 3rd"}

Ohio State University

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Applications of Integration

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