Hey guys. I could some help. This problem caused an issue in my class and I just want to see how everyone goes about doing it and the answer they get:
Find the surface of area of the solid generated from rotating on the interval of 0 to around the -axis.
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I think that the method of cylindrical shells would work best here. The volume V will be
V=∫0π2πxydx=2π∫0πxsin(x)dx.
Evaluating by parts, we let u=x,dv=sin(x)dx⟹du=dx,v=−cos(x), and so
V=2π(−xcos(x)+∫cos(x)dx)=2π(−xcos(x)+sin(x)),
which when evaluated from x=0 to x=π comes out to V=2π(−πcos(π)−0)=2π2.
We could also have used the disc method but that would have been a bit more complicated, as the integral would be of the arcsin function. What was the issue within your class?
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I'm sorry -- but we were attempting to solve for the surface area, not the volume!
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Oh! Sorry. I don't know why I read "surface area" as "volume" the first go around. For the part of the solid generated that is strictly above the xz-plane the surface area integral is
A=∫0π2πx1+cos2(x)dx,
which I believe can only be solved using elliptic integrals, which would be beyond the scope of an AP course. So now I understand what the issue in your class was. (WolframAlpha gives a value of 37.7038.)
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