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 \(\sin(x)\) on the interval of 0 to \(\pi\) around the \(y\)-axis.

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TopNewestI think that the method of cylindrical shells would work best here. The volume \(V\) will be

\(V = \displaystyle\int_{0}^{\pi} 2\pi xy dx = 2\pi \int_{0}^{\pi} x\sin(x) dx\).

Evaluating by parts, we let \(u = x, dv = \sin(x) dx \Longrightarrow du = dx, v = -\cos(x)\), and so

\(V = 2\pi(-x\cos(x) + \displaystyle\int \cos(x) dx) = 2\pi(-x\cos(x) + \sin(x)\)),

which when evaluated from \(x = 0\) to \(x = \pi\) comes out to \(V = 2\pi (-\pi \cos(\pi) - 0) = \boxed{2\pi^{2}}\).

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? – Brian Charlesworth · 10 months, 3 weeks ago

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– Andrew Tawfeek · 10 months, 3 weeks ago

I'm sorry -- but we were attempting to solve for the surface area, not the volume!Log in to reply

\(A = \displaystyle\int_{0}^{\pi} 2\pi x\sqrt{1 + \cos^{2}(x)} dx\),

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.) – Brian Charlesworth · 10 months, 3 weeks ago

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– Andrew Tawfeek · 10 months, 3 weeks ago

Ahh yes, thank you! There was a whole debacle about how to go about solving this and I didn't post the steps that were taken in the case that they were inaccurate. :)Log in to reply