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# Surface Area Problem caused dilemma in my AP Calculus BC class

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.

Note by Andrew Tawfeek
1 year, 6 months ago

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I 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? · 1 year, 6 months ago

I'm sorry -- but we were attempting to solve for the surface area, not the volume! · 1 year, 6 months ago

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

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

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.) · 1 year, 6 months ago