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Calculus/calculus question: revolving a trig function

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Expert: - 5/21/2011


Question
How do I find the volume of the solid generated by revolving the region bounded by the graphs of the equation about the line y=8 of f(x) = sin x, y=0, from 0 to pi/2.

Thank you!

Answer
The radius of a circle at any particular point is 8 - sin(x).
The area of this circle is pi*r^2, so it is pi(8-sin(x))^2.
The answer is then the integral from 0 up to pi/2 of pi(8-sin(x))^2 dx.

The pi in the integrak can be moved out front, and what is left can be squared.
This gives 64 - 16sin(x) + sin^2(x).

The first two terms are easy to integrate,
for the integral of 64 is 64x and the integral of 16sin(x) is -16cos(x).
However, sin^2(x) has to use the trig identity sin^2(x) = (1 - cos(2x))/2.
That can be integrated to x/2 - sin(2x))/4.

So, just put the integral all together, then evaluate at pi/2 and subtract the value gotten at 0.

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