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Question: find the volume generated when the region bounded by $y = \sin x \cos x, 0\le x \le \frac{\pi}{2}$, is revolved about the x-axis.

This question appeared quite tricky, and the book that suggested it didn't append any answers so that I could check my result. However, I realized that I could use the half-angle formulas here (twice), and hence the volume generated is calculated to be the integral:

$\pi \int_0^{\frac{\pi}{2}} (\sin x \cos x)^2 dx = \pi \int_0^{\frac{\pi}{2}} \frac{1– \cos 2x}{2} \frac{1 + \cos 2x}{2}dx = \frac{\pi}{4}\int_0^{\frac{\pi}{2}} (1 – \cos^2 2x) dx$

$= \frac{\pi}{4} \int_0^{\frac{\pi}{2}} \frac{1 – \cos 4x}{2} dx = \frac{\pi}{8} \int_0^{\frac{\pi}{2}} (1 – \cos 4x) $ $dx = \frac{\pi}{8} [x – \frac{\sin 4x}{4}]_0^{\frac{\pi}{2}} = \frac{\pi}{8} \frac{\pi}{2} = \frac{\pi ^2}{16}$

Is this result correct? Thank you in advance.

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    $\begingroup$ It looks good to me! $\endgroup$
    – K. Jiang
    Commented Jun 18 at 22:16
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    $\begingroup$ Correct. Although, to make it slightly shorter, I would have used the identity $2\sin{x}\cos{x}=\sin{2x}$, which surely leads to the same results. $\endgroup$
    – Egor Larionov
    Commented Jun 18 at 22:19
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    $\begingroup$ Seems to be correct. $\endgroup$
    – Simon
    Commented Jun 19 at 5:32

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