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I am stuck on the following problem.

Find the volume of the solid, generated by revolving the region bounded by $y = \sqrt{\sin(4x)}, y = 0$ and $0 \le x \le \frac{\pi}{4}$ about the $x$-axis.

To solve for volume, I have the following integral $$ V = \int_0^{\pi/4} \pi r(x)^2 dx = \pi \int_0^{\pi/4} \sin (4x) dx = \pi \left[- \frac{\cos(4x)}{4}\right]_0^{\pi/4} = \frac{\pi}{4}. $$ That is not one of the solution options. What did I do wrong?

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    $\begingroup$ Did you consider the fact that $cos0$ "contributes" to the answer? $\endgroup$
    – imranfat
    Commented Jan 21, 2020 at 18:11
  • $\begingroup$ oh! dang, yes I see $\endgroup$
    – PineNuts0
    Commented Jan 21, 2020 at 18:13

2 Answers 2

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$$ V = \pi \int_0^{\pi/4} r(x)^2 dx = \pi \int_0^{\pi/4} \sin(4x) dx = \left. -\frac{\pi}{4} \cos (4x) \right|_0^{\pi/4} = -\frac{\pi}{4} [-1 -1] = \frac{\pi}{2} $$

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  • $\begingroup$ thank you! where do you go to type out your math like that? $\endgroup$
    – PineNuts0
    Commented Jan 28, 2020 at 1:56
  • $\begingroup$ @PineNuts0 I learned LaTeX in school and here are the local MathJaX instructions $\endgroup$
    – gt6989b
    Commented Jan 28, 2020 at 4:38
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If you evaluate $\pi[-1/4\cos (4x)]_0^{\pi/4}=-\pi/4(-1-1)=\pi/2$.

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