Calculus 2: Finding Volume of object in 3D space rotating on x-axis

Question

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?

Answer 1

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

Answer 2

If you evaluate $\pi[-1/4\cos (4x)]_0^{\pi/4}=-\pi/4(-1-1)=\pi/2$.