Setting up this volume integral

Question

I just have a quick question on setting up a volume integral of a region bounded by $y = \sin(x)$ and $y = -\sin(x)$ for $0\leq x\leq\pi$ about the x-axis. I keep getting zero when I set it up as $x = \sin^{-1}(y)$ and similar for the outer radius, but I don't know why. Any help is appreciated, thanks, should just be a small problem.

Answer 1

Remember that $\arcsin(x)$ have image $y\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, but you are searching for $0\leq x\leq \pi$.

If you take $0\leq x\leq \frac{\pi}{2}$, then, we have the radius of the bigger disk (in respect to the line $x=5$) is: $$5-\arcsin(y)$$

The smaller disk, instead has radius: $$5-\pi-\arcsin(y)$$

Note that here, I have used $y$ because the disk will have heigth $dy$. So: $$V=\int_{0}^{1}\pi\left((5-\arcsin(y))^2-(5-\pi+\arcsin(y))^2\right)dy=20\pi-4\pi^2$$