What is the volume of the flying saucer?

Question

What is the volume of the flying saucer that comes from rotating $y=\sin x\;(0\le x\le\pi)$ around the $y$ axis?

Here is a diagram for visualization aid: We will work with horizontal circular slices, so we have $$\begin{align} \int_0^1\pi x^2\,dy&=\int_0^1\pi\left(\sin^{-1} y\right)^2\,dy\\ &=\pi\left(y\left(\sin^{-1}y\right)^2+2\sin^{-1}y\sqrt{1-y^2}-2y\right)\biggr|_0^1\\ &=\frac{\pi^3}4-2\pi \end{align}$$

However, the answer key says $V=\int_0^\pi2\pi x\sin x\,dx=2\pi^2$. Where was I wrong?

Answer 1

I figured out why the answer could be $\int_0^\pi2\pi x\sin x\,dx=2\pi^2$. The flying saucer can be visualized by looking at the following longitudinal section: By using cylindrical shells, we have $V=\int_a^b2\pi xh\,dx$, where $a=0$, $b=\pi$ and $h=\sin x$. This leads us to the answer in the key $\int_0^\pi2\pi x\sin x\,dx=2\pi\cdot\pi=2\pi^2$.

Answer 2

I get frustrated if it comes to integrating inverses of trig functions. So I tried something simpler. Take a look at the following figure:

The volume of the red hollow cylinder of inner radius $r$, of outer radius $r+\Delta r$, and of height of $\sin(r)$ is

$$\pi((r+\Delta r)^2-r^2)\sin(r)=\pi(2r\Delta r\sin(r)+\Delta^2 r\sin(r)).$$

So, the total volume of the "negative" of the flying saucer is

$$2\pi\int_0^{\frac{\pi}2}r\sin(r)\ dr=2\pi.$$

The volume of the cylinder of radius $\frac{\pi}2$ and height $1$ is

$$\frac{\ \pi^3}4.$$

So the volume is the flying soucer is

$$\frac{\ \pi^3}4-2\pi.$$