The revolution across the y-axis, and the bounded area is between $y=\sin(x)$, and $y=0$ for ${0\le x \le \pi}$.
I did: $$V_{shells}= 2 \pi \int_\limits{0}^\pi x\sin(x)dx=2\pi^2$$ I am trying to do this via slicing: $$V_{slicing}=\pi \int_\limits{0}^1 \arcsin(y)dy$$ I end up getting: $$\frac{\pi^2}{2}-1$$ and times two the answer doesn't even mirror the first.
For the function $y = \sin x$ in the range $0\le x \le \pi$, the image is $0\le y \le 1$. For such given $y$, there are up to two $x$'s that satisfies $y=\sin x$:
$$\begin{align*} y &= \sin x\\ x &= n\pi + (-1)^n\arcsin y\\ &= \arcsin y \quad \text{ or }\quad \pi - \arcsin y \end{align*}$$
For disk integration, each slice is a "washer" or a cylinder with a hole. The outer radius is $(\pi - \arcsin y)$ and the inner radius of the hole is $\arcsin y$:
$$\begin{align*} V_{slicing} &= \int_0^1\pi\left[\left(\pi - \arcsin y\right)^2 - \left(\arcsin y\right)^2\right]dy\\ &= \int _0^1 \pi\left(\pi^2 - 2\pi\arcsin y\right)dy\\ &= \pi^2 \int _0^1 \left(\pi - 2\arcsin y\right)dy\\ &= \pi^2\left[\pi y -2y\arcsin y-2\sqrt{1-y^2}\right]_0^1\\ &= \pi^2\left[\left(\pi - 2\cdot\frac\pi2\right) - \left(-2\sqrt{1-0^2}\right)\right]\\ &= 2\pi^2 \end{align*}$$