Solution
= Solution
Integrate the horizontal-disk expression over $\pi/2<\theta<\pi$. With $u=\cos\theta$ and $\sin\theta\,d\theta=-du$,
$$
V=\pi a^3\int_{-1}^0(1-u)^2(1-2u)(1-u^2)du=\pi a^3\int_{-1}^0(1-4u+4u^2+2u^3-5u^4+2u^5)du.
$$
Evaluating the <integral> of this <polynomial> gives
$$
\boxed{V=\frac{5\pi a^3}{2}.}
$$