= Solution
In <spherical coordinates>, the <cone> condition forces $z\ge0$ and becomes $0\le\vartheta\le\alpha$. The unit <ball> gives $0\le\rho\le1$. Hence
$$
V=\int_0^{2\pi}\int_0^\alpha\int_0^1\rho^2\sin\vartheta\,d\rho\,d\vartheta\,d\varphi
=\boxed{\frac{2\pi}{3}(1-\cos\alpha).}
$$
For an independent calculation in <cylindrical coordinates>, the lower boundary is $z=r\cot\alpha$ and the upper boundary is $z=\sqrt{1-r^2}$. They meet at $r=\sin\alpha$, since $r^2\cot^2\alpha=1-r^2$. Thus
$$
\begin{aligned}
V&=2\pi\int_0^{\sin\alpha}r\left(\sqrt{1-r^2}-r\cot\alpha\right)\,dr\\
&=\frac{2\pi}{3}\left[1-\cos^3\alpha-\sin^3\alpha\cot\alpha\right]
=\frac{2\pi}{3}(1-\cos\alpha).
\end{aligned}
$$
The equality uses $\sin^3\alpha\cot\alpha=\sin^2\alpha\cos\alpha$. The positive <cosine> on the stated range removes any square-root sign ambiguity.
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