= Solution
Orient the common axis of the wedge faces along $z$ and its azimuthal bisector along positive $x$. Reflection symmetries give $\bar y=\bar z=0$. In <spherical coordinates>, use $0\leq r\leq a$, $0\leq\theta\leq\pi$, and $-\beta/2\leq\varphi\leq\beta/2$. The <volume> is
$$
V=\int_0^a r^2\,dr\int_0^\pi\sin\theta\,d\theta\int_{-\beta/2}^{\beta/2}d\varphi
=\frac{2\beta a^3}{3}.
$$
Since $x=r\sin\theta\cos\varphi$, the first moment is
$$
\int_Vx\,dV=\left(\frac{a^4}{4}\right)
\left(\int_0^\pi\sin^2\theta\,d\theta\right)
\left(\int_{-\beta/2}^{\beta/2}\cos\varphi\,d\varphi\right)
=\frac{\pi a^4}{4}\sin\frac\beta2.
$$
Dividing by the <volume> locates the <centre of mass>:
$$
\boxed{\bar{\mathbf x}=\frac{3\pi a}{8\beta}\sin\frac\beta2\,\mathbf e_x},\qquad
\boxed{f(\beta)=\frac{3\pi}{8\beta}\sin\frac\beta2}.
$$
This is the centroid formula for a <spherical wedge>.
For the sketch, $f$ is positive and decreases from $\lim_{\beta\to0^+}f=3\pi/16$ to $\lim_{\beta\to2\pi^-}f=0$, with $f(\pi)=3/8$. Indeed, on putting $q=\beta/2$, the sign of $f'$ is the sign of $q\cos q-\sin q$. The negative of this expression has <derivative> $q\sin q>0$ for $0<q<\pi$ and starts at zero, proving strict decrease.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2-spherical-wedge-centroid.png]
{title=Distance of the centre of mass of a spherical wedge from its sphere centre}
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