Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 12

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 $0\le r\le a$, $0\le \theta \le \pi$, and $-\beta /2\le \phi \le \beta /2$. The volume isSince $x=r\sin \theta \cos \phi$, the first moment isDividing by the volume locates the centre of mass:This is the centroid formula for a spherical wedge.

For the sketch, $f$ is positive and decreases from ${\lim }_{\beta \to 0^{+}}f=3\pi /16$ to ${\lim }_{\beta \to 2{\pi }^{-}}f=0$, with $f(\pi )=3/8$. Indeed, on putting $q=\beta /2$, the sign of $f^{\prime }$ 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.

Use the same axes as in (a) and choose the outward orientation. On the curved spherical lune, $\mathbf{n}={\mathbf{e}}_r$ and $dA=a^2\sin \theta d\theta d\phi$. Its vector area isThe $y$ and $z$ components cancel. The $x$ component is $a^2(\pi /2)2\sin (\beta /2)$, henceHere ${\mathbf{e}}_x$ points along the azimuthal bisector; another coordinate choice rotates this vector accordingly. Its magnitude is the signed projected area, not the scalar curved area $2\beta a^2$. Reversing the orientation reverses its sign.