Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2002 / Paper 2 / 3

Use spherical polar coordinates with $\theta$ measured from the positive $z$ axis and $\varphi$ around that axis:The standard full-space ranges are $r\ge 0$, $0\le \theta \le \pi$, $0\le \varphi <2\pi$. The factor $r^2\sin \theta$ is the absolute Jacobian determinant and gives the volume element.

The mass density becomes $\rho =\mu (2+\cos \theta )$. Integrate it over the sphere using the spherical polar coordinates and volume element above:The angular integral is $4$, since ${\int }_0^{\pi }\cos \theta \sin \theta d\theta =0$. ThereforeThe term proportional to $z/r$ contributes zero by reflection across the equatorial plane. Thus the answer is also the volume $4\pi a^3/3$ times the average mass density $2\mu$. The undefined direction at the origin affects a single point and has no effect on this volume integral.

In the first octant, spherical polar coordinates have $0\le \theta \le \pi /2$ and $0\le \varphi \le \pi /2$. The charge density is $\lambda xy=\lambda r^2{\sin }^2\theta \cos \varphi \sin \varphi$. Hence the total electric charge isThe three factors are $a^5/5$, $2/3$, and $1/2$, respectively. For the middle integral, set $u=\cos \theta$ to obtain ${\int }_0^1(1-u^2)du=2/3$. Thus