Triple Integrals in Spherical Coordinates

Spherical coordinates describe a point the way you'd give directions on a globe: how far from the center ($\rho$), how far down from the north pole ($\varphi$, from $0$ at the top to $\pi$ at the bottom) and which way around ($\theta$, the same angle as in polar coordinates). In this language a ball of radius $a$ is just $\rho \le a$, and a cone opening up from the origin is just $\varphi \le$ some angle. The factor $\rho^2\sin\varphi$ in $dV$ looks mysterious, but it's only the volume of a tiny curved box, and we can build it from two arc lengths. Bear with it for one paragraph.

Where the factor comes from. A small spherical box has three edges. One is $d\rho$, straight out from the center. Moving $d\varphi$ along a meridian (a line of longitude) travels along a circle of radius $\rho$, so that edge is $\rho\,d\varphi$. Moving $d\theta$ around a circle of latitude travels $(\rho\sin\varphi)\,d\theta$, because that circle has radius $\rho\sin\varphi$, its distance from the $z$-axis. Multiply the three edges: $\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$. It makes sense that $\sin\varphi$ shows up: circles of latitude shrink to a point at the poles, where $\sin\varphi = 0$, so boxes up there are thin slivers.

Ranges: $0 \le \rho$, $0 \le \varphi \le \pi$, $0 \le \theta \le 2\pi$ cover all of space.

A common mistake. It's tempting to let $\varphi$ run from $0$ to $2\pi$ like $\theta$ does. It doesn't need to: $\varphi$ only has to go from the north pole to the south pole, and $\theta$ already sweeps all the way around. Running $\varphi$ to $2\pi$ covers everything twice, and $\sin\varphi$ goes negative on the second half and cancels the first, giving $0$ for the volume of a ball.

Ball of radius $a$: $\displaystyle\int_0^{2\pi}\!\int_0^\pi\!\int_0^a\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta = 2\pi\cdot2\cdot\frac{a^3}{3} = \frac43\pi a^3.$ Notice the limits are all constants, so the integral splits into three easy single integrals. That's the payoff of choosing coordinates that match the shape. In Cartesian coordinates the same ball needs limits like $\pm\sqrt{a^2 - x^2 - y^2}$.

The gravitational pull of planets, the charge in atomic orbitals, and how much of a radio signal a dish antenna collects are computed with spherical integrals.