Triple Integrals in Spherical Coordinates — Question 9

Question 9 – Solution

Strategy. Approximate a small spherical cell by a box with one radial side and two angular arc-length sides.

Step 1: Volume element The local side lengths are $$d\rho,\qquad \rho\,d\phi,\qquad \rho\sin\phi\,d\theta.$$

Their product is $$(d\rho)(\rho\,d\phi)(\rho\sin\phi\,d\theta) =\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$

Step 2: Apply $$\begin{align*} I&=\int_0^{2\pi}\int_0^\pi\int_1^3 \rho^{-2}\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=(2\pi)(2)(3-1)=\boxed{8\pi}. \end{align*}$$

Verification The shell excludes $\rho=0$, so the integrand is bounded. Its units are length$^{-2}$; multiplying by volume produces length, matching the radial factor $3-1$ in the result. Even if the inner radius tends to $0$, the radial integral remains finite: $4\pi\int_0^3d\rho=12\pi$. Thus the singularity is integrable.