Triple Integrals in Spherical Coordinates — Question 10

Question 10 – Solution

Strategy. Divide the first radial moment by volume; the common angular factors cancel, leaving a linear function of $a$.

Step 1: Volume $$\begin{align*} V(a)&=\int_0^{2\pi}\int_0^\pi\int_a^{2a} \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=4\pi\left[\frac{\rho^3}{3}\right]_a^{2a} =\frac{28\pi a^3}{3}. \end{align*}$$

Step 2: Average radius The first radial moment is $$M_\rho(a)=4\pi\int_a^{2a}\rho^3d\rho=15\pi a^4.$$ Hence $$\rho_{\mathrm{avg}}=\frac{M_\rho(a)}{V(a)} =\frac{45a}{28}.$$

Step 3: Recover the shell Setting $45a/28=45/7$ gives $\boxed{a=4}$. Therefore $$\boxed{\text{outer radius}=8},\qquad \boxed{V=\frac{1792\pi}{3}}.$$

Verification The average $45/7\approx 6.43$ lies strictly between the recovered radii $4$ and $8$. Substituting $a=4$ into $28\pi a^3/3$ gives the stated volume.