Triple Integrals in Spherical Coordinates — Question 3

Question 3 – Solution

Strategy. The hemisphere contributes the full azimuthal interval but only $0\le\phi\le\pi/2$; the radial density adds one power of $\rho$.

Step 1: Mass $$\begin{align*} M&=\int_0^{2\pi}\int_0^{\pi/2}\int_1^2 \rho\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=(2\pi)(1)\left[\frac{\rho^4}{4}\right]_1^2 =\boxed{\frac{15\pi}{2}}. \end{align*}$$

Step 2: Average density The region volume is half the difference of two ball volumes: $$V=\frac 12\frac{4\pi}{3}(2^3-1^3)=\frac{14\pi}{3}.$$ Therefore $$\boxed{\delta_{\mathrm{avg}}=\frac{M}{V}=\frac{45}{28}}.$$

Verification Since $1\le\delta=\rho\le 2$ throughout the shell, the value $45/28\approx 1.61$ is admissible. The larger outer spherical layers correctly pull it above the midpoint $3/2$.