Triple Integrals in Spherical Coordinates — Question 7

Question 7 – Solution

Strategy. Resolve the horizontal inequalities into a $\theta$-wedge and the vertical inequality into a $\phi$-hemisphere.

Step 1: Angular bounds The conditions $x\ge 0$ and $y\ge x$ select $\pi/4\le\theta\le\pi/2$. The condition $z\le 0$ selects $\pi/2\le\phi\le\pi$, while the shell gives $2\le\rho\le 3$.

Step 2: Volume $$\begin{align*} V&=\int_{\pi/4}^{\pi/2}\int_{\pi/2}^{\pi}\int_2^3 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\\ &=\left(\frac\pi 4\right)(1)\left(\frac{27-8}{3}\right) =\boxed{\frac{19\pi}{12}}. \end{align*}$$

Verification The azimuthal wedge is $(\pi/4)/(2\pi)=1/8$ of a full turn, and $z\le 0$ supplies half the polar solid angle. Thus $E$ occupies $1/16$ of the full shell; indeed $\frac 1{16}\frac{4\pi}{3}(27-8)=19\pi/12$.