Surface Integrals — Question 5

Question 5 – Solution

Strategy. In spherical parameters, $z$ and $dS$ separate into a polar factor and an azimuthal factor.

Step 1: Parametrize For radius $2$, use $$\mathbf r(\phi,\theta)=\langle 2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi\rangle,$$ where $0\le\phi\le\pi$ and $0\le\theta\le 2\pi$. Then $$z=2\cos\phi,\qquad dS=4\sin\phi\,d\phi\,d\theta.$$

Step 2: Integrate $$\begin{align*} \iint_S z^2\,dS &=\int_0^{2\pi}\int_0^\pi 16\cos^2\phi\sin\phi\,d\phi\,d\theta\\ &=32\pi\int_0^\pi\cos^2\phi\sin\phi\,d\phi\\ &=32\pi\left[\frac{-\cos^3\phi}{3}\right]_0^\pi =\boxed{\frac{64\pi}{3}}. \end{align*}$$ Because $\operatorname{area}(S)=4\pi(2^2)=16\pi$, the average is $$\boxed{\operatorname{Avg}_S(z^2)=\frac 43}.$$

Verification Rotational symmetry gives equal averages of $x^2,y^2,z^2$. Their sum is always $4$, so each average must be $4/3$.