$T(x,y,z)=x$, $S=\{(x,y,z): x^2+y^2+z^2=1\}$, I also got a formula for heat flow $\int \int_S (-k \nabla T)\circ dS$ for some parameter $k$.
My attempt:
$x=sin\phi cos\theta$, $y=sin\phi sin\theta$, $z=cos\phi$ for $0\leq \theta \leq 2\pi$ and $0\leq \phi \leq \pi$,
$\nabla T=\hat{i}$, $S(\phi, \theta)=\hat{i}sin\phi cos\theta +\hat{j}sin\phi sin\theta+\hat{k}cos\theta$,
$dS=\frac{\partial S}{\partial \phi}d\phi \times \frac{\partial S}{\partial \theta}d\theta=(\hat{i}cos\phi cos\theta+\hat{j} cos\phi sin\theta – \hat{k}sin\phi)d\phi \times (-\hat{i}sin\phi sin\theta+\hat{j}sin\phi cos\theta)d\theta=$
$=(\hat{i}sin^2\phi cos\theta + \hat{j} sin^2 \phi sin\theta+\hat{k}sin\phi cos\phi)d\phi d\theta$.
$\nabla T \circ dS=sin^2\phi cos\theta d\phi d\theta$, $\int \int_S (-k \nabla T)\circ dS=k\int_0^{2\pi}\int_0^{\pi}sin^2\phi cos\theta d\phi d\theta=k\int_0^{2\pi}\frac{\pi}{2} cos\theta d\theta=0$
Is this correct? Thanks in advance.
Best Answer
Yes there are two ways to look at it without direct computation
$\textbf{Option 1}$: Surface integral
Using $n = (x,y,z)$ for the unit sphere we have that
$$\iint -k\nabla T \cdot n dS = -k \iint x dS$$
$x$ is an odd function and the sphere has $x$ symmetry i.e. $(-x)^2 + y^2 +z^2 = 1 \iff x^2+y^2+z^2=1$ so the integral is $0$ automatically.
$\textbf{Option 2}$: Divergence theorem
By Divergence theorem we have that
$$-k\iint \nabla T \cdot n dS = -k \iiint \Delta T dV = -k \iiint 0 \:dV = 0$$