Question

Let M be the closed surface that consists of the hemisphere

Answer 1

Since $M$ is closed, you can use the divergence theorem: The flux of $\vec E(x,y,z)$ across $M$ is

$\displaystyle\iint_{\partial M}\vec E\cdot\mathrm d\vec S=\iiint_M(\nabla\cdot\vec E)\,\mathrm dV=54\iiint_M\mathrm dV$

which is 54 times the volume of the hemisphere centered at (0, 0, 0) with radius 1, $\boxed{36\pi}$.

Judging by the question content, you're supposed to find this value by computing the the integral of $\vec E$ across $M_1$ and $M_2$.

• Across $M_1$:

Parameterize the hemisphere by

$\vec r(u,v)=(\cos u\sin v,\sin u\sin v,\cos v)$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$. Take the normal vector to $M_1$ to be

$\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}=(\cos u\sin^2v,\sin u\sin^2v,\sin v\cos v)$

The flux of $\vec E$ across $M_1$ is

$\displaystyle\iint_{M_1}\vec E\cdot\mathrm d\vec S=18\int_0^{\pi/2}\int_0^{2\pi}(\cos u\sin v,\sin u\sin v,\cos v)\cdot\left(\dfrac{\partial\vec r}{\partial v}\times\dfrac{\partial\vec r}{\partial u}\right)\,\mathrm du\,\mathrm dv$

$=\displaystyle18\int_0^{\pi/2}\int_0^{2\pi}\sin v\,\mathrm du\,\mathrm dv=36\pi$

• Across $M_2$:

Parameterize the disk by

$\vec s(u,v)=(u\cos v,u\sin v,0)$

with $0\le u\le1$ and $0\le v\le2\pi$. Take the normal to $M_2$ to be

$\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec u}{\partial v}=(0,0,-u)$

Then the flux across $M_2$ is

$\displaystyle\iint_{M_2}\vec E\cdot\mathrm d\vec S=18\int_0^{2\pi}\int_0^1(u\cos v,u\sin v,0)\cdot\left(\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right)\,\mathrm du\,\mathrm dv=0$

Then the total flux across $M$ is $36\pi$, as expected.