Mathematics: Let m be the closed surface that consists of the hemisphere m1: x2 + y2 + z2 = 1, z = 0, and its base m2: x2 + y2 = 1, z = 0 . let e be the electric field defined by e = (18 x, 18 y, 18 z). find the electric flux across m. write the integral over the hemisphere using spherical coordinates, and use the outward pointing normal. (( a c ))m1 e ·ds = (( )) f) d? df b d where a = pi/2, b =0 , c =2pi , d =0 , using t for ? and p for f, f) = ? ? (( ))m1 e ·ds = (( ))m2 e ·ds = 0 , so (( ))m e ·ds =

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Let m be the closed surface that consists of the

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Seaisnowblue

11.01.2022

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

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

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 is $36\pi$ , as expected.

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