Question

And its base

where
ed surface that consists of the hemisphere
Let E be the electric field defined by E = (18x, 18y, 182). Find the electric flux across M. Write the integral over the hemisphere using spherical
coordinates, and use the outward pointing normal.
a=0.b = c = 0.d = 2n
-0.₁=
Using t for and p for .
f(0,4)=
=36x
M, E-ds:
JM, E-dS=
0 SO
M₁: x² + y² +2²=1, z20,
E-ds 36л
M₂: x² + y² ≤ 1, z=0.
dS=
18 = [" ["F(0.4) do dq.
de
Eds

Answer 1

Parameterize
`M_1` by

`\mathbf s(\theta,\phi) = \cos(\theta)\sin(\phi)\,\mathbf i + \sin(\theta)\sin(\phi)\,\mathbf j + \cos(\phi)\,\mathbf k`

with
`0\le \theta\le2\pi` and
`0\le \phi\le\frac\pi2`.

The outward-pointing normal vector is

`\mathbf n = (\partial\mathbf s)/(\partial \phi) * (\partial\mathbf s)/(\partial \theta) = \cos(\theta)\sin^2(\phi)\,\mathbf i + \sin(\theta)\sin^2(\phi)\,\mathbf j + \cos(\phi)\sin(\phi)\,\mathbf k`

Evaluate
`\mathbf E` at
`\mathbf s`.

`\mathbf E(\mathbf s(\theta,\phi)) = 18 \left(\cos(\theta)\sin(\phi)\,\mathbf i + \sin(\theta)\sin(\phi)\,\mathbf j + \cos(\phi)\,\mathbf k\right)`

Then the integrand reduces to

`\mathbf E \cdot \mathbf n = 18\cos^2(\theta)\sin(\phi)^3 + 18\sin^2(\theta)\sin^3(\phi) \\ ~~~~+ 18\cos(\phi)(\cos^2(\theta)\cos(\phi)\sin(\phi) + \sin^2(\theta)\cos(\phi)\sin(\phi)) \\\\ = 18\sin^3(v) + 18\cos^2(v)\sin(v) \\\\ = 18\sin(v) (\sin^2(v)+\cos^2(v)) = 18\sin(v)`

The flux of
`\mathbf E` across
`M_1` is then

`\displaystyle \iint_(M_1) \mathbf E \cdot d\mathbf S = \int_0^(\pi/2) \int_0^(2\pi) 18\sin(v) \, du \, dv \\\\ = 36\pi \int_0^(\pi/2) \sin(v) \, dv = 36\pi`

and so the missing piece is
`f(\theta,\phi)=18\sin(\phi)`, or in your case
`\boxed{18\sin(p)}`.