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

[tex]\mathbf n = \dfrac{\partial\mathbf s}{\partial \phi} \times \dfrac{\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[/tex]

Evaluate [tex]\mathbf E[/tex] at [tex]\mathbf s[/tex].

[tex]\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)[/tex]

Then the integrand reduces to

[tex]\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)[/tex]

The flux of [tex]\mathbf E[/tex] across [tex]M_1[/tex] is then

[tex]\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[/tex]

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