Find the flux of the curl of field F through the shell S.
-$\mathbf { F } = ( 4 - \mathrm { y } ) \mathbf { i } + ( 3 + \mathrm { x } ) \mathbf { j } + \mathrm { z } ^ { 5 } \mathbf { k } ; \mathrm { S }$ is the upper hemisphere of $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } = 16$
A) $64 \pi$
B) $2 \pi$
C) $32 \pi$
D) $- 4 \pi$

Question

Quizplus · Accepted Answer

Use Stokes' theorem, which states:
$\oint_C \vec{F}\cdot d\vec{r} = \iint_S 
abla 	imes \vec{F} \cdot d\vec{S}$
Where $C$ is the boundary of $S$ (in this case, just the circle at the bottom of the hemisphere).
First, we need to find $
abla 	imes \vec{F}$:
$$
abla 	imes \vec{F} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \
(4-y) & (3+x) & z^5
\end{vmatrix} = -5z^4 \vec{i} + \vec{j} - \vec{k}$$
Then, we can set up the surface integral to find the flux through the hemisphere:
$$\iint_S 
abla 	imes \vec{F} \cdot d\vec{S} = \iint_S (-5z^4 \vec{i} + \vec{j} - \vec{k}) \cdot d\vec{S}$$
Using the equation for the upper hemisphere, we can write $d\vec{S} = \vec{r}_\mathrm{s} \cdot d\Omega$, where $\vec{r}_\mathrm{s}$ is the position vector for the upper hemisphere and $d\Omega$ is a differential solid angle. 
$$\vec{r}_\mathrm{s} = x\vec{i} + y\vec{j} + \sqrt{16-x^2-y^2}\vec{k}$$
$$d\vec{S} = \begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \
x & y & \sqrt{16-x^2-y^2}
\end{vmatrix} = -\frac{x}{\sqrt{16-x^2-y^2}}\vec{i} -\frac{y}{\sqrt{16-x^2-y^2}}\vec{j} + \vec{k} \, dA$$
The limits for the double integral are $0\leq x^2+y^2\leq 16$ (the upper hemisphere) and $0\leq 	heta \leq 2\pi$, $0\leq \phi \leq \frac{\pi}{2}$ (spherical coordinates). Using some calculus, we can simplify $-\frac{x}{\sqrt{16-x^2-y^2}}\vec{i} -\frac{y}{\sqrt{16-x^2-y^2}}\vec{j}$ as $-4\cos\phi \vec{k}$, so we don't need to include $x$ and $y$ in the integral. We can also write $dA$ as $4\pi\sin\phi\,d\phi\,d	heta$. Putting everything together, we get:
$$\iint_S (-5z^4 \vec{i} + \vec{j} - \vec{k}) \cdot d\vec{S} = \int_0^{2\pi} \int_0^{\frac{\pi}{2}} (-5(16\cos\phi)^4)(-4\cos\phi) + (0)(0) + (-1) \cdot 4\pi\sin\phi\,d\phi\,d	heta$$
Evaluating this integral, we get $\boxed{32\pi}$ as the answer.

Find the Flux of the Curl of Field F Through F=(4−y)i+(3+x)j+z5k;S\mathbf { F } = ( 4 - \mathrm { y } ) \mathbf { i } + ( 3 + \mathrm { x } ) \mathbf { j } + \mathrm { z } ^ { 5 } \mathbf { k } ; \mathrm { S }F=(4−y)i+(3+x)j+z5k;S

Find the Flux of the Curl of Field F Through $\mathbf { F } = ( 4 - \mathrm { y } ) \mathbf { i } + ( 3 + \mathrm { x } ) \mathbf { j } + \mathrm { z } ^ { 5 } \mathbf { k } ; \mathrm { S }$