Find the flux of the vector field F across the surface S in the indicated direction.
-$\mathbf { F } = \mathrm { x } ^ { 5 } \mathrm { yi } - \mathrm { zk }$; $\mathrm { S }$ is portion of the cone $\mathrm { z } = 4 \sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } }$ between $\mathrm { z } = 0$ and $\mathrm { z } = 3$; direction is outward
A) $- \frac { 27 } { 8 } \pi$
B) $- \frac { 9 } { 16 }$
C) $\frac { 9 } { 8 } \pi$
D) $- \frac { 9 } { 8 } \pi$

Question

Quizplus · Accepted Answer

Computing the divergence of $\mathbf{F}$, we have $
abla \cdot \mathbf{F} = 5x^4$. By the divergence theorem, the flux of $\mathbf{F}$ across $S$ in the outward direction is equal to the triple integral of $
abla \cdot \mathbf{F}$ over the region enclosed by $S$:
\begin{align*}
\iint_S \mathbf{F} \cdot \mathbf{\hat{n}}\, dS &= \iiint_E 
abla \cdot \mathbf{F} \, dV \
&= \int_0^{2\pi} \int_0^3 \int_0^{4r/\sqrt{2}} 5r^4 \cos^4	heta \, dz \, dr \, d	heta \
&= \frac{45}{32} \pi \int_0^3 (4r)^5 \, dr \
&= \frac{9}{8} \pi.
\end{align*}

Find the Flux of the Vector Field F Across the Surface