Find the flux of the vector field F across the surface S in the indicated direction.
-$\mathbf { F } = x \mathbf { i } + y \mathbf { j } + z ^ { 3 } \mathbf { k }$; S is portion of the cone $z = 3 \sqrt { x ^ { 2 } + y ^ { 2 } }$ between $z = 3$ and $z = 6$; direction is outward
A) $- \frac { 1604 } { 5 } \pi$
B) $\frac { 1604 } { 5 } \pi$
C) $\frac { 802 } { 5 } \pi$
D) $- \frac { 802 } { 5 }$

Question

Quizplus · Accepted Answer

First, we need to parameterize the surface S. From the equation of the cone, we have $z=3\sqrt{x^2+y^2}$, so $x^2+y^2=z^2/9$. This suggests the change of variables $u=x$, $v=y$, $w=z/3$, so that $u^2+v^2=w^2$. Then we can parameterize $S$ as follows:
$$\mathbf{r}(u,v) = \langle u,v,3\sqrt{u^2+v^2}angle\quad	ext{for}\ 0\le w\le 2.$$
The normal to the surface is given by
\begin{align*}
\mathbf{N}(u,v) &= \frac{
abla w}{\|
abla w\|} = \frac{\langle u,v,wangle}{\sqrt{u^2+v^2+w^2}} \
&= \frac{1}{\sqrt{u^2+v^2+9u^2+9v^2}}\langle u,v,3\sqrt{u^2+v^2}angle \
&= \frac{1}{\sqrt{10(u^2+v^2)}}\langle u,v,3\sqrt{u^2+v^2}angle.
\end{align*}
The integral of $\mathbf{F}$ over $S$ in the outward direction is then given by
\begin{align*}
\iint_S \mathbf{F}\cdot\mathbf{N}\,dS &= \iint_D \mathbf{F}(\mathbf{r}(u,v))\cdot\mathbf{N}(u,v)\|\mathbf{r}_u	imes\mathbf{r}_v\|\,du\,dv \
&= \int_0^{2\pi}\int_0^3 \langle u,v,w^3angle \cdot\frac{1}{\sqrt{10(u^2+v^2)}}\frac{1}{3\sqrt{u^2+v^2}}\sqrt{10(u^2+v^2)}\,du\,dv \
&= \frac{1}{15}\int_0^{2\pi}\int_0^3 (u^2+v^2)^{3/2}\,du\,dv \
&= \frac{\pi}{2}\int_0^3 r^5\,dr \
&= \frac{\pi}{2}\cdot\frac{3^6}{6} \
&= \frac{729\pi}{4}.
\end{align*}
Since we want the flux in the outward direction, which is opposite to the direction of $\mathbf{N}$, we need to negate the above result to get the final answer:
$$-\frac{729\pi}{4} = -\frac{5832}{20}\pi = \boxed{-\frac{1604}{5}\pi}.$$

Find the Flux of the Vector Field F Across the Surface