Find the flux of the vector field F across the surface S in the indicated direction.
-$\mathbf { F } ( \mathrm { x } , \mathrm { y } , \mathrm { z } ) = 7 \mathrm { x } \mathbf { i } + 7 \mathrm { yj } + 2 \mathbf { k } , \mathrm { S }$ is the surface cut from the bottom of the paraboloid $\mathrm { z } = \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 }$ by the plane $z = 2$, direction is outward
A) $120 \pi$
B) $- 208 \pi$
C) $24 \pi$
D) $104 \pi$

Question

Quizplus · Accepted Answer

First, we need to find the normal vector of the surface S. The gradient of the paraboloid $\mathrm { z } = \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 }$ is 
$$\nabla (\mathrm { z } ) = 2 \mathrm { x } \mathbf { i } + 2 \mathrm { y }\mathbf { j } - \mathbf { k }$$
At $(x,y,z)=(x,y,2)$, the normal vector points outward from the surface S, and it is given by 
$$\mathbf{n} = \frac{\nabla (\mathrm{z})}{|\nabla (\mathrm{z})|} = \frac{2\mathrm{x}\mathbf{i}+2\mathrm{y}\mathbf{j}-\mathbf{k}}{\sqrt{4\mathrm{x}^2+4\mathrm{y}^2+1}}$$

Now, we can calculate the flux of the vector field F across the surface S by taking the surface integral: 
$$\iint _ { S } \mathbf { F } \cdot \mathbf { n } \mathrm { d } S$$
Since the surface S is given by $\mathrm { z } = 2$ and the paraboloid $\mathrm { z } = \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 }$, we can find the bounds for the integral: 
$$0 \leq \theta \leq 2\pi, \quad 0 \leq r \leq \sqrt{2}, \quad 2 \leq z \leq 4-r^2$$

Then we can evaluate the integral:
\begin{align*}
\iint _ { S } \mathbf { F } \cdot \mathbf { n } \mathrm { d } S &= \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 2 } } \mathbf { F } ( r \cos \theta , r \sin \theta , 2 ) \cdot \mathbf{n}( r \cos \theta , r \sin \theta ) r \mathrm { d } r \mathrm { d } \theta \
&= \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 2 } } \left( 14 r ^ { 3 } \cos \theta + 14 r ^ { 3 } \sin \theta \right) \frac { 2 \mathrm{x} \mathbf{i} + 2 \mathrm{y} \mathbf{j} - \mathbf{k} }{ \sqrt { 4 \mathrm{x} ^ { 2 } + 4 \mathrm{y} ^ { 2 } + 1 } } \cdot \left( - \frac { 2 \mathrm{x} } { \sqrt { 4 \mathrm{x} ^ { 2 } + 4 \mathrm{y} ^ { 2 } + 1 } } \mathbf{i} - \frac { 2 \mathrm{y} } { \sqrt { 4 \mathrm{x} ^ { 2 } + 4 \mathrm{y} ^ { 2 } + 1 } } \mathbf{j} - \frac { 1 } { \sqrt { 4 \mathrm{x} ^ { 2 } + 4 \mathrm{y} ^ { 2 } + 1 } } \mathbf{k} \right) r \mathrm { d } r \mathrm { d } \theta \
&= \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { \sqrt { 2 } } \left( -2r^4 + r^2\sqrt{4r^2+1} \right) \mathrm{d}r \mathrm { d } \theta \
&= \int _ { 0 } ^ { 2 \pi } \left[ -\frac{2}{5}r^5 + \frac{1}{3}r^3\sqrt{4r^2+1}\right]\Big|_{0}^{\sqrt{2}} \mathrm{d} \theta \
&= \int _ { 0 } ^ { 2 \pi } \left( -\frac{32\sqrt{2}}{15} + \frac{4\sqrt{10}}{3} \right) \mathrm{d} \theta \
&= 24\pi
\end{align*}

Therefore, the answer is (C) $24\pi$.

Find the Flux of the Vector Field F Across the Surface