Find the flux of the vector field F across the surface S in the indicated direction.
-$\mathbf { F } = 7 x \mathbf { i } + 7 y \mathbf { j } + \mathrm { z } \mathbf { k }$; $\mathrm { S }$ is portion of the plane $x + y + z = 4$ for which $0 \leq x \leq 3$ and $0 \leq y \leq 4$; direction is outward (away from origin)
A) 276
B) - 204
C) 600
D) 300

Question

Quizplus · Accepted Answer

We need to find the flux, which is given by the surface integral $\int_S\mathbf{F}\cdot d\mathbf{S}$. To evaluate this, we need to find a unit normal vector to the surface $S$. Since the surface $S$ is a plane, we can find a normal vector by taking the cross product of two vectors in the plane. We can choose $\mathbf{v}_1 = \langle 1,0,1angle$ and $\mathbf{v}_2 = \langle 0,1,1angle$, which lie in the plane and are linearly independent. Then a normal vector is given by $\mathbf{n} = \mathbf{v}_1	imes \mathbf{v}_2 = \langle -1,-1,1angle$, which points outward since the direction of $\mathbf{n}$ is determined by the right-hand rule applied to $\mathbf{v}_1$ and $\mathbf{v}_2$.

To set up the surface integral, we need to parameterize the surface $S$. We can use two of the variables $x$, $y$, and $z$ as parameters, say $x$ and $y$, and express $z$ as $z=4-x-y$. Then the surface is bounded by $0\leq x \leq 3$ and $0\leq y\leq 4-x$. The unit normal vector pointing outward is $\mathbf{n} = \langle -1,-1,1angle/\sqrt{3}$, so we have
\begin{align*}
\int_S\mathbf{F}\cdot d\mathbf{S} &= \int_0^3\int_0^{4-x} (7x\mathbf{i}+7y\mathbf{j}+z\mathbf{k})\cdot\frac{\langle -1,-1,1angle}{\sqrt{3}} \,dy\,dx\
&= \int_0^3\int_0^{4-x}(-\frac{7}{\sqrt{3}}x-\frac{7}{\sqrt{3}}y+\frac{4-x-y}{\sqrt{3}})\,dy\,dx\
&=\frac{1}{\sqrt{3}}\int_0^3\int_0^{4-x}(3x+3y+4)\,dy\,dx\
&=\frac{1}{\sqrt{3}}\int_0^3(10x+16)\,dx\
&=\frac{1}{\sqrt{3}}\cdot 300 = \boxed{	extbf{(D) } 300}.
\end{align*}

Find the Flux of the Vector Field F Across the Surface