Use a computer algebra system to compute the flux of F across S. S is the surface of the cube cut from the first octant by the planes $x = \frac { \pi } { 2 } , y = \frac { \pi } { 2 } , z = \frac { \pi } { 2 }$ $\mathbf { F } ( x , y , z ) = 3 \sin x \cos ^ { 2 } y \mathbf { i } + 3 \sin ^ { 3 } z \cos ^ { 4 } z \mathbf { j } + 3 \sin ^ { 5 } z \cos ^ { 6 } x \mathbf { k }$
A) 3
B) 4
C) $6$
D) 0.67
E) 1

Question

Quizplus · Accepted Answer

We will use the Divergence Theorem to solve the problem: 
$$
\int_{S}{\mathbf{F} \cdot \mathrm{d}\mathbf{S}} = \int_{V}{\mathrm{div}\,\mathbf{F}\:\mathrm{d}V}
$$
where $V$ is the solid region enclosed by $S$.

First, let's find the divergence of $\mathbf{F}$: 
$$
\mathrm{div}\,\mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}
$$
$$
= 3\cos ^ {2}y\cos x + 9\sin ^ {2}z\cos ^ {4}z - 30\sin ^ {5}z\cos ^ {6}x
$$

Next, we need to determine the limits of integration for the volume integral $\int_{V}{\mathrm{div}\,\mathbf{F}\:\mathrm{d}V}$. Since $S$ is the surface of the cube cut from the first octant by the planes $x=\frac{\pi}{2}$, $y=\frac{\pi}{2}$, and $z=\frac{\pi}{2}$, the solid region $V$ is the part of the first octant that lies inside the cube. Therefore, the limits of integration are: 
$$
0 \leq x \leq \frac{\pi}{2},\quad 0 \leq y \leq \frac{\pi}{2},\quad 0 \leq z \leq \frac{\pi}{2}
$$

Using a computer algebra system (such as Mathematica) to evaluate the triple integral, we get: 
$$
\int_{V}{\mathrm{div}\,\mathbf{F}\:\mathrm{d}V} = 6
$$

Therefore, the flux of $\mathbf{F}$ across $S$ is: 
$$
\int_{S}{\mathbf{F} \cdot \mathrm{d}\mathbf{S}} = 6
$$

So, the best choice is (C).

Use a Computer Algebra System to Compute the Flux of F