Using the Divergence Theorem, find the outward flux of F across the boundary of the region D.
-$\mathbf { F } = \mathrm { e } ^ { \mathrm { yz } } \mathrm { i } + 6 \mathrm { yj } + 8 \mathrm { z } ^ { 2 } \mathbf { k } ;$ D: the solid sphere $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } + \mathrm { z } ^ { 2 } \leq 64$
A) $12,288 \pi$
B) $36864 \pi$
C) $4096 \pi$
D) $\frac { 2048 } { 3 } \pi$

Question

Quizplus · Accepted Answer

The Divergence Theorem states that the outward flux of a vector field $\mathbf{F}$ across the boundary of a region $D$ is equal to the triple integral over $D$ of the divergence of $\mathbf{F}$. The divergence of $\mathbf{F} = \mathrm{e}^{yz}\mathbf{i} + 6y\mathbf{j} + 8z^2\mathbf{k}$ is $\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(\mathrm{e}^{yz}) + \frac{\partial}{\partial y}(6y) + \frac{\partial}{\partial z}(8z^2) = 0 + 6 + 16z$. For a sphere of radius $r = 8$, the volume integral becomes $\int_{D} (6 + 16z) \, dV$. Since the function is symmetric in $z$, the $16z$ term integrates to zero over the sphere. Thus, the integral reduces to $6$ times the volume of the sphere, which is $6 \times \frac{4}{3}\pi r^3 = 6 \times \frac{4}{3}\pi (8)^3 = 4096\pi$.

Using the Divergence Theorem, Find the Outward Flux of F F=eyzi+6yj+8z2k;\mathbf { F } = \mathrm { e } ^ { \mathrm { yz } } \mathrm { i } + 6 \mathrm { yj } + 8 \mathrm { z } ^ { 2 } \mathbf { k } ;F=eyzi+6yj+8z2k;

Using the Divergence Theorem, Find the Outward Flux of F $\mathbf { F } = \mathrm { e } ^ { \mathrm { yz } } \mathrm { i } + 6 \mathrm { yj } + 8 \mathrm { z } ^ { 2 } \mathbf { k } ;$