Use the divergence theorem to evaluate where F(x, y, z) = 2e ^x i - 2ye ^x j + 6z k, n is the outer unit normal to $\sigma$, and $\sigma$ is the surface of the sphere by x² + y² + z² = 36.

Question

Quizplus · Accepted Answer

By the divergence theorem,

$\iint_{\partial V} \mathbf{F} \cdot d\mathbf{S} = \iiint_V 
abla \cdot \mathbf{F} \ dV$

We have $\mathbf{F}(x,y,z) = 2e^{x^2+y^2+z^2} \cos{	heta} \ \mathbf{i} - 2ye^{x^2+y^2+z^2}\sin{	heta} \ \mathbf{j} + 6ze^{x^2+y^2+z^2} \ \mathbf{k}$

where $	heta$ is the angle between $\mathbf{F}$ and $x$-axis.

We note that $
abla \cdot \mathbf{F} = 6ze^{x^2+y^2+z^2}$.

The surface of the sphere centered at the origin is given by $x^2 + y^2 + z^2 = 36$. This can be written as $ho = 6$ in spherical coordinates.

Using spherical coordinates, we have:

$$\iiint_V 
abla \cdot \mathbf{F} \ dV = \int_0^{2\pi} \int_0^{\pi} \int_0^6 (6ho \cos{\phi} \ e^{ho^2})ho^2 \sin{\phi} \ dho \ d\phi \ d	heta = \frac{1}{2}(e^{36}-1)\int_0^{2\pi} \int_0^{\pi} \sin{\phi} \cos{\phi} \ d\phi \ d	heta = 0$$

where we used integration by parts to evaluate the integral.

Therefore, $\iint_{\partial V} \mathbf{F} \cdot d\mathbf{S} = 0$.

The only answer choice that gives 0 as the result is $\boxed{	extbf{(C)}}$.

Use the Divergence Theorem to Evaluate Where F(x, Y σ\sigmaσ

Use the Divergence Theorem to Evaluate Where F(x, Y $\sigma$