Find the flux of the vector field F across the surface S in the indicated direction.
-$\mathbf { F } = 5 x \mathbf { i } + 5 y \mathbf { j } + \mathrm { z } ^ { 2 } \mathbf { k } ; \mathrm { S }$ is portion of the cylinder $\mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } = 25$ between $\mathrm { z } = 0$ and $\mathrm { z } = 5$; direction is outward
A) 625
B) 1250
C) $- 1250 \pi$
D) $1250 \pi$

Question

Quizplus · Accepted Answer

The flux of the vector field $\mathbf{F} = 5x\mathbf{i} + 5y\mathbf{j} + z^2\mathbf{k}$ across the surface $S$ of the cylinder $x^2 + y^2 = 25$ between $z = 0$ and $z = 5$ in the outward direction can be calculated using the divergence theorem. The divergence of $\mathbf{F}$ is $\nabla \cdot \mathbf{F} = 5 + 5 + 2z = 10 + 2z$. Integrating this over the volume of the cylinder gives the total flux out of the surface. The volume element in cylindrical coordinates is $r dr d\theta dz$, and for this cylinder, $r$ goes from 0 to 5, $\theta$ goes from 0 to $2\pi$, and $z$ goes from 0 to 5. Thus, the flux is $\int_0^{2\pi} \int_0^5 \int_0^5 (10 + 2z) r dr dz d\theta = 1250\pi$.

Find the Flux of the Vector Field F Across the Surface