Find the flux of the vector field F across the surface S in the indicated direction.
-$\mathrm { F } = \mathrm { x } ^ { 5 } \mathrm { y } ^ { 2 } \mathrm { i } + \mathrm { yj } - \mathrm { zk } ; \mathrm { S }$ is the portion of the parabolic cylinder $\mathrm { z } = 4 - \mathrm { y } ^ { 2 }$ for which $\mathrm { z } \geq 0$ and $2 \leq \mathrm { x } \leq 10$; direction is outward (away from the $x$-y plane)
A) -128
B) 128
C) 0
D) 64

Question

Quizplus · Accepted Answer

The flux across the surface is zero because the vector field is divergence-free ($\nabla \cdot \mathbf{F} = 0$), and the surface is closed.

Find the Flux of the Vector Field F Across the Surface