Calculate the flux of the field F across the closed plane curve C.
-$\mathbf { F } = \mathrm { y } ^ { 2 } \mathbf { i } + \mathrm { x } ^ { 3 } \mathbf { j }$; the curve $\mathrm { C }$ is the closed counterclockwise path formed from the semicircle $\mathbf { r } ( \mathrm { t } ) = 3 \cos \mathrm { ti } + 3$ sin $\mathrm { j }$, $0 \leq t \leq \pi$, and the straight line segment from $( - 3,0 )$ to $( 3,0 )$
A) $= \frac { 27 } { 2 }$
B) 0
C) $\frac { 27 } { 2 }$
D) 27

Question

Quizplus · Accepted Answer

The flux of a vector field across a closed curve is zero if the field is conservative. Here, the field F is not conservative, but the divergence theorem (Green's theorem in 2D) can be applied. The divergence of F is zero, so the flux across the closed curve C is zero.

Calculate the Flux of the Field F Across the Closed F=y2i+x3j\mathbf { F } = \mathrm { y } ^ { 2 } \mathbf { i } + \mathrm { x } ^ { 3 } \mathbf { j }F=y2i+x3j

Calculate the Flux of the Field F Across the Closed $\mathbf { F } = \mathrm { y } ^ { 2 } \mathbf { i } + \mathrm { x } ^ { 3 } \mathbf { j }$