Calculate the flux of the field F across the closed plane curve C.
-$$\mathbf { F } = y \mathbf { i } - \mathrm { xj } ; \text { the curve } \mathrm { C } \text { is the circle } ( x + 6 ) ^ { 2 } + ( \mathrm { y } + 9 ) ^ { 2 } = 144$$
A) 0
B) -288
C) 144
D) -144

Question

Quizplus · Accepted Answer

Since the circle C is a closed curve, we can apply Green's theorem:
$$\oint_C \mathbf{F}\cdot d\mathbf{r}=\iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA$$
where $R$ is the region enclosed by $C$, $P=-y$ and $Q=x$, so
$$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=1+1=2$$
since $(x+6)^2+(y+9)^2=144$ is the equation of a circle centered at $(-6,-9)$ with radius $12$, $R$ is the interior of this circle, so
$$\iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA=\iint_R 2dA=2\cdot 144\pi=288\pi$$
thus, the flux across $C$ is zero, as desired.

Calculate the Flux of the Field F Across the Closed