Use the Green's Theorem to calculate the work and the flux

Question

Use the Green's Theorem to calculate the work and the flux for the closed anti-clockwise direction that consists of the square which is determined by the lines $x=0$, $x=1$, $y=0$ and $y=1$ if $\overrightarrow{F}=2xy\hat{i}+3x^2y\hat{j}$. $$$$ I have done the following: $$\text{Work }: \oint_C{\overrightarrow{F}}dR=\int \int_R {\nabla \times \overrightarrow{F} \cdot \hat{n}}dA=\iint_R {\nabla \times \overrightarrow{F} \cdot \hat{k}}dA=\iint_R {(6xy-2x)}dA=\int_0^1 \int_0^1(6xy-2x)dxdy=\int_0^1(3y-1)dy=\frac{1}{2}$$ $$\text{Flux }: \oint_C{\overrightarrow{F} \cdot \hat{n}}ds=\iint_R{\nabla \cdot \overrightarrow{F}}dA=\iint_R{(2y+3x^2)}dA=\int_0^1 \int_0^1{(2y+3x^2)}dxdy=\int_0^1{(2y+1)}dy=2$$ Is this correct???

Answer 1

The work is correct, but the flux is not. Notice that it is conceptually incorrect to talk about the flux for a closed counter-clockwise circuit. Flux is not related to the circulation of the field. What makes sense is the flux through a surface. Therefore, the theorem that has some relevance to what you are trying to calculate is

$\displaystyle \int_S \vec{F} d\vec{s} = \int_V \left(\vec{\nabla}\vec{F}\right) dV$,

where $d\vec{s}$ represents the differential surface vector perpendicular to the surface and $dV$ is the differential volume. However, this is valid for a closed surface that encloses a volume. This is not the case for your exercise, where you have just a flat surface.

To evaluate your flux you have to integrate the field $\vec{F}$ over the surface. Here, you are dealing with the simplest case, because your $d\vec{s}$ vector is always perpendicular to your field $\vec{F}$, so the integral is zero.