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 $\overrightarrow{F}=2xy\hat{i}+3x^2y\hat{j}$.
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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???
[Math] Use the Green’s Theorem to calculate the work and the flux
vector analysis
Related Solutions
The region in question:
The boundary of $D$ with the natural orientation (in preparation for the use of Green's Theorem) is the union $(−C_1) \cup C_2$, where $C_1$ and $C_2$ are traversed in the counterclockwise direction. See oenamen's answer for the proper justification. Writing $−C_1$ means "traverse $C_1$ in the opposite direction". So, Green's theorem gives us
$$ \begin{align} \iint_{D} N_x - M_y \,dx\,dy &= \oint_{(−C_1) \cup C_2} M\,dx + N\,dy \\ &= \oint_{−C_1} M\,dx + N\,dy + \oint_{C_2} M\,dx + N\,dy \\ &= -\oint_{C_1} M\,dx + N\,dy + \oint_{C_2} M\,dx + N\,dy. \end{align} $$
Just show that the integral over $D$ is zero to get your equality.
Yes, this is right. Also, you don't need to do the integral explicitly, because from $\displaystyle \int_S{2\hat{k} \cdot \hat{k}}\,dx\,dy = 2\int_S\,d\sigma$ you can conclude that the result you are looking for is twice the area of the ellipse.
Best Answer
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.