I have the following exercise:
"Use the Stokes' Theorem to find the work of the vector field $ \overrightarrow{F}=x^2 \hat{i}+2x \hat{j}+z^2\hat{k}$ along the anti-counterclockwise oriented area of the ellipse $ 4x^2+y^2=4$."
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I have done the following:
Since the Ellipse is on the $ xy $ plane, $ \hat{n}=\hat{k}$.
$ \nabla \times \overrightarrow{F}=2 \hat{k}$
$d \sigma= dxdy $
$\oint_C{\overrightarrow{F}}dR=\iint_S{\nabla \times \overrightarrow{F} \cdot \hat{n} }d \sigma=\iint_S{2\hat{k} \cdot \hat{k}}dxdy=\int_{-1}^1 \int_{-2 \sqrt{1-x^2}}^{2 \sqrt{1-x^2}}{2}dydx=\int_{-1}^1{8 \sqrt{1-x^2}}dx=4 \pi $.
Is my idea correct?
Best Answer
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.