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I have the answer to a problem and am trying to understand the steps to get to that answer. The problem is $\oint_C(x+2y)dx+(y-2x)dy$ around the ellipse C, defined by $x=4cos\theta, y=3sin\theta, 0\leq \theta < 2\pi$ and C is defined counterclockwise. The answer is $-48\pi$.

Applying Green's Theorem, this is what I have done: $-4\leq x\leq4, -3 \leq y \leq 3$. $$\int_{y=-3}^{y=3}\int_{x=-4}^{x=4}(-2-2)dxdy=\int_{y=-3}^{y=3}-4x\Big|_{-4}^4dy=\int_{y=-3}^{y=3}-32dy=-32y\Big|_{-3}^3=-192$$

So, I know that this is wrong, I just don't know why.

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    $\begingroup$ You need to integrate over the region bounded by an ellipse. You’ve integrated over a rectangle. $\endgroup$ – Ahmed S. Attaalla Oct 6 '18 at 23:03
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That's because, the double integral is over a square and not and ellipse, you have to use the equation of the ellipse:

$$\frac{x^{2}}{16}+\frac{y^{2}}{3}=1$$

You find that the curve is between:

$$y=\pm\sqrt{1-\frac{x^{2}}{16}}$$

Then you're x is between $-4$ and $4$, that is where you get your $\pi$

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    $\begingroup$ $y=\pm\sqrt{9-\frac{9x^{2}}{16}}$ $\endgroup$ – AdamK Oct 7 '18 at 5:50

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