# Evaluate using Green's Theorem over ellipse

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.

• You need to integrate over the region bounded by an ellipse. You’ve integrated over a rectangle. – Ahmed S. Attaalla Oct 6 '18 at 23:03

## 1 Answer

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 $$\pi$$

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