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I am given the integral $$I=\int_{C} \frac{y}{4x^2+7y^2} dx - \frac{x}{4x^2+7y^2}dy$$ where C is the rectangle with vertices $$A=(4, 7), B=(-4, 7), C=(-4, -7), D=(4, -7)$$ oriented in the counterclockwise direction. I am also asked to consider the ellipse with equation $$4x^2+7y^2=1$$ and use Green's Theorem to solve this integral.

I am unable to figure out how to solve this integral, let alone start this question off.

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  • $\begingroup$ I'm not sure if it comes from a book, as it is an online assignment. It is the ellipse part that throws me off completely. $\endgroup$ – greeneggs Apr 11 '16 at 18:29
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Note that for $(x,y)\ne (0,0)$, we have

$$\frac{\partial }{\partial y}\left(\frac{y}{4x^2+7y^2}\right)+\frac{\partial }{\partial x}\left(\frac{x}{4x^2+7y^2}\right)=0$$

Therefore, for any smooth closed contours $C_1$ and $C_2$ that encircle the origin, Green's Theorem guarantees that

$$\oint_{C_1}\left(\frac{y}{4x^2+7y^2}\,dx-\frac{x}{4x^2+7y^2}\,dy\right)=\oint_{C_2}\left(\frac{y}{4x^2+7y^2}\,dx-\frac{x}{4x^2+7y^2}\,dy\right)$$

If we choose $C_2$ to be the ellipse defined by $4x^2+7y^2=1$, then on the ellipse $\frac{dy}{dx}=-\frac{4x}{7y}$ and therefore

$$\begin{align}\oint_{C_2}\left(\frac{y}{4x^2+7y^2}\,dx-\frac{x}{4x^2+7y^2}\,dy\right)&=\oint_{C_2}\frac{1}{7y}\,dx\\\\ &=-\frac{4\sqrt{7}}{7}\int_{0}^{1/2}\frac{1}{\sqrt{1-4x^2}}\,dx\\\\ &=-\frac{\pi}{\sqrt 7} \end{align}$$

Since this is equal to the integral over $C_1$, we can choose $C_1$ to be the rectangular contour as defined in the OP. Therefore, the answer is

$$\bbox[5px,border:2px solid #C0A000]{I=-\frac{\pi}{\sqrt{7}}}$$

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