# Homework Help: Evaluate this double integral

1. Nov 9, 2014

### utkarshakash

1. The problem statement, all variables and given/known data
Evaluate the integral by changing into polar coordinates.
$\displaystyle \int_0^{4a} \int_{y^2/4a}^y \dfrac{x^2-y^2}{x^2+y^2} dx dy$

3. The attempt at a solution
Substituting x=rcos theta and y=rsin theta , the integrand changes to $cos 2 \theta r dr d \theta$ . I know that the region of integration is the area bounded by parabola y^2=4ax and y=x. However, I don't know the limits of r and theta.

2. Nov 9, 2014

### Ray Vickson

Perhaps you should not use $x = r \cos \theta, \: y = r \sin \theta$. Why would you not just do the inner x-integration first, then do the y-integration?

3. Nov 9, 2014

### LCKurtz

Have you drawn a picture of that region? Have you written the equation of that parabola in polar coordinates? You need to do both. Then think of a point inside the region. It will move from $r=0$ to $r$ on the parabola. Once you draw that you should be able to see what values $\theta$ must vary between.

4. Nov 10, 2014

### utkarshakash

Because I'm required to solve it by changing into polar coordinates.

5. Nov 10, 2014

### utkarshakash

The polar equation of given parabola will be $r=-2a/(1+cos \theta)$. The limit of theta is pi/4 to pi/2. Integrating the expression wrt r i'm left with $\int_{\pi/4}^{\pi/2} \dfrac{-2a cos 2 \theta}{1+cos \theta} d \theta$

6. Nov 10, 2014

### LCKurtz

It looks like you have figured out the limits now, but I don't get the same equation for the parabola. Surely $r$ shouldn't be negative. And let's see the double integral after you have put it in polar coordinates and before you integrate it.