Integral in Cartesian plane: $$\int_0^{4a}\int_{\frac{y^2}{4a}}^y \frac{x^2-y^2}{x^2+y^2} dx.dy $$
My progress so far: $$\int\int r(\cos 2\theta) dr.d\theta$$
I want to know the limits of integration of this curve in polar format.
I know that the region of integration will be between a line $y-x=0$ cutting the parabola $y^2=4ax$. But I am having trouble in defining them in terms of $r$ and $\theta$.
Please help.
On
The region is between the parabola $x=\frac{y^2}{4a}$ and the line $x=y$
In polar coordinates the arc of parabola becomes $r \cos t=\frac{r^2 \sin ^2 t }{4 a}$ which simplifies to
$r=4 a \cot t \csc t$ with $t\in\left[\frac{\pi}{4},\frac{\pi}{2}\right]$
The integral should become
$$\int_{\frac{\pi }{4}}^{\frac{\pi }{2}} \int_{4 a \cot t \csc t}^0 r \cos (2 t) \, dr \, dt=\frac{4}{3} (10-3 \pi ) a^2$$
If you would plot the domain, and plot a line going through the origin, you might be able to see that the range for $\theta$ is from $\pi/4$ to $\pi/2$. The $\pi/4$ corresponds to $y=x$. The $\pi/2$ value corresponds to the slope at $x=y=0$ (parabola $y^2=4ax$ is "open" to the right). Now choose any value $\theta$ in this range. You can write $$\tan\theta=y/x$$ and calculate the $x$ and $y$ values at the intersection $$\tan^2\theta=\frac{y^2}{x^2}=\frac{4ax}{x^2}$$ where you get $$x^2\tan^2\theta-4ax=x(x\tan^2\theta-4a)=0$$ One solution is $x=0$ corresponding to $r=0$. The other one is given by $$r\cos\theta\tan^2\theta-4a=0$$