$$\displaystyle \int_{0}^{4a}\int_{y^2/4a}^{y}\frac{x^2-y^2}{x^2+y^2}\ \mathrm dx\ \mathrm dy$$
I wanted to solve this double integral using polar coordinates but unfortunately, I am not able to get the right limits. Please help
My Try:
The integral gets reduced to somehow $$\large \int_{?}^{?}\int_{0 ?}^{4a ?}r\cos2\theta\ \mathrm dr\ \mathrm d\theta $$
Can someone explain this via graph ? I am not sure why 90 degrees upper limit
We should write restrictions for set on $xy$ plane in polar coordinates, which gives
$$\begin{cases}0 \leqslant r\sin\alpha \leqslant 4a \\ \frac{r^2\sin^2 \alpha}{4a} \leqslant r \cos \alpha \leqslant r\sin\alpha \end{cases}$$ So, additionally to being in first quadrant, we obtain $\tan \alpha \geqslant 1$, from right side of second line, i.e. $\alpha \geqslant \frac{\pi}{4}$, and $r \leqslant \frac{4a \cos \alpha}{\sin^2 \alpha}$ from left side. For integral we have $$\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{4a \cos \alpha}{\sin^2 \alpha}}$$