Solution:

Let $I=\iint \frac{\left(x^{2}+y^{2}\right)^{2}}{x^{2} y^{2}} d x d y$

Region of integration is : Area common to the circle

$x^{2}+y^{2}=a x$ and $x^{2}+y^{2}=b y$

To change the Cartesian coordinates to polar coordinates

Put $x=r \cos \theta \quad$ and $\quad y=r \sin \theta$

Circles: $r=a \cos \theta$ and $r= asin\theta$

The function becomes : $f(x, y)=\frac{\left(x^{2}+y^{2}\right)^{2}}{x^{2} y^{2}}=\frac{r^{4}}{r^{4} \sin ^{2} \theta \cos^2\theta}=\frac{4}{\sin ^{2} 2 \theta}=f(r, \theta)$

Intersection of both circles is at angle $=\tan ^{-1} \frac{a}{b}$ .

Divide the region into two equal halves.

$\begin{array}{ll}{\text { For one region, }} & {0 \leq r \leq b \sin \theta} \\ &{0\leq \theta \leq \alpha}\end{array}$

$\begin{array}{ll}{\text { For another region, }} & {0 \leq r \leq a \cos \theta} \\ &{\alpha\leq \theta \leq \frac{\pi}{2}}\end{array}$

$\therefore I=\int_{0}^{\alpha} \int_{0}^{b \sin \theta} \frac{4 r \ d r \ d \theta}{\sin ^{2} 2 \theta}+\int_{0}^{a \cos \theta} \int_{\alpha}^{\frac{\pi}{2}} \frac{4 r \ d r \ d \theta}{\sin ^{2} 2 \theta}$

$\therefore 1=\int_{0}^{\alpha} \frac{4}{\sin ^{2} 2 \theta}\left[\frac{r^{2}}{2}\right]_0^{bsin \theta}{0} d \theta+\int_{0}^{\frac{\pi}{2}} \frac{4}{\sin ^{2} 2 \theta}\left[\frac{r^{2}}{2}\right]_{0}^{\operatorname{acos} \theta} d t$

$=\frac{1}{2} b^{2} \int_{0}^{\alpha} \sec ^{2} \theta d \theta+\frac{a^{2}}{2} \int_{\alpha}^{\frac{\pi}{2}} cosec ^{2} \theta d \theta$

$\begin{aligned} &=\frac{1}{2} b^{2} \tan \alpha+\frac{a^{2}}{2} \cot \alpha \\ &=\frac{a b}{2}+\frac{a b}{2} \\ \therefore I &=a b \end{aligned}$