Solution:

let $\quad I=\int_{0}^{\frac{a}{\sqrt{2}}} \int_{y}^{\sqrt{a^{2}-y^{2}}} \log \left(x^{2}+y^{2}\right) d x \ d y$

$\begin{aligned} \text { Region of integration: } & y \leq x \leq \sqrt{a^{2}-y^{2}} \\ & 0 \leq y \leq \frac{a}{\sqrt{2}} \end{aligned}$

The line $x=y$ is inclined at $45^{\circ}$ to the +ve x -axis.

$\begin{aligned} \text { Curves: } & \text { (i) } \mathrm{x}=\mathrm{y}, \mathrm{y}=0, \mathrm{y}=\frac{a}{\sqrt{2}} \quad \text { lines } \\ & \text { (ii) } x=\sqrt{a^{2}-y^{2}} \\ & x^{2}+y^{2}=a^{2} \quad \text { circle with centre }(0,0) \text { and radius a. } \end{aligned}$

$(\mathrm{x}, \mathrm{y}) \quad \longrightarrow \quad(\mathrm{r}, \theta)$

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

$f(x, y)=\log \left(x^{2}+y^{2}\right)=\log r^{2}=2 \log r=f(r, \theta)$

Limits changes to : $0 \leq r \leq a$

$0 \leq \theta \leq \frac{\pi}{4}$

$\begin{aligned} \therefore & I=\int_{0}^{\frac{\pi}{4}} \int_{0}^{a} 2 \log r . r \ d r \ d \theta \\ &=2 \int_{0}^{\frac{\pi}{4}}\left[\log r \cdot \frac{r^{2}}{2}-\frac{r^{2}}{4}\right]_{0}^{a} d \theta \\=& 2 \int_{0}^{\frac{\pi}{4}}\left[\log a \cdot \frac{a^{2}}{2}-\frac{a^{2}}{4}\right] d \theta \\ \therefore I &=\left[\log a \cdot \frac{a^{2}}{2}-\frac{a^{2}}{4}\right] \times \frac{\pi}{4} \end{aligned}$