Using integration, find the area of the region bounded by the circle $\displaystyle x^{2}+y^{2}=16$, line $\displaystyle y=x$ and $\displaystyle y$-axis, but lying in the $\displaystyle 1^{\text {st }}$ quadrant.

Required area $\displaystyle =\int_{0}^{2 \sqrt{2}} \boldsymbol{y}$ dy $\displaystyle +\int_{2 \sqrt{2}}^{\mathbf{4}} \sqrt{\mathbf{1 6}-\boldsymbol{y}^{\mathbf{2}}}$ dy\begin{aligned} & =\left|\frac{y^{2}}{2}\right|_{0}^{2 \sqrt{2}}+\left|\frac{y}{2} \sqrt{16-y^{2}}+\frac{16}{2} \sin ^{-1}\left(\frac{y}{4}\right)\right|_{2 \sqrt{2}}^{4} & =4+\left\{8 \sin ^{-1}(1)-\sqrt{2} \cdot \sqrt{8}-8 \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right\} & =4+8\left(\frac{\pi}{2}\right)-4-8\left(\frac{\pi}{4}\right) \end{aligned} $$