Given that y = x and $x^2 + y^2$ = 32 Put y = x in $x^2 + y^2$ = 32 $x^2 + y^2$ = 32 ⇒ $2x^2$ = 32 ⇒ x = ± 4 = y Hence, point s of intersection are A (4,4) and B (4,-4) . Required shaded area = $\int\limits_{0}^{4}$ x dx + $\int\limits_{4}^{\sqrt{32}}$ $\sqrt{32-x^2}$ dx (Since Region in the first quadrant) = $\left(\frac{x^2}{4}\right)_{0}^{4}$ + = 4π sq . units