Find the area bounded by the curves lying in third and fourth quadrant $x^{2}+y^{2}=4, x^{2}=-\sqrt{

Question

Find the area bounded by the curves lying in third and fourth quadrant $x^{2}+y^{2}=4, x^{2}=-\sqrt{2} y, x=y$.

Professor PC Thomas Classes · Answer

Curves $x^{2}+y^{2}=4, x^{2}=-\sqrt{2} y$ and $x=y$ could be shown as\ Thus, the required area\ $=\left|\int_{-\sqrt{2}}^{\sqrt{2}} \sqrt{4-x^{2}} d xight|-\left|\int_{-\sqrt{2}}^{0} x d xight|-\left|\int_{0}^{\sqrt{2}} \frac{-x^{2}}{\sqrt{2}} d xight|$\ $=2 \int_{0}^{\sqrt{2}} \sqrt{4-x^{2}} d x-\left|\left(\frac{x^{2}}{2}ight)ight|_{-\sqrt{2}}^{0}-\left|\frac{x^{3}}{3 \sqrt{2}}ight|_{0}^{\sqrt{2}}$\ $=2\left|\frac{x}{2} \sqrt{4-x^{2}} d x+\frac{4}{2} \sin ^{-1} \frac{x}{2}ight|_{0}^{\sqrt{2}}-1-\frac{2}{3}$\ $=(2+\pi)-\frac{5}{3}=\frac{1}{3}+\pi$.\