Question

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

Answer 1

The given curves are

$x^2+y^2=4$ (circle)

$x^2=-\sqrt{2}y$ (parabola , concave downward)

$x=y$ (straight line through origin)

Solving equations (1) and (2), we get

$y^2-\sqrt{2}y-4=0$

$\Rightarrow y=\frac{4\sqrt{2}}{2}$ or $\frac{-2\sqrt{2}}{2}$

$\Rightarrow y=2\sqrt{2}$ or $-\sqrt{2}$

$\Rightarrow x^2=2$ (rejecting $y=2\sqrt{2}$ as $x^2$ is positive)

$\Rightarrow x=\pm \sqrt{2}$

$\therefore$ Points of intersection of (1) and (2) are $B(\sqrt{2},-\sqrt{2}),A(-\sqrt{2},-\sqrt{2})$

Solving (1) and (3), we get

$2x^2=4\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}\Rightarrow y=\pm \sqrt{2}$

$\therefore$ Point of intersection are $(-\sqrt{2},-\sqrt{2}),(\sqrt{2},\sqrt{2})$

Thus, all the three curves pass through the same point $A(-\sqrt{2},-\sqrt{2})$.

Now the required area = the shaded area

$={\int }_{-\sqrt{2}}^0(x-(-\sqrt{4-x^2}\left)\right)dx+{\int }_0^{\sqrt{2}}(-\frac{x^2}{\sqrt{2}}-(-\sqrt{4-x^2}\left)\right)dx$

$=2{\int }_0^{\sqrt{2}}\sqrt{4-x^2}dx+{\int }_{-\sqrt{2}}^{0}xdx-{\int }_0^{\sqrt{2}}\frac{x^2}{\sqrt{2}}dx$

$=2{[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_0^{\sqrt{2}}+{\left[\frac{x^2}{2}\right]}_{-\sqrt{2}}^0-{\left[\frac{x^3}{3\sqrt{2}}\right]}_0^{\sqrt{2}}$

$=2[\frac{\sqrt{2}}{2}\sqrt{4-2}+2{\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right)]+\left[\frac{-2}{2}\right]-\left[\frac{2\sqrt{2}}{3\sqrt{2}}\right]$

$=2[1+2\frac{\pi }{4}]-1-\frac{2}{3}=\pi +\frac{1}{3}sq.units$