Question

Using integration, find the area in the first quadrant bounded by the curve $y=x|x|$, the circle $x^2+y^2=2$ and the y-axis.

Answer 1

Consider the given expressions.

$y=x\left|x\right|$

$x^2+y^2=2$

In the first quadrant,

$y=x\left|x\right|$

$y=x^2$

Intersection point in the first quadrant,

$x^2=\sqrt{2-x^2}$

$x^4=2-x^2$

$x^4+x^2-2=0$

$\Rightarrow x^2=\frac{-1\pm \sqrt{1+8}}{2}=\frac{-1\pm 3}{2}$

$\Rightarrow x^2=1(-2\text{is}\text{not}\text{possible})$

Therefore, required area $A$ is,

$A=\underset{0}{\overset{1}{\int }}(\sqrt{2-x^2}-x^2)dx$

$A={[{\sin }^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{x\sqrt{2-x^2}}{2}-\frac{x^3}{3}]}_0^1$

$A={\sin }^{-1}\frac{1}{\sqrt{2}}+\frac{1}{2}-\frac{1}{3}$

$A={\sin }^{-1}\frac{1}{\sqrt{2}}+\frac{1}{6}$

Hence, this is the required area.