Question

If the area of the circle $x^2+y^2=2$ is divided into two parts by the parabola $y=x^2$ then the area (in sq. units) of the larger part is

• A. $\frac{3\pi }{2}-\frac{1}{3}$
• B. $6\pi -\frac{4}{3}$
• C. $\frac{4\pi }{3}-\frac{2}{3}$
• D. $4\pi -\frac{1}{4}$

Answer 1

Answer: (A)

Given equation of curves
$x^2+y^2=2$ and $y=x^2$

For point of intersection, on solving the given curves, we get
$<br/>\begin{array}{l}<br/>y^2+y-2=0\Rightarrow y^2+2y-y-2=0\\ <br/>\Rightarrow y(y+2)-1(y+2)=0\\ <br/>\Rightarrow y=1[\because y>0]<br/>\end{array}<br/>$
So, the required area
$<br/>\begin{array}{l}<br/>=\pi +2{\int }_0^1\left(\sqrt{2-y^2}\right)-\sqrt{y})dy\\ <br/>=\pi +2{\left[\frac{y}{2}\sqrt{2-y^2}+\frac{2}{2}{\sin }^{-1}\frac{y}{\sqrt{2}}-\frac{2}{3}{y}^{3/2}\right]}_0^1\\ <br/>=\pi +2\left[\frac{1}{2}+\frac{\pi }{4}-\frac{2}{3}\right]=\frac{3\pi }{2}+1-\frac{4}{3}\\ <br/>=\frac{3\pi }{2}-\frac{1}{3}\text{ sq. units }<br/>\end{array}<br/>$