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

• Correct answer
  A

  \(\frac{3 \pi}{2}-\frac{1}{3}\)
• B

  \(6 \pi-\frac{4}{3}\)
• C

  \(\frac{4 \pi}{3}-\frac{2}{3}\)
• D

  \(4 \pi-\frac{1}{4}\)

Solution

Given equation of curves \(x^2+y^2=2 \text { and } y=x^2\)
For point of intersection, on solving the given curves, we get \(\begin{aligned} & y^2+y-2=0 \Rightarrow y^2+2 y-y-2=0 \\ \Rightarrow & y(y+2)-1(y+2) =0 \\ \Rightarrow & y =1 \qquad {[\because y > 0] } \end{aligned}\) So, the required area \(\begin{aligned} & =\pi+2 \int_0^1\left(\sqrt{\left.2-y^2\right)}-\sqrt{y}\right) d y \\ & =\pi+2\left[\frac{y}{2} \sqrt{2-y^2}+\frac{2}{2} \sin ^{-1} \frac{y}{\sqrt{2}}-\frac{2}{3} y^{3 / 2}\right]_0^1 \\ & =\pi+2\left[\frac{1}{2}+\frac{\pi}{4}-\frac{2}{3}\right]=\frac{3 \pi}{2}+1-\frac{4}{3} \end{aligned}\) \(=\frac{3 \pi}{2}-\frac{1}{3} \text { sq. units }\) Hence, option (1) is correct.

Hence, the correct answer is option: 'A'.