Question

The area of the portion of the circle $x^2+y^2=1$, which lies inside the parabola $y^2=1-x$ is

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

Answer 1

The correct option is C $\frac{\pi }{2}+\frac{4}{3}$

The given curve intersect at $(0,1)$ and $(0,-1)$
Therefore, the required area $={\left[\frac{1}{2}\pi r^2\right]}_0^1+2{\int }_0^1(1-y^2)dy$
$=\frac{1}{2}\pi {\left(1\right)}^2+2{[y-\frac{y^3}{3}]}_0^1=\frac{\pi }{2}+\frac{4}{3}$