Question

The area of the region above X-axis included between the parabola $y^2=x$ and the circle $x^2+y^2=2x$ in square units is

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

Answer 1

Answer: (C)

$y^2=x\to \left(1\right)$
$x^2+y^2=2x\to \left(2\right)$
Equation (2) is a circle with centre (1, 0) and radius 1.
Solving (1) and (2), we get the points of intersection (0, 0) and (1, 1)
${\left(x-1\right)}^2+y^2=1$
$y^2=x$
${\left(x-1\right)}^2+x=1$
$x^2-x=0$
$x\left(x-1\right)=0$
$x=0,x=1$
$area=\underset{0}{\overset{1}{\int }}\left\{\sqrt{1-{\left(x-1\right)}^2}-\sqrt{x}\right\}dx$
$=-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}{]}_0^1+{\left[\frac{x-1}{2}\sqrt{1-{\left(x-1\right)}^2}+\frac{1}{2}si{n}^{-1}\left(x-1\right)\right]}_0^1$
$=-\frac{2}{3}+\left\{0+\frac{\pi }{4}\right\}=-\frac{2}{3}+\frac{\pi }{4}$