Q. 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

KCET KCET 2019Application of Integrals

Solution:

$y^{2}=x \quad\to\left(1\right)$ $x^{2}+y^{2}=2x\quad\to\left(2\right)$ 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) $\left(x-1\right)^{2}+y^{2}=1$ $y^{2}=x$ $\left(x-1\right)^{2}+x=1$ $x^{2}-x=0$ $x\left(x-1\right)=0$ $x=0, x=1$ $area =\int\limits_{0}^{1} \left\{\sqrt{1-\left(x-1\right)^{2}}-\sqrt{x}\right\}dx$ $=-\frac{x^{\frac{3}{2}}}{\frac{3}{2}}]_{0}^{1}+\left[\frac{x-1}{2}\sqrt{1-\left(x-1\right)^{2}}+\frac{1}{2}sin^{-1}\left(x-1\right)\right]_{0}^{1}$ $=-\frac{2}{3}+\left\{0+\frac{\pi}{4}\right\}=-\frac{2}{3}+\frac{\pi}{4}$

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