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

• A

  \( \frac{3}{2}-\frac{\Pi}{4} \)
• Correct answer
  B

  \( \frac{I}{4}-\frac{2}{3} \)
• C

  \( \frac{I}{4}-\frac{3}{2} \)
• D

  \( \frac{2}{3}-\frac{\Pi}{4} \)

Solution

(B)

\[ \begin{array}{l} y^{2}=x \rightarrow(1) \\ x^{2}+y^{2}=2 x \rightarrow(2) \end{array} \]
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 \( (1,1) \)
\( (x-1)^{2}+y^{2}=1 \)
\( y^{2}=x \)
\( (x-1)^{2}+x=1 \)
\( x^{2}-x=0 \)
\( x(x-1)=0 \)
\( x=0, x=1 \)
area \( =\int_{0}^{1}\left\{\sqrt{1-(x-1)^{2}}-\sqrt{x}\right\} d x \)
\( =-\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{1}+ \)
\( \left[\frac{x-1}{2} \sqrt{1-(x-1)^{2}}+\frac{1}{2} \sin ^{-1}(x-1)\right]^{1} \)

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