Question

The area of the region lying above x-axis, and included between the circle $x^2+y^2=2ax$ & the parabola $y^2=ax,a>0$ is

• A. $8\pi a^2$
• B. $a^2(\frac{\pi }{4}-\frac{2}{3})$
• C. $\frac{16\pi a^2}{9}$
• D. $\pi (\frac{27}{8}+3a^2)$

Answer 1

The correct option is B $a^2(\frac{\pi }{4}-\frac{2}{3})$


$x^2+y^2=2ax$ and $y^2=ax$
$\Rightarrow x^2+ax=2ax$
$\Rightarrow x^2=ax\Rightarrow x=0,x=a$
Point of intesection are: $(0,0),(a,a)$

Required area
$=\frac{\text{Area of circle}}{4}-\underset{0}{\overset{a}{\int }}\sqrt{ax}dx$
$=\frac{1}{4}\left(\pi a^2\right)-\sqrt{a}{\left[\frac{x^{3/2}}{3/2}\right]}_0^a$
$=\frac{\pi a^2}{4}-\frac{2a^2}{3}$
$=a^2(\frac{\pi }{4}-\frac{2}{3})$