Question

The area of the region lying above x-axis and included between the circle $x^2+y^2=2ax$ and 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

Answer: (B)

$a^2(\frac{\pi }{4}-\frac{2}{3})$

${(x-a)}^2+y^2=a^2$
$\Rightarrow x^2+ax=2ax$
$y^2=ax$
$x^2=ax$
$x^2+ax=2ax$
$(0,0),(a,a)$
Area $=\frac{1}{4}\left(Areaofcircle\right)-{\int }_0^a\sqrt{ax}dx=\frac{1}{4}\left(\pi a^2\right)\sqrt{a}{\left(\frac{x^{3/2}}{3/2}\right)}_0^a=\frac{\pi a^2}{4}-\frac{\pi {2a}^2}{3}=a^2(\frac{\pi }{4}-\frac{2}{3})$