Question

Find area bounded by $x^2+y^2\le 2ax,y^2\ge ax$, $x\ge 0$, $y\le 0$

Answer 1

We have to find shaded area in the $4^{th}$ quadrant that is same as area in $1^{st}$ quadrant.

$A={\int }_0^a\sqrt{2ax-x^2}-\sqrt{ax}dx$

A= Area of quater circle - ${\int }_0^a\sqrt{ax}dx$

$A=\frac{\pi a^2}{4}-\sqrt{a}{\left[\frac{x\frac{3}{2}}{\frac{3}{2}}\right]}_0^a$

$A=\frac{\pi a^2}{4}-\sqrt{a}\left(\frac{2}{3}{a}^{\frac{3}{2}}\right)$

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