Question

Using integration find the area of the region $\left\{\right(x,y):x^2+y^2\le 2ax,y^2\ge ax,x\ge 0,y\ge 0\}$

Answer 1

$x^2+y^2-2ax+a^2=a^2$ is a circle, which can also be written as $(x-a{)}^2+y^2=a^2$

Center : $(a,0)$ and radius = $a$

The region given is that inside the circle.

$y^2=ax$ is a parabola with its face opening towards positive infinity.

$y^2\ge ax$ is the area outside the parabola.

The common area is thus inside the circle, outside the parabola and in the first quadrant.

The intersection point is obtained by $x^2+ax-2ax+a^2=a^2$

i.e. $x=a,0$

Area $={\int }_0^a[x^2+y^2-2ax-y^2+ax]dx$

Area $={\int }_0^a[x^2-2ax+ax]dx$

$={}_0^a[\frac{x^3}{3}-\frac{a{x}^2}{2}]$

$=\frac{a^3}{3}-\frac{a^3}{2}$

Taking its modulus, Area$=\frac{a^3}{6}$