Question

Find the area of the region lying above X-axis and included between the circle 

x2 + y2 = 2ax and the parabola y2 = ax lying above X-axis

Answer 1

Solution.

$y^2+x^2=2ax,\newline y^2=ax.$

Let be $S$ is finding area.

The required area at a = 1 (for a example):

Find the points of intersection:

$(x-a)^2+y^2-a^2=y^2-ax,$

$x^2-ax=0,$

$x(x-a)=0.$

From here $(0,0)$ and $(a,a)$ are the points of intersection.

$$S=\frac{1}{4}πr^2-\int_0^a \sqrt{ax}dx=\newline =\frac{1}{4}πa^2-\frac{2}{3}a^2=\newline =\frac{3π-8}{12}a^2.$$

Answer. $S=\frac{3π-8}{12}a^2.$