Find the area of the region `{(x,y): x^2 +y^2 le 2ax` , `y^2 le ax` , `x le 0` , `y le 0 }`

Explanation

Clearly we have to find the area of the region lying in the first quadrant `(x ge 0, y ge 0 )` included between the circle `x^(2)+y^(2)=2ax` and the parabola `y^2 =ax`
Thus , the equations of the given curves are
`x^2+y^2 =2ax…(i)`
and `y^2 =ax` ...(ii)
Now , clearly `x^2+y^2=2ax ` is a circle with its center `B(a,0)` and radius `=a` units
And `y^2=ax ` is parabola with `O(0,0)` as its vertex and the x-axis as its axis. We can draw its figure as shown .
Their points of intersection may be obtained by solving `(i)` and `(ii)` and keeping in view that `x ge 0 and y gt 0`
Using `(ii)` in `(i)` ,we get
`x^2-ax =0 rArr x(x -a) =0`
`rArr x=0 or x =a`
Now `(x=0 rArr y=0 )` and `(x=a rArr y=a)`
Thus, two curves intesect at O(0,0) and A(a,a)
`therefore " required area " =underset(0)overset(a)int sqrt(2ax -x^2)dx - underset(0)overset(a)int sqrt(ax)dx`
`=underset(0)overset(a)intsqrt(a^2-(x-a)^2)dx -sqrt(a).underset(0)overset(a)int sqrt(x)dx `
`=[((x-a)sqrt(a^2-(x-a)^2))/(2)+a^2/2 sin^(-1)""((x-a)/a)]_0^a-sqrt(a)[2/3x^(3//2)]_0^a`
`={a^2/2sin^(-1)(0)-a^2/2 sin^(-1)(-1)-2/3a^2}`
`=((pia^2)/4-2/3a^2)`sq units
Hence, the required area is `=((pia^2)/4-2/3a^2)` sq units .