The area of the region included between the curves `x^2+y^2=a^2 and sqrt|x|+sqrt|y|=sqrta(a > 0)` is

Explanation

Given curves are ` sqrt(|x|) + sqrt(|y|) = sqrta " " … (i)`
and `x^(2) + y^(2) = a^(2) " "…. (ii)`
Curve (ii) is a circle .
Now we have to draw the graph of curve (i)
(1) Curve (i) cuts x-axis at (-a,0) and (a,0) and y-axis at (0,-a) and (0,a)
(2) Curve (i) is symmetrical about x-axis as well as y-axis
(3) In first quadrant from (i) we have
`sqrtx + sqrty = sqrta implies (dy)/(dx) = - (sqrty)/(sqrtx) lt 0`
`implies ` y is decreasing
(4) `(dy)/(dx) = - ((sqrta - sqrtx)/(sqrtx)) = 1 - (sqrta)/(sqrtx)`
`therefore (d^(2)y)/(dx^(2)) = (1)/(2) sqrta . x^(-(3)/(2))= (sqrta)/(2.x^((3)/(2))) gt 0`
`therefore` Curve is convex downward .
Now , required area = 4[shaded area in the first quadrant]
=` 4[(pi a^(2))/(4) - underset(0)overset(a)(int) (sqrta - sqrtx)^(2) dx ] = 4[(pi a^(2))/(4) - underset(0)overset(a)(int)(a + x - 2sqrta sqrtx) dx`]
`= 4 {(pia^(2))/(4) - [ax + (x^(2))/(2) - (4)/(3) sqrta x^((3)/(2)) ]_(0)^(a)}`
`= (pi - (2)/(3)) a^(2)` sq. units