Let us first look at the graph of the region:
As shown, the area can be divided into two sub-regions A1 and A2. By equating graphs, the points of intersection can be found to be x = a and x = a*sqrt(2). Let y1 = sqrt(3a2 - x2), y2 = sqrt( 2ax ) and y3 = x2/2a. Then, the area of A1 is the integral of
y2 - y3 from x = 0 to x = a. And the area of A2 is equivalent to the integral of y1 - y3 from x = a to x = a*sqrt(2).
Now these integrals have to be evaluated and simplified. The first integral can be solved simply by exponents. The second integral is more tricky, it requires a trigonometric substitution, where we let x = a*sqrt(3)*sin(t). The solution, after adding the integrals, is equal to a2*sqrt(2)/3 + 3a2*(sin-1 sqrt((2/3)1/2) - sin-11/sqrt(3))/2.