First, let's get an idea of what the domain being integrated over is for the original function:

D = {(x,y) | 0 <= y <= a, -sqrt(a^2 - y^2) <= x <= 0}

x = -sqrt(a^2 - y^2) --> this is the left half of a circle with radius a

y = 0 -> a --> on the left half of the circle, this is the upper portion

From these, I can conclude that we are dealing with 1/4 of a circle. To be precise, we are dealing with the 1/4 of the circle that is in the 2nd quadrant. From this, I can conclude that the polar domain becomes:

D = {(r,theta) | 0 <= r <= a, pi/2 <= theta <= pi}

Can you do the problem from here?