Find the value of the double integral: $\int\limits_0^{2a}\int\limits_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)xdxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$ Evaluate $$\int_0^{2a}\int_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)xdxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$$
Here I changed the order of integration but after that I am not able to think of the relevant substitution to proceed with the integration.
I have tried to substitute it with polar co-ordinates but didn't succeed in solving that either.
I have checked the solution provided here Evaluate the integral $\int_0^{2a}\int_0^\sqrt{2ax-x^2}\frac{\phi'(y)(x^2+y^2)x dxdy}{\sqrt{4a^2x^2-(x^2+y^2)^2}}$ , but this solution is very lengthy.
I am looking for a shorter way of solving this question which is more obvious to hit during exam and also considers the space constraint.
If someone could suggest what substitutions to make to solve it in lesser steps.

Changing the order and taking $x^2+y^2=t,dt=2xdx$ gives,$$\frac12\int_0^a\phi'(y)\int_{2a^2-2a\sqrt{a^2-y^2}}^{2a^2+2a\sqrt{a^2-y^2}}\frac{t~dt}{\sqrt{4a^2(t-y^2)-t^2}}~dy$$Now$$\frac t{\sqrt{4a^2(t-y^2)-t^2}}=\frac{-1}2\left[\frac{4a^2-2t}{\sqrt{4a^2(t-y^2)-t^2}}\right]+\frac{2a^2}{\sqrt{4a^2(a^2-y^2)-(t-2a^2)^2}}$$and so$$\int\frac{t~dt}{\sqrt{4a^2(t-y^2)-t^2}}=-\sqrt{(4a^2-t)t-4a^2y^2}+2a^2\sin^{-1}\left(\frac{t-2a^2}{2a\sqrt{a^2-y^2}}\right)$$Applying the limits should give$$\frac12\int_0^a\phi'(y)[0+2a^2\pi]dy=\pi a^2[\phi(a)-\phi(0)]$$
