Change the order of integration & evaluate it.
$$ \int_{0}^{2a} \int_{0}^{\sqrt{2ax-x^2}} \phi'(y) \frac{(x^2+y^2)x}{\sqrt{(4a^2x^2-(x^2+y^2)^2})} dxdy $$
I changed the order but could'nt solve it. Is there any 'tricky' variable substitution involved ? I couldn't solve it even with polar co-ordinates due to the presence of $ \phi'(y) $ factor in the integrand? Is there a way to circumvent this?
We use substitutions $x' = x/a$ and $y=y'/a$ and omit primes:
$$ a^2\int_{0}^{2} \int_{0}^{\sqrt{2x-x^2}} \phi'(ay) \frac{(x^2+y^2)x}{\sqrt{4x^2-(x^2+y^2)^2}} dxdy $$
The integration happens in top half-circle with center $(1,0)$ and radius $1$. We swap order of integration: $$ \frac{a^2}2\int_0^1\phi'(ay)dy\int_{1-\sqrt{1-y^2}}^{1+\sqrt{1-y^2}}\frac{x^2+y^2}{\sqrt{4x^2-(x^2+y^2)^2}}(2xdx) $$
We make substitution $r = x^2+y^2-2$, $dr = 2xdx$. $$ \frac{a^2}2\int_0^1\phi'(ay)dy\int_{-2\sqrt{1-y^2}}^{2\sqrt{1-y^2}}\frac{r+2}{\sqrt{4(1-y^2)-r^2}}dr $$
We see, that there is odd ($r$) and even ($2$) component in integral over symmetric region. Odd component will vanish, so we left with: $$ 2a^2\int_0^1\phi'(ay)dy\int_{0}^{2\sqrt{1-y^2}}\frac{dr}{\sqrt{4(1-y^2)-r^2}} $$
Substitution $r=2t\sqrt{1-y^2}$: $$ 2a^2\int_0^1\phi'(ay)dy\int_{0}^{1}\frac{dt}{\sqrt{1-t^2}}=\pi a^2\int_0^1\phi'(ay)dy = \pi a \Big(\phi(a)-\phi(0)\Big) $$