In polar coordinates, the inequality changes to
[tex]x^2+y^2\le4x\implies r^2\le4r\cos\theta\implies r\le4\cos\theta[/tex]
which is a circle of radius 2 and centered at (2, 0). The set D is then
[tex]D=\left\{(r,\theta)\mid0\le r\le4\cos\theta\land0\le\theta\le\pi\right\}[/tex]
The integral is then
[tex]\displaystyle\iint_D\frac{y^2}{x^2+y^2}\,\mathrm dx\,\mathrm dy=\int_0^\pi\int_0^{4\cos\theta}\frac{r^2\sin^2\theta}{r^2}r\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\int_0^\pi\int_0^{4\cos\theta}r\sin^2\theta\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\frac12\int_0^\pi((4\cos\theta)^2-0^2)\sin^2\theta\,\mathrm d\theta[/tex]
[tex]=\displaystyle8\int_0^\pi\cos^2\theta\sin^2\theta\,\mathrm d\theta[/tex]
There are several ways to compute the remaining integral; one would be to invoke the double-angle formula,
[tex]\sin(2\theta)=2\sin\theta\cos\theta[/tex]
so that the integral is
[tex]=\displaystyle8\int_0^\pi\frac{\sin^2(2\theta)}4\,\mathrm d\theta[/tex]
[tex]=\displaystyle2\int_0^\pi\sin^2(2\theta)\,\mathrm d\theta[/tex]
Then invoke another double-angle formula,
[tex]\sin^2\theta=\dfrac{1-\cos(2\theta)}2[/tex]
to change the integral to
[tex]=\displaystyle\int_0^\pi1-\cos(4\theta)\,\mathrm d\theta[/tex]
[tex]=(\pi-\cos(4\pi))-(0-\cos0)=\boxed{\pi}[/tex]
