Recall that ew pass to polar coordinates using $x=r\cos(\theta)$ and $r\sin(\theta)$.

a) Plug the values above into $\frac{y^2}{\sqrt{x^2 + y^2}}$ we get 

$$\frac{r^2\sin^2(\theta)}{\sqrt{r^2\cos^2(\theta) + r^2\sin^2(\theta)}} = \frac{r^2\sin^2(\theta)}{r\sqrt{\cos^2(\theta) + \sin^2(\theta)}} = r\sin^2(\theta).$$

b) The area we are interested is the area between the concentric circles below (this shape is called and annulus)

This are can be described by the inequalities 

$$0<\theta\leq 2\pi \quad \mbox{ and }\quad \var{r1}<r<\var{r2}$$  

in polar coordinates.

c) Note that the order the integration does not matter as boundaries for both variables are constants. The integral thus can be written as 

$$\int_{\var{r1}}^{\var{r2}}\int_{0}^{2\pi} r^2\sin^2(\theta)\, d\theta\,dr.$$

d) 

$$\int_{\var{r1}}^{\var{r2}}\int_{0}^{2\pi} r^2\sin^2(\theta)\, d\theta\,dr = \int_{\var{r1}}^{\var{r2}}r^2 \int_{0}^{2\pi} \sin^2(\theta)\, d\theta\,dr  = -\int_{\var{r1}}^{\var{r2}}r^2 \left[\frac{\sin(2\theta) - 2\theta}{4}\right]_0^{2\pi}\,dr  = \\[3mm ]\int_{\var{r1}}^{\var{r2}}r^2 \pi\,dr = \pi\left[\frac{r^3}{3}\right]_{\var{r1}}^{\var{r2}}  = \pi\left(\frac{\var{r2}^3 - \var{r1}^3}{3}\right) = \var{ans}.$$