Any easy way to evaluate this double integral? $$ \iint_S \frac{x^2+xy+3y^2}{2x^2+xy+2y^2}dxdy$$
where 
$$S=\left\{(x,y)\vert x^2+y^2\leq1 \right\}$$
I tried 
$$x=r\cos \theta, \quad y=r\sin\theta$$ substitution but it still not clear.
Is polar-coordinates the only solution for this?
Or how can I solve this easier?
 A: 
I tried 
  $$x=r\cos \theta, \quad y=r\sin\theta$$ substitution but it still not clear.

This is a good idea. After simplifying (see below), you get:
$$\frac{x^2+xy+3y^2}{2x^2+xy+2y^2} \xrightarrow{\color{red}{(\star)}} 1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}$$
Don't forget the extra $\color{blue}{r}$ when converting to polar coordinates and you can split the variables:
$$\begin{align}
\iint_S \frac{x^2+xy+3y^2}{2x^2+xy+2y^2} \,\mbox{d}x\,\mbox{d}y 
& = \iint_S \left(1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}\right) \color{blue}{r} \,\mbox{d}r\,\mbox{d}\theta \\[6pt]
& = \underbrace{\int_0^1 r \,\mbox{d}r}_{\frac{1}{2}} \; \underbrace{\int_0^{2\pi} \left(1 - \frac{2 \cos(2 \theta)}{4 + \sin(2 \theta)}\right) \,\mbox{d}\theta}_{2\pi \; + \; 0}
\end{align}$$
$$$$

$$\begin{align}
\color{red}{(\star)} \quad\frac{x^2+xy+3y^2}{2x^2+xy+2y^2}
& = \frac{2x^2+xy+2y^2+y^2-x^2}{2x^2+xy+2y^2} \\[5pt]
& = 1+\frac{y^2-x^2}{2\left(x^2+y^2\right)+xy} \\[5pt]
& = 1+\frac{r^2\sin^2\theta-r^2\cos^2\theta}{2r^2+r^2\cos\theta\sin\theta} \\[5pt]
& = 1-\frac{2\cos(2\theta)}{4+\sin(2\theta)}
\end{align}$$
