multiple integral finding limit integration question : 
$\iint_D \ln(x^2 + y^2)dxdy$ ,
 domain : $4\le x^2+y^2 \le 9$ 
im having hard times to find the limit for integration  because it is in polar coordinates, i got that $2 \le r\le 3$ but i dont know how to find the $\theta$ limit. 
and also i dont know the graph actually, is it the same as logarithm graph? is it better to draw the graph? is there another way beside drawing the graph? thanks!

You make the change of variables: 
$$x=r\cos{\theta},y=r\sin{\theta}.$$
Hence: 
$$4\le x^2+y^2 \le 9 \Rightarrow 2^2\le r^2\le 3^2 \Rightarrow 2\le r\le 3.$$
The domain is the region between the cocentric circles with the center at the origin and the radii $2$ and $3$, therefore:
$$0\le \theta \le 2\pi.$$
Note: If the domain was $4\le x^2+y^2\le 9, x\ge 0, y\ge 0$, then it would be the region between the concentric circles with the center at the origin and the radii $2$ and $3$ that lies in the first quadrant, implying $0\le \theta \le \frac{\pi}{2}$.

The formula of double integration in polar coordinates is:
  $$\iint_{D} f(x,y)dxdy = \int_{\theta_1}^{\theta_2} \int_{r_1}^{r_2}f(r\cos{\theta},r\sin{\theta})\cdot Jdrd{\theta},$$
  where $J$ is the Jacobian:
  $$J=\begin{vmatrix}
        x_r & x_{\theta} \\
        y_r & y_{\theta} \\
        \end{vmatrix} = \begin{vmatrix}
        \cos{\theta} & -r\sin{\theta} \\
        \sin{\theta} & r\cos{\theta} \\
        \end{vmatrix} = r.$$

Hence the integral:
$$\iint_{4\le x^2+y^2\le 9} \ln(x^2+y^2)dxdy =\int_0^{2\pi} \int_2^3 \ln{r^2}\cdot rdrd{\theta}=$$
$$\left(\int_0^{2\pi}d{\theta}\right) \left(\int_2^3 \underbrace{\ln{r}}_{u} \cdot \underbrace{2rdr}_{dv}\right)=(2\pi)\left(\ln r \cdot r^2 \bigg{|}_2^3-\int_2^3 r^2\cdot \frac1r dr\right)=$$
$$(2\pi)\left(9\ln 3-4\ln 2-\frac52 \right)\approx 28.996.$$
