multiple integral finding limit integration

Question

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!

Answer 1

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

Answer 2

With polar coordinates we get

$\iint_D \ln(x^2 + y^2)dxdy= \int_{0}^{2 \pi} \int_2^3\ln (r^2)r dr d \theta=4 \pi \int_2^3 r\ln (r) dr$.