integral is :

\(\displaystyle \int\int \ln{x^2+y^2}\,dx\,dy\)

by the region D defined by:

\(\displaystyle x^2+y^2=e^2\)

\(\displaystyle x^2+y^2=e^4\)

so I used :

\(\displaystyle x=\rho\cos\varphi\)

\(\displaystyle y=\rho\sin\varphi\)

\(\displaystyle J=\rho\)

and I got that my limits now are for \(\displaystyle \rho\) from \(\displaystyle e\) to \(\displaystyle e^2\), and for \(\displaystyle \varphi\) from \(\displaystyle 0\) to \(\displaystyle 2\pi\).

now my integral is:

\(\displaystyle I=\displaystyle\int_0^{2\pi} \,d\varphi \int_e ^{e^2} \ln(\rho^2) \rho \,d\rho = 2\int_0^{2\pi} \,d\varphi \int_e ^{e^2} \ln(\rho) \rho \,d\rho\)

so I now do one first...

\(\displaystyle I_1= \displaystyle \int_e ^{e^2} \ln(\rho)\rho\,d\rho\)

\(\displaystyle u=\ln(\rho), \,du=\frac{1}{\rho} \,d\rho\)

\(\displaystyle \,dv=\rho\,d\rho , v=\frac{\rho^2}{2}\)

so I get that one like this...

\(\displaystyle \displaystyle I_1=\frac{\rho^2}{2}\ln(\rho) \mid _e ^{e^2} - \int_e ^{e^2} \frac {\rho^2}{2} \frac{1}{\rho} \,d\rho\)

\(\displaystyle \displaystyle I_1=e^4-\frac{e^2}{2}-\frac{1}{4} (e^4-e^2)=\frac{3e^4}{4}-\frac{e^2}{4}\)

so my complete integral is :

\(\displaystyle \displaystyle I=2\int_0 ^{2\pi} [\frac{3e^4}{4}-\frac{e^2}{4}] \,d\varphi= \frac {3e^4-e^2}{2} 2\pi = \pi (3e^4-e^2)\)

(Thinking)(Worried)(Thinking)