Just need someone to verify integral, please

I got this one, and tree times I check it, and found few mistakes, and I corrected them all, but now I'm not so sure if i got it all corrected. So if anybody could check it and tell me where do I do it wrong, I'll bee grateful :D

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)