How do you integrate f(x,y$\displaystyle )=\frac{ln(x^{2}+y^{2})}{x^{2}+y^{2}}\$ over the region $\displaystyle 1{\le}x^{2}+y^{2}{\le}e^{2}$?

Printable View

- Apr 5th 2009, 04:46 PMantmanConverting to a Polar Integral
How do you integrate f(x,y$\displaystyle )=\frac{ln(x^{2}+y^{2})}{x^{2}+y^{2}}\$ over the region $\displaystyle 1{\le}x^{2}+y^{2}{\le}e^{2}$?

- Apr 5th 2009, 04:48 PMMush
- Apr 5th 2009, 04:48 PMmr fantastic
- Apr 5th 2009, 04:50 PMJhevon
following your lead, and with the aid of

**Mush**'s post we convert to polar coordinates. we obtain the integral $\displaystyle \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r^2}{r^2} \cdot r ~dr~d \theta = 2 \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r}r ~dr~d \theta$

you should be able to handle that - Apr 5th 2009, 04:53 PMMush
- Apr 5th 2009, 04:58 PMJhevon
- Apr 5th 2009, 04:59 PMMush
- Apr 5th 2009, 05:17 PMantman
Would is be $\displaystyle 2ln(e^{2})^{2}\pi$?

- Apr 5th 2009, 05:37 PMJhevon
- Apr 5th 2009, 05:57 PMantman
You're right. So then it would simply be $\displaystyle 8\pi$.. correct?

- Apr 5th 2009, 06:06 PMJhevon