# Thread: Converting to a Polar Integral

1. ## Converting to a Polar Integral

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

2. Originally Posted by antman
How do you integrate f(x,y $)=\frac{ln(x^{2}+y{2})}{x^{2}+y{2}}\$ over the region $1{\le}x^{2}+y^{2}{\le}e^{2}$?
$x = r \cos(\theta)$

$y = r \sin(\theta)$.

Hence $x^2 + y^2 = r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2(\sin^2(\theta) + \cos^2(\theta)) = r^2$

3. Originally Posted by antman
How do you integrate f(x,y $)=\frac{ln(x^{2}+y{2})}{x^{2}+y{2}}\$ over the region $1{\le}x^{2}+y^{2}{\le}e^{2}$?
Switch to polar coordinates.

4. Originally Posted by antman
How do you integrate f(x,y $)=\frac{ln(x^{2}+y{2})}{x^{2}+y{2}}\$ over the region $1{\le}x^{2}+y^{2}{\le}e^{2}$?
following your lead, and with the aid of Mush's post we convert to polar coordinates. we obtain the integral $\int_0^{2 \pi} \int_1^{e^2} \frac {\ln r^2}{r^2} \cdot r ~dr~d \theta = 2 \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r}r ~dr~d \theta$

you should be able to handle that

5. Originally Posted by Jhevon
following your lead, and with the aid of Mush's post we convert to polar coordinates. we obtain the integral $\int_0^{2 \pi} \int_1^{e^2} \frac {\ln r^2}{r^2} \cdot r ~dr~d \theta = 2 \int_0^{2 \pi} \int_1^{e^2} \frac {\ln r}r ~dr~d \theta$

you should be able to handle that
Should still be $r^2$ on the bottom?

6. Originally Posted by Mush
Should still be $r^2$ on the bottom?
no, i don't think so. the $r$ from the $r~dr ~d \theta$ cancels one of the $r$'s at the bottom

7. Originally Posted by Jhevon
no, i don't think so. the $r$ from the $r~dr ~d \theta$ cancels one of the $r$'s at the bottom
Oh, indeed.

8. Would is be $2ln(e^{2})^{2}\pi$?

9. Originally Posted by antman
Would is be $2ln(e^{2})^{2}\pi$?
i suppose you mean $2 (\ln e^2)^2 \pi$. then yes. but you left the answer in a "clumsy" form. of course, $\ln e^2 = 2$...

10. You're right. So then it would simply be $8\pi$.. correct?

11. Originally Posted by antman
You're right. So then it would simply be $8\pi$.. correct?
yes