Converting to a Polar Integral

• Apr 5th 2009, 04:46 PM
antman
Converting 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 PM
Mush
Quote:

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

$\displaystyle x = r \cos(\theta)$

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

Hence $\displaystyle x^2 + y^2 = r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = r^2(\sin^2(\theta) + \cos^2(\theta)) = r^2$
• Apr 5th 2009, 04:48 PM
mr fantastic
Quote:

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

Switch to polar coordinates.
• Apr 5th 2009, 04:50 PM
Jhevon
Quote:

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

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 PM
Mush
Quote:

Originally Posted by Jhevon
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

Should still be $\displaystyle r^2$ on the bottom?
• Apr 5th 2009, 04:58 PM
Jhevon
Quote:

Originally Posted by Mush
Should still be $\displaystyle r^2$ on the bottom?

no, i don't think so. the $\displaystyle r$ from the $\displaystyle r~dr ~d \theta$ cancels one of the $\displaystyle r$'s at the bottom
• Apr 5th 2009, 04:59 PM
Mush
Quote:

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

Oh, indeed.
• Apr 5th 2009, 05:17 PM
antman
Would is be $\displaystyle 2ln(e^{2})^{2}\pi$?
• Apr 5th 2009, 05:37 PM
Jhevon
Quote:

Originally Posted by antman
Would is be $\displaystyle 2ln(e^{2})^{2}\pi$?

i suppose you mean $\displaystyle 2 (\ln e^2)^2 \pi$. then yes. but you left the answer in a "clumsy" form. of course, $\displaystyle \ln e^2 = 2$...
• Apr 5th 2009, 05:57 PM
antman
You're right. So then it would simply be $\displaystyle 8\pi$.. correct?
• Apr 5th 2009, 06:06 PM
Jhevon
Quote:

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

yes