# Transforming integral in cartesian coordinates to polar coordinates

• Sep 23rd 2010, 02:46 PM
gralla55
Transforming integral in cartesian coordinates to polar coordinates
x^2 + y^2 is of course a circle, but my book is very unclear on how to switch the coordinate system of an integral. Thanks for any help on this one!
• Sep 23rd 2010, 04:07 PM
Ackbeet
Right. So you have to transform the limits, the integrand, and the differentials. As for the limits, you have to visualize the region enclosed by the limits as given, then try to see how the magnitude/angle description would fit that shape, and what the limits would be.

Your integrand is fairly straight-forward to change over. Just remember that cartesian in terms of polar look like this:

$\displaystyle \displaystyle{x=r \cos(\theta)}$
$\displaystyle \displaystyle{y=r \sin(\theta).}$

Furthermore, polar in terms of cartesian look like this:

$\displaystyle \displaystyle{r=\sqrt{x^{2}+y^{2}}}$
$\displaystyle \displaystyle{\theta=\tan^{-1}(y/x).}$

The differential area is just $\displaystyle \displaystyle{dx\,dy=r\,dr\,d\theta.}$

Can you finish from here?
• Sep 23rd 2010, 04:51 PM
gralla55
Thanks alot! All right, then I guess I just substitute the x and y for r cos 0 and r sin 0. The (root)a^2 - y^2 looks like the upper half of a circle, so perhaps the upper limit on the first integral should be pi. I'm still not sure if that's correct or what to do with the a's though. Any further hints? Thanks again.
• Sep 23rd 2010, 05:39 PM
Ackbeet
Your integrand and differentials are correct (you can simplify the integrand, if you use a certain well-known trig identity). But your limits aren't correct yet. Remember how $\displaystyle r$ and $\displaystyle \theta$ work, right? Your angle is measured from where? And the distance $\displaystyle r$ is measured from where? How does this correspond to your semicircle?
• Sep 23rd 2010, 06:32 PM
gralla55
Thank you so much for helping out! Could this be correct?
• Sep 23rd 2010, 06:38 PM
Ackbeet
Your $\displaystyle r$ limits are correct. Your $\displaystyle \theta$ limits are incorrect, but you're on the right track. Think about the orientation of the semicircle relative to where you measure $\displaystyle \theta$ from.

Also, you didn't quite simplify the integrand correctly. Again, you're on the right track, though.
• Sep 23rd 2010, 06:42 PM
gralla55
Perhaps this is it? Thanks again!
• Sep 24th 2010, 02:37 AM
Ackbeet
Excellent! Your angle limits are correct now. Don't they make more sense? However, I must confess to misleading you about the $\displaystyle r$ limits. You don't want to integrate from $\displaystyle 0$ to $\displaystyle r$, but instead you want to integrate from $\displaystyle 0$ to what?

Your integrand still is not correct, though it's oh-so-close. Look at this expression (the exponent):

$\displaystyle (r\cos(\theta))^{2}+(r\sin(\theta))^{2}=r^{2}\cos^ {2}(\theta)+r^{2}\sin^{2}(\theta)=r^{2}(\cos^{2}(\ theta)+\sin^{2}(\theta)).$

How does this simplify?

Comment: learn LaTeX! That's how I can post my questions and have the math appear right alongside text. Unfortunately, when I go into the Advanced Editing mode to do some serious math typing, I can't see the pictures you've scanned in. So I have to have an extra tab open in order to navigate to the problem so I can see your scanned images. It also takes longer to scan than to type LaTeX, once you've learned it a bit. Double-click expressions that are in LaTeX in order to see how to do things. For example: here's the code for the last integral you posted:

Code:

\displaystyle{\int_{-\pi/2}^{\pi/2}\int_{0}^{r}e^{2r^{2}}r\,dr\,d\theta},
which produces

$\displaystyle \displaystyle{\int_{-\pi/2}^{\pi/2}\int_{0}^{r}e^{2r^{2}}r\,dr\,d\theta}$ when enclosed with the math tags (available in the Advanced Editing mode).
• Sep 24th 2010, 02:43 AM
Prove It
Actually I think since $\displaystyle x^2 + y^2 = r^2$, the integrand should become

$\displaystyle r\,e^{r^2}\,dr\,d\theta$.

Also if $\displaystyle x = \sqrt{a^2 - y^2}$ is your upper limit, then $\displaystyle x^2 + y^2 = a^2$, which is a circle of radius $\displaystyle a$.

Therefore we are integrating from $\displaystyle 0 \leq r \leq a$.
• Sep 24th 2010, 04:20 AM
HallsofIvy
Quote:

Originally Posted by Prove It
Actually I think since $\displaystyle x^2 + y^2 = r^2$, the integrand should become

$\displaystyle r\,e^{r^2}\,dr\,d\theta$.

Also if $\displaystyle x = \sqrt{a^2 - y^2}$ is your upper limit, then $\displaystyle x^2 + y^2 = a^2$, which is a circle of radius $\displaystyle a$.

Therefore we are integrating from $\displaystyle 0 \leq r \leq a$.

No, $\displaystyle x^2+ y^2= a^2$ is a circle but $\displaystyle x= \sqrt{a^2- y^2}$ is a semi-circle. x is positive so it is the semi-circle on the right of the y-axis. That's why the limits given before, $\displaystyle -\pi/2$ and $\displaystyle \pi/2$, are correct.
• Sep 24th 2010, 04:33 AM
Prove It
Quote:

Originally Posted by HallsofIvy
No, $\displaystyle x^2+ y^2= a^2$ is a circle but $\displaystyle x= \sqrt{a^2- y^2}$ is a semi-circle. x is positive so it is the semi-circle on the right of the y-axis. That's why the limits given before, $\displaystyle -\pi/2$ and $\displaystyle \pi/2$, are correct.

Correct, but I didn't say anything about the limits of $\displaystyle \theta$, I was just trying to fix a mistake in the integrand and help simplify the limits on $\displaystyle r$.

Anyway, I think everything that was needed to be said has been said, and the integral is

$\displaystyle \displaystyle{\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\int_0^a{r\,e^{r^2} \,dr}\,d\theta}}$