# Integration in Polar Coordinates (double integrals)

1. Oct 18, 2009

### CloudTenshi

1. The problem statement, all variables and given/known data

We define the improper integral (over the entire plane R^2) I as a double integral [-inf,inf]x[-inf,inf] of e^-(x^2+y^2)dA as equal to the lim as a-> inf of the double integral under Da of e^-(x^2+y^2)dA where Da is the disk with the radius a and center at the origin.

Show that the I (the original double integral) equals pi.

Sorry, this is rather difficult to type.

2. Relevant equations

Other than what I've said, I know that the equation of a disk is x^2 + y^2 = r^2

3. The attempt at a solution

With what I was given, I was able to substitute (x^2 + y^2) with r^2 resulting in the double integral of e^-(r^2). I also found the domain of r to be between 0 and a, and guessed that theta would range from 0 to 2pi. The problem I run into here is how to integrate the function. I first integrated the function with respect to theta because I thought that would be easier, but if I do that, I get the double integral [0,a]x[0,2pi] which doesn't seem right because of the a. Even if it was right, I don't know how to integrate the function of e^-(r^2). Working backwards, I deduced that I need to get an r into the equation somehow so I could integrate the function, but I don't know where to pull the other r from.

2. Oct 18, 2009

### Staff: Mentor

When you change from a rectangular (or Cartesian) iterated integral to a polar iterated integral, your differentials change from dx dy or dy dx to r dr d$\theta$. Your integral will look like this:
$$\int_{\theta = 0}^{2\pi} \lim_{a \rightarrow \infty}\int_{r = 0}^{a} e^{-r^2} r dr d\theta$$

Click on the integral to see the LaTeX script I used.

3. Oct 18, 2009

### CloudTenshi

Thanks a bunch. I realized my mistake with the r shortly after, but I had made a careless mistake because e^0 doesn't equal 0, but rather 1. I didn't realize that till I retried the problem with the integral you provided. Where the limit sign actually is doesn't really matter right? I could have it outside both integrals and the answer I get should be the same as if it was inside the integral right? By the way, the answer I got was pi.

But just wondering, would the value of $$\int_{x = -infinity}^{infinity} e^{-x^2} dx$$ be the sqrt of pi. I assumed it would be so, but I don't really get the concept behind it. I tried doing the integral in terms of polar coordinates but by substituted x with rcos$$theta[tex] I still get a function I have no idea how to integrate. Are polar coordinates necessary? Last edited: Oct 18, 2009 4. Oct 19, 2009 ### Mark44 ### Staff: Mentor Yes, the polar form is necessary, because e^(-x^2) does not have an antiderivative that is expressible in terms of elementary functions. Changing to polar form allows you to carry out a "trick" to let you evaluate the integral. [tex]\int_{y = 0}^{\infty}\int_{x = 0}^{\infty} e^{-x^2 - y^2} dx dy~=~\int_{\theta = 0}^{2\pi} \lim_{a \rightarrow \infty}\int_{r = 0}^{a} e^{-r^2} r dr d\theta$$

Carry out the inner integration (a simple substitution will do), then take the limit, and finally carry out the outer integration. What you end up with will be equal to the value of the iterated integral in Cartesian form. Because the x and y variables are uncoupled, the iterated integral is equal to $$\left(\int_{x = 0}^{\infty} e^{-x^2} dx\right)\left(\int_{y = 0}^{\infty} e^{-y^2} dy\right)$$