Double Integration in Polar Coordinates

Date: 03/27/2003 at 11:54:58
From: Vij
Subject: Double integration in polar coordinates

Evaluate double integral x-y/x*2+y*2 over x*2+y*2 <= 1

One integral varies from 0 to 2phi; the other varies from 0 to 
2cos(theta). The problem is with the limits.

Date: 03/27/2003 at 12:20:50
From: Doctor Douglas
Subject: Re: Double integration in polar coordinates

Hi, Vij,

You can perform the integration either using the Cartesian coordinates 
(x,y), or using the polar coordinates (r,theta). Depending on the 
problem, one of these approaches may be much easier than the other. I 
believe that what you've said above means that the domain of 
integration is the interior of the unit circle (which is usually 
written as x^2+y^2 < 1).

Assuming that that is indeed the case, then if you perform the
integration in polar coordinates, the limits will be

   r:      0 to 1
   theta:  0 to 2*pi          (or -pi to pi), etc.

Do you see how every point in the circle is covered by this set of 
coordinates?

Now, if you do the integrals in the Cartesian coordinates x and y, the 
limits may look something like the following, depending on which order 
of integration you choose:

   x:      -1 to 1
   y:      -sqrt(1-x^2) to +sqrt(1-x^2)    [the range of y varies
                                            depending on x]

Again, this set of coordinates (x,y) will handle every point in the
unit circle. However, you can see that the limits of integration are 
much simpler in the polar case. This is one of the reasons that the 
integration using polar coordinates may be simpler.  

I hope this helps. Please write back if I misunderstood your
question.

- Doctor Douglas, The Math Forum
  http://mathforum.org/dr.math/