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/
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