Understanding what integrating in polar gives you

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    Integrating Polar
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SUMMARY

This discussion focuses on the integration of the function \(x^2 - y^2\) over a specified region in polar coordinates, specifically from the first quadrant between circles of radius 1 and 2. The integral, calculated using Wolfram Alpha, yields a result of 0, which confuses the user due to the visual representation of the area being a quarter washer in the xy-plane. The key conclusion is that the symmetry of the function across the line \(x = y\) leads to equal positive and negative contributions to the integral, resulting in a net value of 0.

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tnmann10
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I am not understanding integration with polar coordinates, or at least visualizing what is happening. Here's the integral calculated in Wolfram:

http://www.wolframalpha.com/input/?i=integrate+(r^2(cost^2-sint^2))r+drdt+t=(0)..(pi/2)+r=(1)..(2)+

the integral before I changed it to polar was just ∫R(x2-y2)dA where R is the first quadrant region between the circles of radius 1 and 2

mathematically it makes sense that the answer is 0.

but when you draw the picture it is a quarter of a washer in the xy plane. This does not seem like 0 to me. Would someone please explain where my thinking is going wrong?

Thanks

First post:smile:
 
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The function x^2 -y^2 is symmetric in that region when you reflect over the line x=y, with the function negative on one half and positive on the other
 
ok that makes sense, but why the line x=y?
 
tnmann10 said:
ok that makes sense, but why the line x=y?

Think about the nature of ^2 term in terms of the solutions (hint: + and -).
 

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