- #1
shaleba
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1. The problem:
I came up to this integration and noticed that it contains singularity at the center (x,y)=(0,0).
[tex] \iint\limits_D \frac{x^2-y^2}{(x^2+y^2)^2} dxdy[/tex]
Note that the integration domain is a unit circle centered at the origin.
I was thinking that the maybe the integral could be evaluated through the polar coordinate. So I used the follows:
[tex] x^2 +y^2 = r^2 [/tex]
[tex] x= r \cdot cos(\theta), y= r \cdot sin(\theta) [/tex]
In that case the integration was transformed to the following integration:
[tex] \int_0^1 \frac{1}{r}dr \cdot \int_0^{2\pi}(cos^2(\theta)-sin^2(\theta))d\theta[/tex]
Evaluation of the first term still includes a singular point with r=0, while the second term gives 0. However, looks like there is a finite value for this integration which is not 0.
Can anybody helps me out on this? Thanks very much.
I came up to this integration and noticed that it contains singularity at the center (x,y)=(0,0).
[tex] \iint\limits_D \frac{x^2-y^2}{(x^2+y^2)^2} dxdy[/tex]
Note that the integration domain is a unit circle centered at the origin.
I was thinking that the maybe the integral could be evaluated through the polar coordinate. So I used the follows:
Homework Equations
[tex] x^2 +y^2 = r^2 [/tex]
[tex] x= r \cdot cos(\theta), y= r \cdot sin(\theta) [/tex]
The Attempt at a Solution
In that case the integration was transformed to the following integration:
[tex] \int_0^1 \frac{1}{r}dr \cdot \int_0^{2\pi}(cos^2(\theta)-sin^2(\theta))d\theta[/tex]
Evaluation of the first term still includes a singular point with r=0, while the second term gives 0. However, looks like there is a finite value for this integration which is not 0.
Can anybody helps me out on this? Thanks very much.
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