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$$ \int_\Gamma \frac{x \; dy-y \; dx}{x^2+y^2} $$ where $\Gamma$ is a circle: $x^2+y^2-2x-2y+1=0$

By "completing the square" I see that the circle has a radius of $1$ and is moved one point to the right and one up. Partial derivatives of $M(x,y)$ and $N(x,y)$ are: $$ \frac{\partial M}{\partial y}=\frac{y^2-x^2}{(x^2+y^2)^2} $$ $$ \frac{\partial N}{\partial x}=\frac{y^2-x^2}{(x^2+y^2)^2} $$ So, I get $$ \int\int \left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right) \; dA = \int\int dA $$ Am I correct up to this point? How do I continue? What should be the bounds for r if I convert it into polar coordinates?

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2 Answers 2

up vote 1 down vote accepted

Your working seems right.

Did you notice that you got the same terms, and now you are trying to integrate their difference?

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Well I saw it, but doesn't that mean I'm actually looking for area of this circle? –  aarnes Mar 10 '12 at 17:04
    
@aarnes: What would you get if you were trying to find the integral of $1$,then? –  Aryabhata Mar 10 '12 at 17:06
    
Oh, that would be $2\pi$. –  aarnes Mar 10 '12 at 17:10
    
@aarnes: No, you are integrating $1$ over the area so you would get the area, i.e. $\pi r^2$. Here $r=1$. What that says is the volume of a cylinder of height $1$ with the base same as the circle is $\pi$. When you integrate $0$, you are trying to find the volume of cylinder of height $0$... –  Aryabhata Mar 10 '12 at 17:19
    
So if volume of a cylinder is $\pi r^2h$ and h=0 then volume would be zero? –  aarnes Mar 10 '12 at 17:27

The integral $\displaystyle\int\int dA$ is just the area of the circle, so if you're allowed to cite other ways of finding that than doing the integral in polar coordinates, then you've got it.

However, if you subtract something from itself, you get $0$, so if you've got $\displaystyle\int\int 0\;dA$, then the integral is $0$.

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