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Let $\gamma$ be a smooth Jordan curve in $\mathbb{R}^2\setminus\{(0,0)\}$ from $(1,0)$ to $(1,0)$, winding about the origin once in the clockwise direction. Compute:

$\int_{\gamma}\frac{y}{x^2+y^2}dx-\frac{x}{x^2+y^2}dy$

I figured this problem suggested to use the Fundamental Theorem of Line Integrals, so if $\gamma(t)=\langle x(t),y(t)\rangle$, then it seems natural to want to write

\begin{align} &\int_{\gamma}\frac{y}{x^2+y^2}dx-\frac{x}{x^2+y^2}dy\\ &\qquad=\int_{0}^{1}\frac{y(t)x'(t)}{x^2(t)+y^2(t)}+\frac{x(t)y'(t)}{y^2(t)+x^2(t)}\ dt\\ &\qquad=\int_{0}^{1}\nabla f(x(t),y(t))\cdot \langle x'(t),y'(t)\rangle dt \end{align}

where $f(x,y)=\frac{-1}{2(x^2+y^2)}$. Then the fundamental theorem would imply that the line integral evaluates to $0$, since the gradient is obviously continuous on $\gamma$.

But - the last inequality is wrong, insofar as the order is wrong. But I thought it might be more useful to show you where I'm stuck, and demonstrate I thought about this before posting.

Sincerely,

Someone who hasn't had to take a line integral in five years.

  • 0
    Does not seems that $\nabla f$ corresponds to the given differential form.2012-09-11

2 Answers 2

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The fundamental/divergence/Green's/Stokes' theorem does not apply if your function has singularity inside the domain. In this case, there is a singularity at $(0, 0)$, and your curve does go around the singularity.

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    It is homotopy invariant if your homotopy does not cross the origin. (You already proved that your vector field is conservative except at the origin. )2012-09-11
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Perhaps you should write your function in the complex form $f(z) dz$, where $dz=dx+idy$. Then use the residue theorem, as this function looks like it's holomorphic outside of zero, so you only have to calculate its residue in 0.