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Consider the vector field:$$\vec G = \left(\frac y{x^2+y^2}, \frac {-x}{x^2+y^2}\right)$$ compute $\int_\Gamma \vec G$ where $\Gamma$ is the proportion of a parabola $y=a(x-1)^2$ from (1,0) to (2,a). Hint: use arctan substitution.

I've tried to parametrize this a million different ways, but I can find anything that simplifies into something remotely integrable. Any ideas?

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I have no idea what this means. And yes, I know line integrals. –  Ron Gordon Feb 27 '13 at 13:00

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Let $$(x,y)\to \phi(x,y):=\arctan{y\over x}\in\ \biggl]{-{\pi\over2}},{\pi\over2}\biggr[\ $$ denote the polar angle in the right half plane. Then your $\vec G$ is nothing else but $-\nabla\phi$ (check it!). Therefore $$\int_\Gamma\vec G\cdot d{\bf z}=-\int_\Gamma \nabla\phi\cdot d{\bf z}=-\bigl(\phi(2,a)-\phi(1,0)\bigr)=-\arctan{a\over2}\ .$$

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