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So I'm practicing a few problems and I can't get this one -

$$P(x,y) = e^x \sin(y) \\ Q(x,y) = e^x \cos(y)$$

$C$ is the right hand loop of the graph of the polar equation $r^2 = 4\cos(\theta)$

I want to evaluate:


Now I tried the right hand side of Green's theorem, but it's difficult because $\frac{dp}{dy}$ in polar has a $\cos(r\sin(\theta))$ term in it.

If I just try parametrizing with $x = a\sin(t)$ and $y = a\cos(t)$, where $-\pi/2 \leq t \leq -\pi/2$, then I get another ludicrous integral with $e^{a\cos(t)}\sin(a\sin(t))a\cos(t)\:dt$ as $P \:dx$, which seems insane to solve.

So I think I may be missing some trick in doing this problem. What am I missing and how should I do this?

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up vote 1 down vote accepted

You apply this theorem with $\partial_{x}Q=e^{x}\cos[y],\partial_{y}P=e^{x}\cos[y]$, and the difference is trivially 0. I do not see any trouble in the computation.

Note $$r^{2}=\cos[\theta]$$ can be simplified as $x^{2}+y^{2}=\cos[\arctan[y/x]]=\frac{1}{\sqrt{\frac{y^{2}}{x^{2}}+1}}=\frac{x}{\sqrt{x^{2}+y^{2}}}\leftrightarrow (x^{2}+y^{2})^{3/2}=x$. A graph can be found in here:^{2}%3D\cos[\theta]

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oh wow- that was stupid - thanks! – praks5432 Nov 12 '12 at 0:51

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