I'm having some trouble understanding this question: what is $\int_C xdy - ydx$, where C is the curve composed of a straight line segment C from $(−2, 0)$ to $(0, 0)$, a straight line segment from $(0, 0)$ to $(0, −2)$, and the part of the circle of radius 2, centered at the origin, traversed counterclockwise starting from $(0,−2)$ and ending at $(−2, 0)$. I tried breaking it down into 2 curves, and integrating separately but I'm not able to get the answer of $6\pi$. Anyone knows what's the correct way to solve this problem?
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By using Green's Theorem, you can turn this into a trivial integral and simultaneously avoid parameterizing the curve. – Travis Willse Jan 16 '15 at 00:33
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Recall Green-Stokes's theorem: if a closed curve $C$ encloses a region $R$, going counterclockwise, then $$\int_C P \,{\rm d}x + Q\,{\rm d}y = \iint_R \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,{\rm d}A,$$ for $\cal C^1$ functions $P$ and $Q$.
So, using Green-Stokes, we have: $$\int_C -y \,{\rm d}x + x \,{\rm d}y = \iint_R \frac{\partial x}{\partial x} - \left(\frac{\partial(-y)}{\partial y}\right) \,{\rm d}A = 2 \iint_R \,{\rm d}A = 2 \,\frac{3}{4}\,\pi \,2^2 = 6\pi,$$ where $R$ is the interior of this pacman (its eye is just me having fun, it is simply connected)
Jokes apart, $\iint_R \,{\rm d}A = \frac{3}{4}\pi r^2$, and no calculus is needed for this part.
Ivo Terek
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Is a requirement of Green's theorem that the velocity field vector must not be conservative? In this case, how do we show that the vector is not conservative? Or does that not matter? – thbcm Jan 16 '15 at 04:42
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