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Problem of the Week #35 - January 28th, 2013

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Chris L T521

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Jan 26, 2012
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Here's this week's problem.

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Problem: Show that Green's theorem for regions in $\mathbb{R}^2$ with boundary consisting of a disjoint union of one or more circles is a consequence of Stoke's theorem.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below:

Pf: Let $C_1\subset\mathbb{R}$ be a circle with counterclockwise (positive) orientation and let $C_i\subset\text{Int}\,C_1$ for $i\geq 2$ be circles with clockwise (negative) orientation. For some region in $\mathbb{R}^2$ who's boundary is (homotopic to) a circle $C$, we know by Green's theorem that
\[\int_C f\,dx+g\,dy = \iint\limits_{R}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,dA.\]
Now, if we take our region $M=M_1\backslash M_2\backslash M_3\backslash\cdots$ where $M_i=\text{Int}\,C_i$ for each $i\geq 1$, then it follows that $\displaystyle\partial M=\bigsqcup_i C_i$ (here, the box cup represents a disjoint union).

By Stoke's theorem, if we take $\omega=f\,dx+g\,dy$, we have that
\[\begin{aligned}\int_{\partial M}\omega &= \int_{\bigsqcup_i C_i} f\,dx+g\,dy\\ &= \sum_i \int_{C_i}f\,dx+g\,dy \\ &= \sum_i\int_{M_i}\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,dA\\ &= \int_M\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,dA\\ &= \int_M \,d\omega.\end{aligned}\]

Thus, Green's theorem for a region who's boundary is a disjoint union of one or more circles is a consequence of Stoke's theorem. Q.E.D.
 
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