Green's Theorem and Area

Show that the area bounded by a simple closed curve is given by \(\frac{1}{2} \oint{xdy-ydx} \).

Solution

We use Green's Theorem, and let P=yP=-y , Q=xQ= x. Thus

Pdx+Qdy=(QxPy)dxdy\oint { Pdx+Qdy } =\iint { \left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) dxdy }

ydx+xdy=((x)x(y)y)dxdy\oint { -ydx+xdy } =\iint { \left(\frac{\partial (x)}{\partial x}-\frac{\partial (-y)}{\partial y} \right)dxdy }

xdyydx=2dxdy\oint { xdy-ydx } = 2\iint {dxdy }

xdyydx=2A\oint { xdy-ydx } = 2\iint {A }

A=12xdyydxA = \frac{1}{2}\oint { xdy-ydx }

Check out my other notes at Proof, Disproof, and Derivation

#Calculus #VectorCalculus

Note by Steven Zheng
6 years, 6 months ago

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Comments

This very question was asked in my semester examination which I could happily answer using the process described here.

Nishant Sharma - 6 years, 6 months ago
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