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I've come across a Green's Theorem proof that has me perplexed.

Using the area formula: $$A = \frac{1}{2}\int_C xdy - ydx $$ Prove that:$$A = \frac{1}{2}\int_a^b r^2d\theta$$for a region in polar coordinates.

I assume a parametrisation is needed, but I'm not sure where to start due to the change in variables. My first thoughts are to change coordinates to $x=rcos\theta$ and $y=rsin\theta$. I also have assumed the $r^2$ is a result of the Jacobian being $r$ and some simplification of the $cos^2\theta + sin^2\theta$ identity.

Any help would be appreciated.

Thanks!

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  • $\begingroup$ where did you find the area formula that you started with? $\endgroup$ – joseph May 1 '19 at 7:45
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As implied by the polar formula, the modulus should be considered a function of the angle $r=r(\theta)$. Indeed, some simple closed curves orbiting the origin can be parameterized this way:

Suppose the boundary of a domain $D$ has the parameterization $r(\theta)(\cos\theta, \sin\theta)$ for $\theta\in[0,2\pi]$. Then \begin{multline} \oint_{\partial D}xdy - ydx = \\ = \int_0^{2\pi} \left[ r(\theta)\cos\theta\cdot \left(r^\prime(\theta)\sin\theta + r(\theta)\cos\theta\right) - r(\theta)\sin\theta\cdot \left(r^\prime(\theta)\cos\theta - r(\theta)\sin\theta\right)\right]d\theta = \\ = \int_0^{2\pi} r(\theta)^2\left(\sin^2\theta+\cos^2\theta\right)d\theta \end{multline} Which yields the integral you described. The same similarly holds over $\theta\in[a,b]$ where $b-a=2\pi$.

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