Area between Radii and Curve in Polar Coordinates

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Theorem

Let $C$ be a curve expressed in polar coordinates $\polar {r, \theta}$ as:

$r = \map g \theta$

where $g$ is a real function.


Let $\theta = \theta_a$ and $\theta = \theta_b$ be the two rays from the pole at angles $\theta_a$ and $\theta_b$ to the polar axis respectively.


Then the area $\AA$ between $\theta_a$, $\theta_b$ and $C$ is given by:

$\ds \AA = \int \limits_{\theta \mathop = \theta_a}^{\theta \mathop = \theta_b} \frac {\paren {\map g \theta}^2 \rd \theta} 2$

as long as $\paren {\map g \theta}^2$ is integrable.


Proof

AreaPolarIntegral.png


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Consider the area of the brown triangle.

This would be:

$a_\triangle = \dfrac 1 2 r^2 \map \sin {\delta \theta}$

We will be using non-standard analysis, so let $\delta \theta = \varepsilon > 0$, an infinitesimal.

Thus:

$a_\triangle = \dfrac 1 2 r^2 \sin \varepsilon$

Using the Power Series Expansion for Sine Function:

\(\ds A_\triangle\) \(=\) \(\ds \frac 1 2 r^2 \paren {\varepsilon - \frac {\varepsilon^3} {3!} + \frac {\varepsilon^5} {5!} - \map \OO {\varepsilon^7} }\) Power Series Expansion for Sine Function
\(\ds \) \(=\) \(\ds \frac 1 2 r^2 \varepsilon\) $\varepsilon ^2 = 0$
\(\ds \AA\) \(=\) \(\ds \int_{\theta_a}^{\theta_b} \frac {\paren {\map g \theta}^2} 2 \rd \theta\) Summing all areas of the triangles

$\blacksquare$


Sources

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