# Area between Radii and Curve in Polar Coordinates

## 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 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 $\ds$ $=$ $\ds \frac 1 2 r^2 \varepsilon$ $\varepsilon ^2 = 0$ $\ds \AA$ $=$ $\ds \int_{\delta_a}^{\delta_b} \frac {\paren {\map g \theta}^2} 2 \rd \theta$ Summing all areas of the triangles

$\blacksquare$