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
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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 nonstandard 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
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 13.16$
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): area
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): area