# Area between Radii and Curve in Polar Coordinates

Jump to navigation
Jump to search

## 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

This article needs to be linked to other articles.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

This article needs to be tidied.In particular: In particular the terms are to be defined.Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Tidy}}` from the code. |

Although this article appears correct, it's inelegant. There has to be a better way of doing it.In particular: We haven't covered NSA on $\mathsf{Pr} \infty \mathsf{fWiki}$, and for a result as trivial as this one it seems like overkill.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by redesigning it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Improve}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

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 |

$\blacksquare$

There is believed to be a mistake here, possibly a typo.In particular: Check the limits of the integralYou can help ProofWiki by reviewing it, and either correcting it or adding some explanatory material as to why you believe it is actually correct after all.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Mistake}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |