The standard method of calculating area of ellipse:(adsbygoogle = window.adsbygoogle || []).push({});

[tex] \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 [/tex]

[tex] Area = \int_C -ydx \hbox { or } \int_C xdy [/tex]

It is more convient to use polar coordinate [itex] x=a cos \theta \; \hbox { and }\; y=b sin \theta [/itex]

[tex]dy = b cos \theta[/tex]

[tex] \hbox{ Using } \int_C xdy = \int_0^{2\pi} ab \; cos^2 \theta =\pi ab[/tex]

I am trying to solve in rectangular coordiantes where:

[tex] \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow \; y \;=\; ^+_- b\sqrt{1-\frac{x^2}{a^2} }[/tex]

Using [itex]\int_C -ydx [/itex]

[tex]\int_C -ydx = ^-_+ \int_{-a}^a b\sqrt{1-\frac{x^2}{a^2} } dx [/tex]

[tex] \hbox{ Let } \; sin \theta = \frac{x}{a} \Rightarrow dx = a cos \theta \hbox{ and } t= -\frac{\pi}{2} \hbox { to } \frac{\pi}{2}[/tex]

[tex]\int_C -ydx = ^-_+\int_{-a}^a b\sqrt{1-\frac{x^2}{a^2} } dx = ^-_+ab \int_{ -\frac{\pi}{2} }^{ \frac{\pi}{2}} cos^2 \theta d \theta = ^-_+\frac{ab}{2}\int_{ -\frac{\pi}{2} }^{ \frac{\pi}{2}} [1+cos (2\theta)] d \theta = ^-_+ \frac{\pi ab}{2}[/tex]

Notice the answer is half of using polar coordinate which show the correct answer. I understand there is + and - on the square root which I don't know how to incorporate in. How do I mathametically incorporate into the equation and get the correct answer?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Finding area of ellipse using line integral.

Loading...

Similar Threads - Finding area ellipse | Date |
---|---|

I How to find a solution to this linear ODE? | Feb 21, 2018 |

I How to find the transfer function of xy, xy' etc? | Dec 30, 2017 |

I Using Complex Numbers to find the solutions (simple Q.) | Dec 29, 2017 |

A Better way to find Finite Difference | Oct 16, 2017 |

Poisson equation: what shape gives largest area average | Jan 21, 2011 |

**Physics Forums - The Fusion of Science and Community**