Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding area of ellipse using line integral.

  1. Jul 29, 2010 #1
    The standard method of calculating area of ellipse:

    [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?
  2. jcsd
  3. Jul 30, 2010 #2
    You have to integrate all the way around C, which is not simply -a to a. For the top half of the ellipse, x goes from -a to a, but then for the bottom half x goes from a to -a.
  4. Jul 30, 2010 #3
    Thanks for you response. That's exactly what I think, but how do you put in as formula?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook