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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?
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