Finding area of ellipse using line integral.

[yungman]

The standard method of calculating area of ellipse:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

Area = \int_C -ydx \hbox { or } \int_C xdy

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

dy = b cos \theta

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


I am trying to solve in rectangular coordiantes where:

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

Using \int_C -ydx

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

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

\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}

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?

 

[g_edgar]

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.

 

[yungman]

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.

Thanks for you response. That's exactly what I think, but how do you put in as formula?