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?