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I would like to find the average distance from center of ellipse centered at the origin to some point on the eclipse.

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

I thought of converting to polar so,

$$r^2((\frac{\cos \theta}{a})^2+(\frac{\sin \theta}{b})^2)=1$$

$$r=\left((\frac{\cos \theta}{a})^2+(\frac{\sin \theta}{b})^2 \right)^{-1/2}$$

So the distance I am looking for is,

$$\frac{1}{2\pi} \int_{0}^{2\pi} \left((\frac{\cos \theta}{a})^2+(\frac{\sin \theta}{b})^2 \right)^{-1/2} d\theta$$

Correct? I don't know how to go from here.

I also thought about alternatively using line integrals so that what I am looking for is,

$$\frac{\int_C \sqrt{x^2+y^2} ds}{\int_C ds}$$

Then using the parametrization $x=a\cos\theta$ and $y=a\sin \theta$ but that still doesn't get me anywhere.

My question is my approaches correct, and what can I do to proceed?

Thanks in advance.

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    "Average" with respect to what measure? Parametrizing by angle will not give you the same result as an arc length parametrization. – 2017-01-11
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    Sorry I don't quite understand. What do you mean? @heropup – 2017-01-11
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    To find an average you need to consider a large number of points (and using algebra rather than statistics can make this infinite). How do you want the points spread out around the ellipse? If you have them equally spaced by angle you will get a different answer than if you have them equally spaced by distance. – 2017-01-11
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    Oh that clears thing up a lot. Would my first approach be average if we take points equally spaced by angle. And my second approach by taking points equally spaced on the ellipse? @IanMiller – 2017-01-11
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    In equal increments of arc length or equal increments of angle? – 2017-01-11

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