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The "average radius" is the average of the distance between the center and the perimeter of the closed shape.

It "appears" correct that a curve parametrized by $\theta$ gives the correct average radius. How do we justify this mathematically?

To clarify, take the ellipse $x^2/4+y^2/9=1$ with respect to the origin. It can be parametrized as $(2\cos(t),2\sin(t))$ or converted in terms of $\theta$

$$r=\frac{6}{\sqrt{4\cos^2(\theta)+9\sin^2(\theta)}}$$

The average distance formula of $(2\cos(t),3\sin(t))$ with respect the origin is $\frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{4\cos^2(t)+9\sin^2(t)}\approx2.525$ but the average distance formula of the polar equation is $\frac{1}{2\pi}\int_{0}^{2\pi}\frac{6}{\sqrt{4\cos^2(\theta)+9\sin^2(\theta)}}\approx 2.425$.

Why is the correct average radius $2.425$ and not $2.525?$

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    $\begingroup$ See my answer about AGM. $\endgroup$ Jan 5, 2018 at 15:13
  • $\begingroup$ The term "average radius" might be what's misleading here. "Average radius", see Ng Chunk Tak's answer, and "average distance to closest point on boundary" are different things. $\endgroup$ Jan 5, 2018 at 15:15
  • $\begingroup$ @ColmBhandal "Average Radius" should be angular average. $\endgroup$
    – Arbuja
    Jan 5, 2018 at 15:22
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    $\begingroup$ Ah right- the answer below of Ng Chung Tak to me is perfect. Basically, the point is that time doesn't vary linearly with the angle, so as you smoothly pass through time, the angle change will speed up or slow down- it won't be constant. This means the integral over time puts more weight on some angles than others. Same arguments apply for time vs. arclength or arclength vs. angular average. $\endgroup$ Jan 5, 2018 at 16:49

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Average radial distances vary according to different aspects, see the summary below:

$$ \begin{array}{|c|c|c|} \hline & \text{Keplerian orbit} & \text{Hooke's law orbit} \\ \hline & & \\ \text{angular average} & \langle r \rangle_{\theta}=b & \langle r \rangle_{\theta}= \dfrac{2b}{\pi} K(e) \\ & &\\ \text{time average} & \langle r \rangle_{t}=a\left( 1+\dfrac{e^2}{2} \right) & \langle r \rangle_{t}=\dfrac{2a}{\pi} E(e) \\ & & \\ \text{arclength average} & \langle r \rangle_{s}=a & \langle r \rangle_{s}= \dfrac{a(2-e^2)}{2E(e)}E\left( \dfrac{e^2}{2-e^2} \right) \\ & &\\ \hline \end{array}$$

where $e=\sqrt{1-\dfrac{b^2}{a^2}}$

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  • $\begingroup$ Say we didn't know the center and we wanted to find the point in the shape with the highest average distance. For all average distances, do we end up with the same point? $\endgroup$
    – Arbuja
    Jan 5, 2018 at 15:37
  • $\begingroup$ See this $\endgroup$
    – Arbuja
    Jan 5, 2018 at 15:38
  • $\begingroup$ That's far beyond the scope here. Anyways, I may interested the minimal value instead. Think about using calculus of variation. $\endgroup$ Jan 5, 2018 at 16:55
  • $\begingroup$ The minimal value rarely exists. There is usually one or more maximums. $\endgroup$
    – Arbuja
    Jan 5, 2018 at 17:31
  • $\begingroup$ In most cases, just one maximum. Otherwise, the weighted average of the maximums. $\endgroup$
    – Arbuja
    Jan 5, 2018 at 19:56

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