Why, mathematically, does a closed curve parametrized by $\theta$ give the correct average of the distance between the center and perimeter?

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