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

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 $2.525?$

• 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. Jan 5, 2018 at 15:15
• @ColmBhandal "Average Radius" should be angular average. Jan 5, 2018 at 15:22
• 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. Jan 5, 2018 at 16:49

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