Post History
#2: Post edited
by
TheCodidacter, or rather ACodidacter
·
2023-06-15T07:13:50Z (over 1 year ago)
Slight notational mistake
- The problem is indeed that you aren't computing the average distance. What you're computing is a *median* distance. Essentially, you are equally weighting points that are just slightly outside $C_2$ and points that are near the boundary of $C_1$, but those contribute different amounts to the average distance. To reinforce this, note that if you did an arbitrary area-preserving transformation that maintained which points were in or out of $C_2$, then you could have half the points be at any arbitrary distance allowing you to increase the average distance arbitrarily. Imagine $C_2$ being the cross-section of a metal cylinder and everything outside of $C_2$ (and in $C_1)$ being goop that you could stretch arbitrarily. You could stretch that goop into a dumbbell shape with a very long and narrow connecting part.
A typical, fairly generic definition of the average value, $\bar f$, of a function, $f$, over a continuous domain, $D$, is $\bar f = \int_D f(x) dx / \int_D dx$. In probabilistic language, this would be the *expected value* of $f$ if $D$ were the sample space with uniform probability for each point. In this case, $D$ is the disc of radius $R$, so the denominator integral is just its area, namely $\pi R^2$. The numerator integral can be written as $\int_0^{2\pi}\int_0^R r^2 d\theta dr$. The perhaps not immediately obvious part is why the integrand is $r^2$ and not just $r$. Intuitively, this is because changing $\theta$ by a small amount changes where you are in direct proportion to the distance from the center of rotation, i.e. $r$. If the $r^2$ isn't obvious to you, you should try to derive it yourself. One, somewhat tedious, approach is to rewrite the integral in Cartesian coordinates and do a change of variables. A more direct way would be to look up what polar coordinates means not just for the coordinates themselves but also the tangent vectors.- The result you quoted then follows through basic integration, $$\int_0^{2\pi}\int_0^R r^2 dr d\theta = 2\pi\int_0^R r^2 dr = 2\pi R^3/3$$ which is divided by $\pi R^2$.
- The problem is indeed that you aren't computing the average distance. What you're computing is a *median* distance. Essentially, you are equally weighting points that are just slightly outside $C_2$ and points that are near the boundary of $C_1$, but those contribute different amounts to the average distance. To reinforce this, note that if you did an arbitrary area-preserving transformation that maintained which points were in or out of $C_2$, then you could have half the points be at any arbitrary distance allowing you to increase the average distance arbitrarily. Imagine $C_2$ being the cross-section of a metal cylinder and everything outside of $C_2$ (and in $C_1)$ being goop that you could stretch arbitrarily. You could stretch that goop into a dumbbell shape with a very long and narrow connecting part.
- A typical, fairly generic definition of the average value, $\bar f$, of a function, $f$, over a continuous domain, $D$, is $\bar f = \int_D f(x) dx / \int_D dx$. In probabilistic language, this would be the *expected value* of $f$ if $D$ were the sample space with uniform probability for each point. In this case, $D$ is the disc of radius $R$, so the denominator integral is just its area, namely $\pi R^2$. The numerator integral can be written as $\int_0^{2\pi}\int_0^R r^2 drd\theta$. The perhaps not immediately obvious part is why the integrand is $r^2$ and not just $r$. Intuitively, this is because changing $\theta$ by a small amount changes where you are in direct proportion to the distance from the center of rotation, i.e. $r$. If the $r^2$ isn't obvious to you, you should try to derive it yourself. One, somewhat tedious, approach is to rewrite the integral in Cartesian coordinates and do a change of variables. A more direct way would be to look up what polar coordinates means not just for the coordinates themselves but also the tangent vectors.
- The result you quoted then follows through basic integration, $$\int_0^{2\pi}\int_0^R r^2 dr d\theta = 2\pi\int_0^R r^2 dr = 2\pi R^3/3$$ which is divided by $\pi R^2$.
#1: Initial revision
by
Derek Elkins
·
2023-06-14T09:44:34Z (over 1 year ago)
The problem is indeed that you aren't computing the average distance. What you're computing is a *median* distance. Essentially, you are equally weighting points that are just slightly outside $C_2$ and points that are near the boundary of $C_1$, but those contribute different amounts to the average distance. To reinforce this, note that if you did an arbitrary area-preserving transformation that maintained which points were in or out of $C_2$, then you could have half the points be at any arbitrary distance allowing you to increase the average distance arbitrarily. Imagine $C_2$ being the cross-section of a metal cylinder and everything outside of $C_2$ (and in $C_1)$ being goop that you could stretch arbitrarily. You could stretch that goop into a dumbbell shape with a very long and narrow connecting part.
A typical, fairly generic definition of the average value, $\bar f$, of a function, $f$, over a continuous domain, $D$, is $\bar f = \int_D f(x) dx / \int_D dx$. In probabilistic language, this would be the *expected value* of $f$ if $D$ were the sample space with uniform probability for each point. In this case, $D$ is the disc of radius $R$, so the denominator integral is just its area, namely $\pi R^2$. The numerator integral can be written as $\int_0^{2\pi}\int_0^R r^2 d\theta dr$. The perhaps not immediately obvious part is why the integrand is $r^2$ and not just $r$. Intuitively, this is because changing $\theta$ by a small amount changes where you are in direct proportion to the distance from the center of rotation, i.e. $r$. If the $r^2$ isn't obvious to you, you should try to derive it yourself. One, somewhat tedious, approach is to rewrite the integral in Cartesian coordinates and do a change of variables. A more direct way would be to look up what polar coordinates means not just for the coordinates themselves but also the tangent vectors.
The result you quoted then follows through basic integration, $$\int_0^{2\pi}\int_0^R r^2 dr d\theta = 2\pi\int_0^R r^2 dr = 2\pi R^3/3$$ which is divided by $\pi R^2$.