# Thread: Average distance from origin to points in the ball

1. ## Average distance from origin to points in the ball

Find the average distance from the origin to the points

(1) in the ball $B(0,a) \subset \Re^2$

(2) in the ball $B(0,a) \subset \Re^3$

I need help setting up the problem. I'm not sure where to begin here.

EDIT: For (1), I worked backwards from the answer, $\frac{2}{3}a$, to find a setup for the problem, but I'm not sure if I understand the meaning of the setup:

$Ave. dist = \frac{1}{Volume(ball)}\int_{\theta=0}^{2\pi}\int_{ r=0}^{a}{r*r dr d\theta}$

Since I'm using polar coordinates, r is the Jacobian. And since I am finding the average distance from the origin to all the points in the ball, I want to add up all of those distances, r, in the integral and then divide by the volume of the ball. Does that explain that the setup?

For part (2), should I be using the same approach, but in spherical coordinates?

That is: $Ave. dist = \frac{1}{Volume(ball)}\int_{\theta=0}^{2\pi}\int_{ \phi=0}^{\pi}\int_{\rho=0}^{a}{\rho*{\rho}^2 \sin\phi d\rho d\phi d\theta}$

2. Originally Posted by Redding1234
Find the average distance from the origin to the points

(1) in the ball $B(0,a) \subset \Re^2$

(2) in the ball $B(0,a) \subset \Re^3$

I need help setting up the problem. I'm not sure where to begin here.

EDIT: For (1), I worked backwards from the answer, $\frac{2}{3}a$, to find a setup for the problem, but I'm not sure if I understand the meaning of the setup:

$Ave. dist = \frac{1}{Volume(ball)}\int_{\theta=0}^{2\pi}\int_{ r=0}^{a}{r*r dr d\theta}$

Since I'm using polar coordinates, r is the Jacobian. And since I am finding the average distance from the origin to all the points in the ball, I want to add up all of those distances, r, in the integral and then divide by the volume of the ball. Does that explain that the setup?

For part (2), should I be using the same approach, but in spherical coordinates?

That is: $Ave. dist = \frac{1}{Volume(ball)}\int_{\theta=0}^{2\pi}\int_{ \phi=0}^{\pi}\int_{\rho=0}^{a}{\rho*{\rho}^2 \sin\phi d\rho d\phi d\theta}$
Do you know that,

$\bar f = \frac{1}{Volume} \iiint_D f(x,y,z) dV$

So you would set the triple integral up as you would for any other problem. This time though, you are simply dividing by volume. In 2D of course volume becomes area.

So if we were to calculate this for part 1 (a 2D figure. It is not actually a ball, rather a disk of radius a) we would need

$\bar f = \frac{1}{area} \iint_D \sqrt{x^2 + y^2}dA = \frac{1}{ \pi a^2 } \int_0^{ 2 \pi } d \pi \int_0^a r^2dr = \frac{ 2a}{3}$

For part 2 (a 3D figure we would need)

$\bar f = \frac{1}{volume} \iiint_D \sqrt{a^2 + x^2 + y^2}dV$

You can do this in spherical or in cylindrical co-ordinates.