What is the average height of a circle?

[qqwref]

The definition of the average value of something which varies continuously (let's say a function f(x) defined over a<x<b) is, According to Wikipedia because I don't have a stats textbook on me:
average value of f(x) = 1/(b-a) * integral from a to b of f(x) dx.

Now the 1/(b-a) part is just 1/(width of interval), and the integral from a to b of f(x) dx is the area f(x) takes up inside that interval. So reduced to simpler language this formula just says that the average height of a 2D object is its area divided by its width.

I dunno, I think it sounds reasonable that the average height*average width of any object should equal the area. Not that that makes it true, though...

This might sound reasonable but unfortunately it does not work. (Average height * width = height * average width = area, though.) Consider a right isosceles triangle that is half of an a by b rectangle. Then the average height of the triangle is a/2 and the average width is b/2; you can see this with just intuition. But now average height * average width is ab/4, not ab/2 which is the correct area of the triangle.

 

[brunson]

If you place the circle of radius r centered on the origin, then I think it's something along the lines of (the integral from -r to r of 2sin(x)dx) divided by 2r. By symmetry, this is equal to int(sin(x)dx, 0, r)/r.

Spoiler

I think the answer is r, but it's been a long time since calc.

Edit: That's not correct, I've expressed the height in terms of theta, not x.

The height of the circle at a given x would be 2sin(arccos(x)), not 2sin(x). Also, I missed doubling my integral when I cut my limits in half to use symmetry. So, I believe the answer is 2*int(sin(arccos(x)), 0, r)/r.

Spoiler

Again, warnings against dodgy math, but I think that integral evaluates to r*pi/2

It's interesting that this result is equal to the area of the circle (pi*r^2) divided by the diameter (2r). So, whoever said that back at the beginning of the thread was correct.