# What is the average height of a circle?



## Nukoca (Sep 27, 2009)

Assume the diameter(height) of a circle is four. 

Make an infinite amount of lines parallel to the diameter, all the way until you get to a line tangent to the circle. 

My calculations tell me this would be the _sum_ of the infinite numbers in between the lengths 0 and 4 divided by the _number_ of the infinite number of chords parallel to the diameter, which would be the average length of the height of the circle as a whole.

So (0 ... 4)/∞, which doesn't make sense. I can tell I need a calculus-knowledgable person to help me out here.

I tried to explain what I'm trying to get on Yahoo Answers and failed miserably.


Help?


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## spdqbr (Sep 27, 2009)

This sounds like a job for calculus! Particularly the mean vale theorem.

Really, have you studied limits at all?


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## fanwuq (Sep 27, 2009)

Edit:


qqwref said:


> The average height would be the area of the circle divided by the horizontal amount of space the circle takes up, right? So (2*2*pi)/4 = pi.


OK. So the average height is not at the midpoint. Guess that was a horrible assumption.


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## qqwref (Sep 27, 2009)

The average height would be the area of the circle divided by the horizontal amount of space the circle takes up, right? So (2*2*pi)/4 = pi.


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## Nukoca (Sep 27, 2009)

spdqbr said:


> Really, have you studied limits at all?



No... but I plan to in the near future. I hope to become a physicist when I'm a bit taller and more educated.



qqwref said:


> The average height would be the area of the circle divided by the horizontal amount of space the circle takes up, right? So (2*2*pi)/4 = pi.



Wait a minute. Can you explain that a bit more thoroughly?

@blah (right below me)


blah said:


> What's being taller got to do with anything?


Well... when I'm older. That was what I was saying.


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## blah (Sep 27, 2009)

What's being taller got to do with anything?


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## gylve (Sep 27, 2009)

The equation of the circle should be x^2 + y^2 = 4
So de length of a given chord parallel to the diameter is 2*sqrt(4-h^2)

Discretize the interval [-2,2] into 2N points, the average lenght of the chords should be:
1/(2N) * Sum(k=-N:N) 2*sqrt(4-(2*k/N)^2) =
1/N * Sum(k=0:N) 4*sqrt(1-(k/N)^2) =
4*Integral(0:1)sqrt(1-x^2)=
4*Pi/4 = Pi


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## spdqbr (Sep 27, 2009)

fanwuq said:


> I don't think it is mean value theorem.



Doh, I meant average value, forgetting what the mean value theorem *actually* is for a minute there. I'm blaming it on sleep deprivation from work 

gylve took the approach I attempted to steer Nukoca toward though, well done!


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## Lucas Garron (Sep 27, 2009)

spdqbr said:


> This sounds like a job for calculus! Particularly the mean vale theorem.
> 
> Really, have you studied limits at all?


Mean Value Theorem? No. Have _you_ studied the mean value theorem at all?

Anyhow, I agree with qq's derivation, which barely needs more explanation.


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## spdqbr (Sep 27, 2009)

Lucas Garron said:


> Have _you_ studied the mean value theorem at all?



Yeah, but it's been a while. Right principle wrong theorem, sorry.


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## Nukoca (Sep 27, 2009)

qqwref said:


> The average height would be the area of the circle divided by the horizontal amount of space the circle takes up, right? So (2*2*pi)/4 = pi.



Wait, so why would the average height be A/d?

Another question. By taking the average height of a circle and squaring it, would you get the area of the circle? If so, then the average height of a 4=d circle could be found by taking the area with pi and taking the square root. Now... assuming that the average height^2 DOES equal the area of the circle... the average height would be 3.5449077.


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## TomZ (Sep 27, 2009)

No, you wouldn't:

A=pi*r^2
average height=pi*r^2/2r
average height squared = pi^2*r^4/4r^2

pi^2*r^4/4r^2=pi*r^2 for r=0


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## Nukoca (Sep 27, 2009)

TomZ said:


> average height=pi*r^2/2r


 
Why? Can someone please explain the logic here to me before I blow a blood vessel?


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## gylve (Sep 27, 2009)

I showed why the average height is area/diameter on my preview post, but you need to understand limits and riemann's integral.


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## TomZ (Sep 27, 2009)

Before you edited your post, you said you understood qqwref. What I did was nothing other than to rephrase him in a formula form.

By definition dividing the area of an object by it's width will give you the height. That's what I did. I divided the area (pi*r^2) by the width (2pi).

Here an alternate explanation:

A circle can be described as x^2+y^2=r^2
I'm going to use r=1 to make it easier.

First, we want to find the height as a function of y. Because the height of the circle is two times the (positive) x-coordinate, we find: h=2*sqrt(1-y^2)

Next, we're going to split the circle into four vertical strips and average out the heights. The strips' x-ranges are -1 to -0.5, -0.5 to 0, 0 to 0.5 and 0.5 to 1. To get better accuracy we are going to measure the height at the middle of each strip:

h=(h(-0.75)+h(-0.25)+h(0.25)+h(0.75))/4 (about 1.63)

Ideally we would want columns of zero with to get an exact answer. To do this, there is a trick called integration. By taking the integral from -r to r (in our case, -1 to 1) we find the average height of a radius 1 circle.

This integral comes out to pi, which means that the average height of a circle is pi*r.


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## Nukoca (Sep 27, 2009)

Thanks for replying to me. And sorry for losing my temper there.




TomZ said:


> Before you edited your post, you said you understood qqwref. What I did was nothing other than to rephrase him in a formula form.



Well, I thought I did. But later I reread it and I realized that I didn't. 




TomZ said:


> By definition dividing the area of an object by it's width will give you the height. That's what I did. I divided the area (pi*r^2) by the width (2pi).



But the width was 4... unless you mean that 2pi is the average width? Sorry, I'm just trying to keep up here.
And earlier you said that the average height^2 wouldn't equal the area... right?




TomZ said:


> Here an alternate explanation:
> 
> A circle can be described as x^2+y^2=r^2
> I'm going to use r=1 to make it easier.
> ...


Kay... makes sense.



TomZ said:


> Next, we're going to split the circle into four vertical strips and average out the heights. The strips' x-ranges are -1 to -0.5, -0.5 to 0, 0 to 0.5 and 0.5 to 1. To get better accuracy we are going to measure the height at the middle of each strip:
> 
> h=(h(-0.75)+h(-0.25)+h(0.25)+h(0.75))/4 (about 1.63)
> 
> ...



Ok... I think it makes sense. That's the sort of idea I had in the first place... I'd read about the Greeks getting approximations of the area of a circle by putting many-sided polygons inside one... and they made more and more sides to it until they got a pretty accurate answer.


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## Stefan (Sep 27, 2009)

TomZ said:


> *By definition* dividing the area of an object by it's width will give you the height.


I'd like to see that definition, please (on let's say mathworld, wikipedia, etc).


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## TomZ (Sep 28, 2009)

Nukoca said:


> TomZ said:
> 
> 
> > By definition dividing the area of an object by it's width will give you the height. That's what I did. I divided the area (pi*r^2) by the width (2pi).
> ...



I made a mistake there. It should have said 2r instead of 2pi.

Now let's see if I can come up with a way to talk me out of it or vandalize wikipedia this afternoon to answer Stefan's question..


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## Nukoca (Sep 28, 2009)

TomZ said:


> Now let's see if I can come up with a way to talk me out of it or vandalize wikipedia this afternoon to answer Stefan's question..



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...


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## Rune (Sep 28, 2009)

Nukoca said:


> TomZ said:
> 
> 
> > Now let's see if I can come up with a way to talk me out of it or vandalize wikipedia this afternoon to answer Stefan's question..
> ...



It should mean that the area of a circle would be some a^2.
And a=b*sqrt(pi) if r=b


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## qqwref (Sep 28, 2009)

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.




Nukoca said:


> 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.


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## brunson (Sep 28, 2009)

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.


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