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Using Triple Integrals 
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#1
Dec1803, 09:44 AM

P: 867

I was wondering if it was possible (just out of curiosity) to show that the circumference of a unitcircle is 2(pi) WITHOUT using the rule C = 2(pi)r?
Is it possible to take a unitsphere (a sphere with r=1) and triple integrate it. Something like... SSS sqrt(x^2 + y^2 + z^2)dxdydz = 2(pi) I thought this might be an interesting little problem. Anyone more than welcome to add to the calculation. Any help would be great too! I'll post back with anything I can find. Cheers. 


#2
Dec1803, 04:59 PM

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P: 6,038

For the first question, where are you starting? When I went to school many years ago, I learnt the definition of pi was the ratio of the circumference to the diameter of circle.
For the triple integral (you didn't specify the domain of integration, but I assume you meant the volume of the sphere.), I am not sure what you are trying to get, but the answer to the problem you stated is pi. 


#3
Dec1803, 07:04 PM

P: 678

The problem is, you have to define [itex]\pi[/itex] as something first. A lot of the time, people define [itex]\pi[/itex] as the ratio of the circumference of a circle to the diameter of a circle. So in that case [itex]C=2\pi r[/itex] pretty much by definition.
You don't need to define [itex]\pi[/itex] that way. For example, you can use: [tex]\pi=2\cdot\int_{1}^{1}\sqrt{1x^2}\,dx[/tex] And then you can use that definition to define what the area of a circle is, and use that to define the sin and cos functions, and eventually you can show that [itex]C=2\pi r[/itex]. But somewhere you have to define [itex]\pi[/itex]. You can't derive the value of [itex]\pi[/itex] without definining it somewhere first. If you don't like [itex]\pi=\frac{C}{2r}[/itex] then you can use something else that's logically equivalent, but the definition isn't really any more rigorous. 


#4
Dec2003, 07:37 AM

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P: 39,339

Using Triple Integrals
[Why] would one want to use a sphere (and triple integral) to prove something about a circle when it is much easier to use the circle itself?
As pointed out before, how you prove specifically that C= πd depends on how you define π! Most people define &pi as the ratio of the circumference of a circle to its diameter from which the equation follows immediately. Of course, one can use the arclength integral to prove that the circumference is a constant times the diameter which you would need to know before you could define π that way. 


#5
Dec2003, 11:09 AM

P: 220

The ancient greeks proved that for ANY circle, the ratio of the circumference to the diameter is a constant. For example: A circle with a diameter of 50 miles, and a circumference C1 (measured in miles), and a circle of diameter 3 inches, and circumference C2 (measured in miles), were related as follows: C1/50 = C2/3 = some mathematical constant It was archimedes who developed the method of exhaustion to figure out this constant to any desired degree of accuracy. Basically, archimedes inscribed and circumscribed regular polygons around acircle of arbitrary radius. It is clear that the circumference of the circle is trapped between the perimeter of the circumscribed polygons, and the inscribed polygons. As the number of sides of the polygons increases without bound, the perimeters of the polygons approach the circumference of the circle. Archimedes method of exaustion is the first known use of the concept of a limit in human history. There is no need to use a triple integral to derive C = 2 (pi) R. And even if you try, your limits of integration will surely involve the constant (pi), so you really aren't proving that C = 2 (pi) R. The most intuitive way to understand the formula for the circumference of a circle, is to study Euclid, in conjunction with real 'circles' whose perimeter is a piece of yarn, or thread. You certainly understand the concept of length, and the length of the yarn can be measured. You can also measure the radius of your circle. The more experimental accuracy you have, the better you will know the value of pi. By measurement, you will get about 3.14. Using archimedes method of exhaustion, you don't have to perform any measurements of 'real circles' at all, and you can compute (pi) to any desired degree of accuracy. 


#6
Dec2203, 11:43 PM

P: 867

Thanks guys. But I didn't want to prove anything I just wondered if it was possible to do that calculation and come up with pi at the end. Something like SSS (x^2 + y^2 + z^2)e^(x^2 + y^2 + z^2)dxdydz = 2(pi).
How is this done? Cheers. 


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