How do you prove that the circumference of a circle is 2pir2πr?

3 Answers
Feb 27, 2015

Basically this is a definition thing.

piπ is defined to be the ratio of the circumference of a circle over its diameter (or 2 times its radius).

This ratio is a constant since all circles are geometrically similar and linear proportions between any similar geometric figures are constant.

If you were looking for how the value of the ratio piπ is calculated or how we know that the same ratio applies to the area of a circle, then those are different questions. (Ask if either of these are what you were looking for).

Feb 28, 2015

If we imagine the circle centered in the origin with radius rr, it has the equation:

x^2+y^2=r^2x2+y2=r2, (in the graph the radius is 2):

graph{x^2+y^2=4 [-10, 10, -5, 5]}

or y=+-sqrt(r^2-x^2)y=±r2x2

And considering the fourth of circle in the first quadrant, we can obtain the lenght of a line with the integral:

L=4int_0^rsqrt(1+(y')^2)dx.

This integral is quite long, so we can parametrize the circle as usual:

x=rcostheta
y=rsintheta

and use this integral:

L=int_a^bsqrt([x'(theta)]^2+[y'(theta)]^2)d theta.

Since:

x'=-rsintheta
y'=rcostheta

So:

L=4int_0^(pi/2)sqrt(r^2sin^2theta+r^2cos^2theta)d theta=

=4int_0^(pi/2)sqrt(r^2(sin^2theta+cos^2theta))d theta=

=4int_0^(pi/2)rd theta=4[rtheta]_0^(pi/2)=4rpi/2=2pir.

Jan 21, 2017

I don't think you can prove it, because that is, or is equivalent, to the definition of pi.

Explanation:

Similarly I don't think you prove that int_1^x 1/t dt=ln |x|, because ln x is defined by precisely that integral. However, in both cases there are alternative definition in terms of limits of infinite series, but if you go down that path you just go round in circles: you can arbitrarily pick any one of the relationships as a definition and show that the others necessarily follow, and that they are all related to the same quantity, be it pi or e. Evaluating pi or e is a different matter entirely.