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

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Since you have asked this question in the calculus section I believe this question deserves a different approach. I believe the previous answer is correct and I am aware of the fact that ##pi## was defined on the basis of the circumference of the circle. Yet I will proceed with my method.

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Assume a circle with radius ##r##. Consider a differential element of length ##dx## along the perimeter (I have used the word ‘perimeter’ since we have not defined circumference yet.).

So, if we can find a way to sum all the ##dx##s along the boundary of the circle that should give us the ‘circumference’ of the circle.

The element ##dx## subtends an angle ##d(theta)## at the centre of the circle. Since ##dx## is very small the curvature of the circle may be ignored and the element may be considered to be a line segment. The triangle thus formed is an isosceles triangle. Dropping a perpendicular from the vertex with ##d(theta)## angle to the base(i.e. the line segment of ##dx## length) will divide the angle and the segment into two equal halves.

Applying trigonometry to the right-angled triangle,
##sin(d(theta)/2)=(dx/2)/r##
So, ##dx=2*r*sin(d(theta)/2)##

Since ##d(theta)/2## is an extremely small quantity we can say that,
##sin(d(theta)/2)~~d(theta)/2## [from the concept of limits]

Hence, ##dx=2*r*d(theta)/2=r*d(theta)##

Now, a circle is made of such infinite line segments attached one after the other. The ##theta## varies from ##0## to ##2pi##. So, if we integrate ##dx## from ##0## to ##2pi## we will get the expression for the circumference of s circle.

##int_0^(2pi)dx=int_0^(2pi)r*d(theta)##

Since ##r## is independent of ##theta## it can be taken out of the integral.

##=r*int_0^(2pi)d(theta)=r*[theta]_0^(2pi)=r*[2pi-0]=r*2pi=2pi*r##

which is the expression for the circumference of the circle.

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