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Calculus/integral calculus

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Expert: - 8/18/2011


Question
priyanka
priyanka
show the circumference of the circle of radius 'r' is 2*pi*r by integration

Answer
Hello Priyanka,

Using the formula for arclength:
integral sqrt(1+[f'(x)]^2), x from a to b, where f(x)=sqrt(r^2-x^2)
So we will calculate the arclength of the semi-circle given by f(x),
then double it to get the entire circle.

f'(x)=-x/sqrt(r^2-x^2), so that 1+[f'(x)]^2=1+x^2/(r^2-x^2)=r^2/(r^2-x^2)

So, the circumference, C, is:
C=2*integral sqrt(r^2/(r^2-x^2)), x from -r to r.
=2*integral r/sqrt(r^2-x^2), x from -r to r
=2*pi*r

OK?

Abe  

Abe Mantell

Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience

Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook

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