[SOLVED]Circumference of a Circle

Member
Hello again,

I'm finding myself stuck on what is probably a simple question, but I believe I am taking the wrong approach.
The section is "Volumes with infinite integrals," and the chapter is "Improper Integrals."
The question, "Use calculus to find the circumference of a circle with radius a."
I've looked at this from the approach of integrating the circle equation in the arc length formula, but I don't think it's looking for a proof like that. Somehow this question should have an integral with infinity in it's limit.

A shove in the right direction would be greatly appreciated.

Thanks again, everyone.

Mac

Sudharaka

Well-known member
MHB Math Helper
Hello again,

I'm finding myself stuck on what is probably a simple question, but I believe I am taking the wrong approach.
The section is "Volumes with infinite integrals," and the chapter is "Improper Integrals."
The question, "Use calculus to find the circumference of a circle with radius a."
I've looked at this from the approach of integrating the circle equation in the arc length formula, but I don't think it's looking for a proof like that. Somehow this question should have an integral with infinity in it's limit.

A shove in the right direction would be greatly appreciated.

Thanks again, everyone.

Mac
Hi MacLaddy, Improper integrals may not contain a "infinity" in their limits. They are defined in terms of the boundedness of the domain of integration and the integrand.

The Cartesian equation of a circle of radius $$a$$ with its center at $$(0,0)$$ is given by,

$x^2+y^2=a^2$

$\Rightarrow y=\pm\sqrt{a^2-x^2}\mbox{ where }-a\leq x\leq a$

$\Rightarrow y=\sqrt{a^2-x^2}\mbox{ where }0\leq x\leq a$

$\Rightarrow y'=\frac{dy}{dx}=\frac{-x}{\sqrt{a^2-x^2}}$

The arc length of a curve defined on $$(c,d)$$ is given by,

$s = \int_{c}^{d} \sqrt { 1 + [f'(x)]^2 }\, dx$

In this case, $$c=0\mbox{ and }d=a$$. But on the interval $$[0,a]$$the integrand becomes unbounded. Therefore, we get the following improper integral if the limit exists.

$s = \int_{0}^{a} \sqrt { 1 + \left[\frac{-x}{\sqrt{a^2-x^2}}\right]^2 }\, dx:=\lim_{k\rightarrow a}\int_{0}^{k} \sqrt { 1 + \left[\frac{-x}{\sqrt{a^2-x^2}}\right]^2 }$

$\Rightarrow L = \lim_{k\rightarrow a}\int_{0}^{k} \sqrt {\frac{a^2}{a^2-x^2}}\, dx$

$\Rightarrow L =\left[a\sin^{-1}\left(\frac{x}{a}\right)\right]_{0}^{k\rightarrow a}$

$\Rightarrow L =\frac{\pi a}{2}$

Since the curve is symmetric around the x and y axes we have to multiply this by 4 in order to get the circumference$$(C)$$ of the curve.

$\therefore C=2\pi a$

Kind Regards,
Sudharaka.

• Jameson