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[SOLVED] Circumference of a Circle

MacLaddy

Member
Jan 29, 2012
52
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
Feb 5, 2012
1,621
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,

\[x^2+y^2=a^2\]

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

In the first quadrant,

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

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

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.
 

MacLaddy

Member
Jan 29, 2012
52
Wow... Thanks Sudharaka, that's quite the reply. I appreciate that.

It looks like I was "somewhat" on the right track before, but I am definitely misunderstanding some of the inner workings in this process. I'll stare at this for a few days and make sure it sinks in fully.

Thanks again,
Mac