Arc length of a circle using integration

[Swallow]

Hello there,
suppose i want to find the arc length of a circle x^2+y^2=R^2 using integration, implicitly differentiating the equation, i find y'=-(x/y)

now,

arc length (circumference)= (\int \sqrt{1+y'^2}dx
putting the value of y'=-(x/y) and substituting for y^2 from the equation of the circle
(y^2= R^2-x^2)
solving, the equation i get
circumference= R*{ sin^-1 [x/R] }^{a}_{b}

where a and b are the limits of integration

whats bugging me here is the limits, when i use the limits [-R,R], i get circumeference=\pi*R

now what i don't get is why do i have to multiply by two to get the actual answer, i mean i didnt use the equation of the upper/lower semicircles ANYWHERE in my calculations, shouldn't these limits be giving me the full circumference, without the need to multiply by two??

 

[arildno]

Well, -R to R is the interval x runs along on the--half circle.

Remember that the inverse sine function is uniquely defined only on intervals of the type 0 to pie.

 

[Swallow]

Thanks for the quick reply arildno.

ButI still don't get it, I mean x runs along both half circles doesn't it?
both upper and lower AND our function (x^2+y^2)=R^2 accounts for both the semi circles, so shouldn't it's integral ALSO account for both half circles?

 

[arildno]

No.

Because a function can only have one function value for each argument value.

The graph of the full circle is not the graph of a function because for practically all x-values, there are are two y-values on the graph.

Thus, representing "y" as a function of x, you have necessarily restricted yourself to the half-circle.

You could do differently, by regarding BOTH "x" and "y" as functions of a single variable, say the angle made to the positive x-axis.

In that case, you have the parametrization:
x(t)=R\cos(t), y(t)=R\sin(t), 0\leq{t}\leq{2\pi}, R\geq{0}.

In this case, your arc-length integral becomes:

L=\int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt=\int_{0}^{2\pi}Rdt=2\pi{R}

 

[Swallow]

OK, I'm starting to get it a bit.

A function, should only have one y value for each x value, so inverse sine is a function if and only if we restrict the range (i.e.we make it the principal inverse sine function).

BUT if we consider sine inverse (the relation; which has an unlimited range), then does the fact (that a FUNCTION can only have one y value for each argument), make any difference in our ability to calculate the arc length of sine inverse (again the relation not the function) in a range of (say) -100pi to 100pi.

can we directly apply the limits in the integral (as i have applied to the circle equation) or do we have to account for the fact that sine inverse is not a function and multiply the answer by some constant (as in the case of the circle) in order to get the "real" arc length?

(If I am not wrong)we do NOT have to do anything of this sort in case if the inverse sine. Why are these two cases different/?

 

[arildno]

Well, you can't integrate a "relation", in your usage of the terms.
Only among functions do you find integrable fellows.

 

[Swallow]

but supposing i integrate sine inverse the function, (i can do that can't i?)
THEN i apply the limits of integration from -100pi to 100pi (which are not Actually in the range of the function), will that give me the arc length?

 

[arildno]

No.
Because those values are not within the domain of your function.

 

[Swallow]

i mean if i integrate with respect to dy. then take the limits -100pi to 100 pi, that would give an answer, what would be the physical meaning of that answer?

 

[arildno]

Sigh.
you have WRONG ideas about what functions&integration are.
Go back to your texgtbolok, there is nothing further to be said on this matter.

 

[timthereaper]

The way I see it, there are several problems with your analysis are:

1. The equation of a circle (x^2 + y^2 = R^2) is not a function in the first place because each x value maps to two separate y values. You either have to break the circle into two halves and then multiply by 2, or use the parametrization (x = R cos t, y = R sin t). To properly use the arc length formula, you have to use the parametrization.

2. You are using the substitution y^2 = R^2 - x^2. Because the arc length formula you're using integrates over dx, you are making y a function of x (y(x) = Sqrt[R^2 - x^2]) which only yields a half circle.

3. The real part of the inverse sine function doesn't have an unlimited range. Both the domain and range are bounded. The domain is [-1,1] and the range is [-Pi/2,Pi/2].



If I'm wrong, please let me know. I am by no means an expert.