Arc length of a circle using integration

  • Thread starter Swallow
  • Start date
21
0
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)= ([tex]\int[/tex] [tex]\sqrt{1+y'^2}[/tex]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] }[tex]^{a}_{b}[/tex]

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=[tex]\pi[/tex]*R

now what i dont 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, shouldnt these limits be giving me the full circumference, without the need to multiply by two??
 
Last edited:

arildno

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
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.
 
21
0
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 shouldnt it's integral ALSO account for both half circles?
 

arildno

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
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:
[tex]x(t)=R\cos(t), y(t)=R\sin(t), 0\leq{t}\leq{2\pi}, R\geq{0}[/tex].

In this case, your arc-length integral becomes:

[tex]L=\int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}dt=\int_{0}^{2\pi}Rdt=2\pi{R}[/tex]
 
21
0
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 Im 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

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
Well, you can't integrate a "relation", in your usage of the terms.
Only among functions do you find integrable fellows.
 
21
0
but supposing i integrate sine inverse the function, (i can do that cant 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

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
No.
Because those values are not within the domain of your function.
 
21
0
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

Science Advisor
Homework Helper
Gold Member
Dearly Missed
9,948
130
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.
 
477
32
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.
 

Related Threads for: Arc length of a circle using integration

  • Posted
Replies
16
Views
5K
  • Posted
Replies
2
Views
2K
  • Posted
Replies
8
Views
3K
Replies
6
Views
2K
  • Posted
Replies
7
Views
2K
Replies
6
Views
4K
Replies
11
Views
994
Replies
3
Views
2K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top