Length of the arc of a quadrant

[brotherbobby]

Homework Statement

Calculate the arc length $s$ of a quadrant of a circle of radius $a$.

Relevant Equations

1. Circumference of a circle of radius $r$ : $C = \pi r^2$.
2. The line element $ds = \sqrt{1+y'^2}dx$

Statement of the problem : Calculate the arc length $s$ of the quadrant in the figure shown alongside.
Attempt : (I have been able to solve the problem integrating with respect to $x$. However, when I integrate relative to $y$, I get a minus $(-)$ sign. Where I am going wrong? I furnish the details below.)
[In truth, we're aware of the answer without a calculation. Given the circumference of a circle, the arc length $s = \dfrac{2\pi a}{4}\Rightarrow\boxed{s = \dfrac{\pi a}{2}}$.]
Writing $ds$ as a function of $y$, we have $ds = \sqrt{1+x'^2}\;dy$, where $x'=\dfrac{dx}{dy}$.
The equation of the arc : $x = \sqrt{a^2-y^2}\Rightarrow x' = -\dfrac{y}{\sqrt{a^2-y^2}}\Rightarrow \sqrt{1+x'^2} = \dfrac{a}{\sqrt{a^2-y^2}}$. Of course we remember that for the curve, $0\le x,y\le a$, so both variables are everywhere positive, when taking square roots.
Thus $\displaystyle{s = \int\limits_a^0\dfrac{a}{\sqrt{a^2-y^2}}dy = a\;\left.\sin^{-1}\dfrac{y}{a}\right|_{a}^{0}=a\left(0-\dfrac{\pi}{2}\right)=\boxed{-\dfrac{\pi a}{2}}}$.

Why am I getting the $-$ sign? Is it because I should have taken the line element $ds = \sqrt{1+x'^2}\;(-dy)$, since $y$ is decreasing?

Many thanks.

 

[BvU]

Why am I getting the $-$ sign? Is it because I should have taken the line element $ds = \sqrt{1+x'^2}\;(-dy)$, since $y$ is decreasing?

Perhaps because y is running from 0 to a and not from a to 0 ?

(ah, you answered your own question )

 

[Mark44]

Perhaps because y is running from 0 to a and not from a to 0 ?

Yes. Although x runs from a to 0 as we run along the curve in counterclockwise fashion, y runs from 0 to a. Switching the limits of integration changes the sign of the result.

 

[erobz]

I would say avoid the $( -dy )$ thinking - it's corrupt. The delta $\Delta$ or $d$ captures change in either direction ( $\pm$), it's going to lead to errors if you start thinking that it is signed.

Use

$$ s = \int_0^a \sqrt{ 1 + ( y' )^2 } ~dx $$

and it will all work out.

 

[brotherbobby]

Yes. Although x runs from a to 0 as we run along the curve in counterclockwise fashion, y runs from 0 to a. Switching the limits of integration changes the sign of the result.

I agree with you, but there's a basic discrepancy that bothers me here.

If you want to find out the arc length $s$ integrating with respect to $x$, you run $x$ from $x = 0\rightarrow a$, you are going clockwise.
If you continue running clockwise but integrate with respect to $y$ where it runs from $a\rightarrow 0$, you get a negative answer for the arc length. Since arc length is a scalar and always positive, we can call this length $-s$.
We know that the line element $ds = \sqrt{dx^2+dy^2}$, which can come with both signs. What is really going on here that brings in the negative sign, when you maintain the direction of motion along the arc but integrate with respect to a different variable?

 

[brotherbobby]

I would say avoid the $( -dy )$ thinking - it's corrupt. The delta $\Delta$ or $d$ captures change in either direction ( $\pm$), it's going to lead to errors if you start thinking that it is signed.

Use

$$ s = \int_0^a \sqrt{ 1 + ( y' )^2 } ~dx $$

and it will all work out.

I have already done it that way, integrate with respect to $x$ and faced no problems. (Please see at the beginning of my post #1, where I start attempting a solution to the problem)
Question is, what happens when you integrate with respect to $y$ and maintain your direction of motion along the curve that results in a negative sign?

 

[A.T.]

Switching the limits of integration changes the sign of the result.

What is really going on here that brings in the negative sign, when you maintain the direction of motion along the arc but integrate with respect to a different variable?

Swapping the roles of x & y in the integration, while keeping the direction along the curve (eg. clockwise) means that you are swapping the limits of integration, which swaps the sign of the result.

In your diagram, draw the direction of integration along the arc as an arrow, and then project it onto the axis of the variable you are integrating with respect to. Note if the projection goes in the negative or positive direction along the axis.

 

[erobz]

I have already done it that way, integrate with respect to $x$ and faced no problems. (Please see at the beginning of my post #1, where I start attempting a solution to the problem)
Question is, what happens when you integrate with respect to $y$ and maintain your direction of motion along the curve that results in a negative sign?

Sorry.

A friend of mine isn't entirely happy about the arguments so far. They point out that the standard formula for arc length is:


$$L=\int_a^b\sqrt{1+[f'(u)]^2}~du,\qquad a<b$$


This suggests that the more general expression for the arc-length element should be


$$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}~|dx|$$

or equivalently:

$$ds=\sqrt{1+\left(\frac{dx}{dy}\right)^2}~|dy|$$


rather than simply


$$ds=\sqrt{1+\left(\frac{dy}{dx}\right)^2}~dx.$$


The absolute value preserves the nonnegative nature of arc length, regardless of the direction of traversal.

The conventional definite integral achieves the same result by taking the limits in increasing order, $(a<b)$, so that $du$ represents a positive increment in the corresponding Riemann sum.

 

[vela]

Question is, what happens when you integrate with respect to $y$ and maintain your direction of motion along the curve that results in a negative sign?

It's what you said at the end of your first post. Generally, you have $\sqrt{c^2} = \lvert c \rvert$, not $c$. When you pulled the $dy^2$ out from the square root, it should have been $\lvert dy \rvert = -dy$ because of the limits you chose for the integral.