In the first example with the unit circle I don't get why x = -cos(teta) isn't the x-coordinate simply cos(teta) as the trigonometric coordinate are given by the point x =(x1,y1)=(cos(teta),sin(teta)) ? :)

By convention, we think of the unit circle as starting at the point (1,0). This corresponds to a x =cos(theta) parametrization. However, since the integral travels from the lower limit (-1,0) to the upper limit (1,0) we negate the cosine to switch direction of travel. Hope this helps

Haven't we learned about arc length in ordinary integral classes previously? Why are we studying this again?

It is a review leading to Arc length of parametric curves which leads to line integrals in scalar fields which leads to line integrals in vector fields and on it goes. Sometimes a little refresher review on a concept helps when a new topic depends on it.

First example, step 3, how to solve the integral using trigonometric substitution. This method is only valid for integrals of arc lengths right? Because if I find that integral in other contexts, I couldn´t think of the function inside the integral as a semicircle, because it is not a semicircle, although in the explanation is treated as a semicircle. So I don´t understand this method for solving these kind of integrals.

This cannot be solved by hand. This article and previous articles show how to set up the integral not how to solve this. The integrala in example 3 can only be approximated

in example 1, step 1, I don't understand where did the 2 come from in the denominator? in the d(1-x^2)/2(square root of /(1-x^2)

It comes from the derivative of sqrt(1-x^(2)). Take a simpler example. Let's say we want the derivative of sqrt(x). This is just x^(1/2). Using the power rule, we get (1/2)x^(-1/2) or 1/(2sqrt(x)). The example in the article is the same, but as the inside of the square is a function, you need the chain rule.

Main content

Course: Multivariable calculus > Unit 4

Lesson 2: Line integrals for scalar functions (articles)

Arc length of function graphs, examples

Google Classroom

Practice finding the arc length of various function graphs.

Background

Arc length of function graphs, introduction

Example 1: Practice with a semicircle

Consider a semicircle of radius

1

, centered at the origin, as pictured on the right. From geometry, we know that the length of this curve is

π

. Let's practice our newfound method of computing arc length to rediscover the length of a semicircle.

By definition, all points

(x, y)

on the circle are a distance

1

from the origin, so we have

$x^{2} + y^{2} = 1$ ‍

Rearranging to write

y

as a function of

x

, we have

$y = \sqrt{1 - x^{2}}$ ‍

As you set up the arc length integral, it helps to imagine approximating this curve with a bunch of small lines.

Writing down the arc-length integral, ignoring the bounds for just a moment, we get:

$\begin{array}{r} \int \sqrt{(d x)^{2} + (d y)^{2}} \end{array}$ ‍

Just as before, we think of the integrand

\sqrt{(d x)^{2} + (d y)^{2}}

as representing the length of one of these little lines approximating the curve (via the Pythagorean theorem).

Now we start plugging in the definition of our particular curve into the integral.

Step 1: Write $d y$ ‍ in terms of $d x$ ‍

Use the fact that

y = \sqrt{1 - x^{2}}

to write

d y

in terms of

d x

$d y =$ ‍
$d x$ ‍

Step 2: Replace $d y$ ‍ in the integral

Plug this expression for

d y

into the integral to write the integrand completely in terms of

x

and

d x

$\int$ ‍
$d x$ ‍

Step 3: Place bounds on the integral and solve

Since the curve is defined between when

x = - 1

and

x = 1

, set these values as the bounds of your integral and solve it.

(Sorry, no entry box with a happy green checkmark. We know from geometry that the arc length is

π

, but the interesting part is to work through it to see how

π

pops out when using an arc length integral.)

Our integral looks like this:

$\begin{array}{r} \int_{- 1}^{1} \sqrt{\frac{1}{1 - x^{2}}} d x \end{array}$ ‍

Now, there are a few ways you might proceed from here.

Plug this integral into a calculator or computer algebra system and watch the answer pop out.
Somehow remember or recognize that this integral is $\arcsin (x)$ ‍
Solve the integral using trigonometric substitution.

All of these are valid, but the interesting one to focus on is the trigonometric substitution. Rather than just plugging and chugging, I want to shed some light on what this substitution actually means geometrically. Hopefully, this makes clear how and why this technique works, which will help you recognize whether or not such a substitution will work in other integrals that you might encounter.

What does it mean that our integral is currently written in terms of

x

? It basically has you walking along the

x

-axis, from

- 1

1

, measuring the snippet of arc length on the circle above your head as you go.

However, in this case, it might be natural to imagine walking along the circle itself. Consider the following diagram:

Each point of the circle can be described with a value of the angle

θ

, ranging from

0

π

. Moreover, since the radius is

1

, the value of

θ

measured in radians actually reflects arc length. A tiny change

d θ

will correspond with a tiny change in arc length. Therefore, it should seem reasonable that an alternative integral describing the desired arc length is

$\begin{array}{r} \int_{0}^{π} d θ \end{array}$ ‍

In a sense, we're done. It's easy to see that this integral evaluates to

π

, which is indeed the arc length of our semi-circle. But what's the relationship between this and our previous integral?

Since the

x

value of any point on the circle is

$x = - \cos (θ)$ ‍

this suggests that substituting

- \cos (θ)

for

x

will simplify our integral. Let's actually walk through that process. When we apply this substitution, we have to do three things

Step 1: Replace

x

with

- \cos (θ)

Step 2: Replace

d x

with

d (- \cos (θ)) = \sin (θ) d θ

Step 3: Write the integral's limits

x = - 1

and

x = 1

in terms of

θ

The relation $x = - 1$ ‍ translates to the relation $- \cos (θ) = - 1$ ‍. This happens when $θ = 0$ ‍.
The relation $x = 1$ ‍ translates to the relation $- \cos (θ) = 1$ ‍. This happens when $θ = π$ ‍.

Applying all of these exchanges, we get

\begin{array}{r} \int_{- 1}^{1} \sqrt{\frac{1}{1 - x^{2}}} d x \to \int_{0}^{π} \sqrt{\frac{1}{1 - \cos^{2} (θ)}} \sin (θ) d θ \end{array}

Simplifying, using the identity

\sin^{2} (θ) + \cos^{2} (θ) = 1

, this becomes

$\begin{aligned} \int_{0}^{π} \sqrt{\frac{1}{1 - \cos^{2} (θ)}} \sin (θ) d θ \\ \int_{0}^{π} \sqrt{\frac{1}{\sin^{2} (θ)}} \sin (θ) d θ \\ \int_{0}^{π} \frac{1}{\sin (θ)} \sin (θ) d θ \\ \int_{0}^{π} d θ \end{aligned}$ ‍

So yes, trigonometric substitution greatly simplifies our integral, but it's not just magic. The reason it makes things simpler is that the curve is more naturally described with the value

θ

shown in the diagram above.

Practice setting up arc length integrals

The actual integral you get for arc length is often difficult to compute. However, the important skill to practice is setting up that integral. So let's practice that a few times without worrying about computing the final integral (you can use a calculator or wolfram alpha once you get a concrete integral).

Example 2: Sine curve

What integral represents the arc length of the graph of

y = \sin (x)

between

x = 0

and

x = 2 π

$\begin{array}{r} \int_{a}^{b} \end{array}$ ‍
$d x$ ‍

a =

b =

As always, the integral starts its life looking like this:

$\begin{array}{r} \int \sqrt{d x^{2} + d y^{2}} \end{array}$ ‍

Step 1: Write

d y

in terms of

d x

$\begin{aligned} y & = \sin (x) \\ d (y) & = d (\sin (x)) \\ d y & = \cos (x) d x \end{aligned}$ ‍

I think it's enlightening to remind yourself of what this means, beyond just throwing the symbol

d

in front of everything. If you are sitting at a point

(x, y)

on the curve, and you take a tiny step

d x

to the right, the amount you must step up to end up back on the curve happens to be

\cos (x) d x

Step 2: Plug this value of

d y

into the integral

$\begin{aligned} \int \sqrt{d x^{2} + d y^{2}} & = \int \sqrt{d x^{2} + (\cos (x) d x)^{2}} \\ = \int \sqrt{1 + \cos^{2} (x)} d x \end{aligned}$ ‍

Step 3: Put bounds on the integral.

In this case, the problem explicitly states that

x

runs from

0

2 π

$\begin{array}{r} \int_{0}^{2 π} \sqrt{1 + \cos^{2} (x)} d x \end{array}$ ‍

Example 3: Up, not right

Consider the curve representing

$y = \pm \sqrt{x}$ ‍

For all values where

x \leq 4

. Find an integral expressing this curve's arc length. But this time, write everything in the integral in terms of

y

, not

x

$\begin{array}{r} \int_{a}^{b} \end{array}$ ‍
$d y$ ‍

a =

b =

Unlike all examples up to this point, the curve is not the graph of a single function of

x

. Instead, it involves two separate functions:

$y = \sqrt{x}$ ‍

$y = - \sqrt{x}$ ‍

However, it is the graph of a function of

y

with respect to

x

$x = y^{2}$ ‍

So rather than constructing two separate integrals in terms of

d x

, each of which could be thought of as a rightward walk along the graph, we construct a single integral with respect to

d y

, which can be thought of as walk along the curve from bottom to top.

The steps to setting up this integral are nearly identical to everything we've done up to this point, but pay careful attention to how we are now writing all expressions with

x

in terms of

y

, rather than vice versa.

Step 1: Write $d x$ ‍ in terms of $d y$ ‍

$\begin{aligned} x & = y^{2} \\ d (x) & = d (y^{2}) \\ d x & = 2 y d y \end{aligned}$ ‍

Step 2: Plug this value of $d x$ ‍ into the integral

$\begin{aligned} \int \sqrt{d x^{2} + d y^{2}} & = \int \sqrt{(2 y d y)^{2} + d y^{2}} \\ = \int \sqrt{((2 y)^{2} + 1) d y^{2}} \\ = \int \sqrt{4 y^{2} + 1} d y \end{aligned}$ ‍

Step 3: Place bounds on the integral

This time, everything in the integral is in terms of

y

, so the bounds must reflect

y

values.

Looking at the graph above, we see that the end points of the curve which correspond with

x = 4

have

y

values

y = - 2

and

y = 2

$\begin{array}{r} \int_{- 2}^{2} \sqrt{4 y^{2} + 1} d y \end{array}$ ‍

The fundamental reason for writing things in terms of

y

instead of

x

is that it's easier to think of moving along this curve from bottom to top, rather than from left to right. In general, and this is an important principle to remember, when you are integrating along a curve, the formulas and symbols you use should reflect how you would actually walk along the curve.

Example 4: Full generality

Suppose you have any arbitrary function

f (x)

, with derivative

f^{'} (x)

. Which of the following represents the arc length of the graph

$y = f (x)$ ‍

between the points

x = a

and

x = b

Oftentimes people will start teaching arc length by presenting this formula. Personally, I think that takes all the fun out of discovering it yourself and getting a genuine feel for what it really represents.

Summary

We use the words arc length to describe how long a curve is. If you imagine the curve as a string, this is the length of that string once you pull it taut.
You can find the arc length of a curve with an integral of the form
$\begin{array}{r} \int \sqrt{(d x)^{2} + (d y)^{2}} \end{array}$ ‍
If the curve is the graph of a function $y = f (x)$ ‍, replace the $d y$ ‍ term in the integral with $f^{'} (x) d x$ ‍, then factor out the $d x$ ‍. The boundary values of the integral will be the leftmost and rightmost $x$ ‍-values of the curve.
When setting up an arc length integral, it helps to think about how you would actually choose to walk along the curve.

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Sort by:

Aryan Jamali
Posted 8 years ago. Direct link to Aryan Jamali's post “In the first example with...”
In the first example with the unit circle I don't get why x = -cos(teta) isn't the x-coordinate simply cos(teta) as the trigonometric coordinate are given by the point x =(x1,y1)=(cos(teta),sin(teta)) ? :)
Button navigates to signup pageButton navigates to signup page
(11 votes)
Answer
- jjakow
  Posted 6 years ago. Direct link to jjakow's post “By convention, we think o...”
  By convention, we think of the unit circle as starting at the point (1,0). This corresponds to a
  x =cos(theta) parametrization. However, since the integral travels from the lower limit (-1,0) to the upper limit (1,0) we negate the cosine to switch direction of travel. Hope this helps
  Button navigates to signup page
  (6 votes)
MichaelSuitang
Posted 8 years ago. Direct link to MichaelSuitang's post “Haven't we learned about ...”
Haven't we learned about arc length in ordinary integral classes previously? Why are we studying this again?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Stefen
  Posted 8 years ago. Direct link to Stefen's post “It is a review leading to...”
  It is a review leading to Arc length of parametric curves which leads to line integrals in scalar fields which leads to line integrals in vector fields and on it goes.
  Sometimes a little refresher review on a concept helps when a new topic depends on it.
  Button navigates to signup page
  (14 votes)
jp2338
Posted a year ago. Direct link to jp2338's post “For the part of x = -cos(...”
For the part of x = -cos(theta), I think a better wway to describe theta is the angle between the radius and the positive side of x-axis ( in the picture it is the angle between the radius and negative side of x-axis). And then the angle between the radius and negative side of x-axis (i.e the theta in current picture) would be pi-theta, then x = cos(pi-theta) = -cos(theta). Changing the definition of theta would be easier for student to know, otherwise it seems x=-cos(theta) comes from nowhere.
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(2 votes)
Answer
Victor Gutierrez
Posted 6 years ago. Direct link to Victor Gutierrez's post “First example, step 3, ho...”
First example, step 3, how to solve the integral using trigonometric substitution. This method is only valid for integrals of arc lengths right? Because if I find that integral in other contexts, I couldn´t think of the function inside the integral as a semicircle, because it is not a semicircle, although in the explanation is treated as a semicircle. So I don´t understand this method for solving these kind of integrals.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- jjakow
  Posted 6 years ago. Direct link to jjakow's post “This cannot be solved by ...”
  This cannot be solved by hand. This article and previous articles show how to set up the integral not how to solve this. The integrala in example 3 can only be approximated
  Button navigates to signup page
  (2 votes)
bluemoon29yume
Posted a year ago. Direct link to bluemoon29yume's post “in example 1, step 1, I d...”
in example 1, step 1, I don't understand where did the 2 come from in the denominator? in the d(1-x^2)/2(square root of /(1-x^2)
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Venkata
  Posted a year ago. Direct link to Venkata's post “It comes from the derivat...”
  It comes from the derivative of sqrt(1-x^(2)). Take a simpler example. Let's say we want the derivative of sqrt(x). This is just x^(1/2). Using the power rule, we get (1/2)x^(-1/2) or 1/(2sqrt(x)). The example in the article is the same, but as the inside of the square is a function, you need the chain rule.
  Button navigates to signup page
  (2 votes)
jimlkport
Posted 4 years ago. Direct link to jimlkport's post “I put in 1+4y^2 instead o...”
I put in 1+4y^2 instead of 4y^2+1 and got wrong answer! Nice going........
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(0 votes)
Answer

Course: Multivariable calculus > Unit 4

Arc length of function graphs, examples

Background

Example 1: Practice with a semicircle

Step 1: Write $d y$ ‍ in terms of $d x$ ‍

Step 2: Replace $d y$ ‍ in the integral

Step 3: Place bounds on the integral and solve

Practice setting up arc length integrals

Example 2: Sine curve

Example 3: Up, not right

Step 1: Write $d x$ ‍ in terms of $d y$ ‍

Step 2: Plug this value of $d x$ ‍ into the integral

Step 3: Place bounds on the integral

Example 4: Full generality

Step 1: Write $d y$ ‍ in terms of $d x$ ‍

Step 2: Plug this expression for $d y$ ‍ into the arc length integral

Step 3: Put bounds on the integral

Summary

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Multivariable calculus

Course: Multivariable calculus > Unit 4

Background

Example 1: Practice with a semicircle

Step 1: Write dy‍ in terms of dx‍

Step 2: Replace dy‍ in the integral

Step 3: Place bounds on the integral and solve

Practice setting up arc length integrals

Example 2: Sine curve

Example 3: Up, not right

Example​ 4: Full generality

Summary

Want to join the conversation?

Step 1: Write $d y$ ‍ in terms of $d x$ ‍

Step 2: Replace $d y$ ‍ in the integral

Example 4: Full generality