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## Multivariable calculus

### Course: Multivariable calculus>Unit 4

Lesson 2: Line integrals for scalar functions (articles)

# Arc length of function graphs, examples

Practice finding the arc length of various function graphs.

## 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 $\pi$. Let's practice our newfound method of computing arc length to rediscover the length of a semicircle.
By definition, all points $\left(x,y\right)$ 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:
$\begin{array}{r}\int \sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}\end{array}$
Just as before, we think of the integrand $\sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}$ 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 $dy$‍  in terms of $dx$‍

Use the fact that $y=\sqrt{1-{x}^{2}}$ to write $dy$ in terms of $dx$.
$dy=$
$dx$

### Step 2: Replace $dy$‍  in the integral

Plug this expression for $dy$ into the integral to write the integrand completely in terms of $x$ and $dx$.
$\int$
$dx$

### 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 $\pi$, but the interesting part is to work through it to see how $\pi$ pops out when using an arc length integral.)

## 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 $y=\mathrm{sin}\left(x\right)$ between $x=0$ and $x=2\pi$?
$\begin{array}{r}{\int }_{a}^{b}\end{array}$
$dx$
$a=$
$b=$

## Example 3: Up, not right

Consider the curve representing
$y=±\sqrt{x}$
For all values where $x\le 4$. Find an integral expressing this curve's arc length. But this time, write everything in the integral in terms of $y$, not $x$.
$\begin{array}{r}{\int }_{a}^{b}\end{array}$
$dy$
$a=$
$b=$

## Example​ 4: Full generality

Suppose you have any arbitrary function $f\left(x\right)$, with derivative ${f}^{\prime }\left(x\right)$. Which of the following represents the arc length of the graph
$y=f\left(x\right)$
between the points $x=a$ and $x=b$?
$\begin{array}{r}\int \sqrt{\left(dx{\right)}^{2}+\left(dy{\right)}^{2}}\end{array}$
• If the curve is the graph of a function $y=f\left(x\right)$, replace the $dy$ term in the integral with ${f}^{\prime }\left(x\right)dx$, then factor out the $dx$. The boundary values of the integral will be the leftmost and rightmost $x$-values of the curve.