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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 parenthesis, x, comma, y, right parenthesis on the circle are a distance 1 from the origin, so we have
x, squared, plus, y, squared, equals, 1
Rearranging to write y as a function of x, we have
y, equals, square root of, 1, minus, x, squared, end square root
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{aligned} \int \sqrt{(dx)^2 + (dy)^2} \end{aligned}
Just as before, we think of the integrand square root of, left parenthesis, d, x, right parenthesis, squared, plus, left parenthesis, d, y, right parenthesis, squared, end square root 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 $dy$d, y in terms of $dx$d, x

Use the fact that y, equals, square root of, 1, minus, x, squared, end square root to write d, y in terms of d, x.
d, y, equals
d, x

### Step 2: Replace $dy$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.
integral
d, x

### Step 3: Place bounds on the integral and solve

Since the curve is defined between when x, equals, minus, 1 and x, equals, 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, equals, sine, left parenthesis, x, right parenthesis between x, equals, 0 and x, equals, 2, pi?
\begin{aligned} \int_a^b \end{aligned}
d, x
a, equals
b, equals

## Example 3: Up, not right

Consider the curve representing
y, equals, plus minus, square root of, x, end square root
For all values where x, is less than or equal to, 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{aligned} \int_a^b \end{aligned}
d, y
a, equals
b, equals

## Example​ 4: Full generality

Suppose you have any arbitrary function f, left parenthesis, x, right parenthesis, with derivative f, prime, left parenthesis, x, right parenthesis. Which of the following represents the arc length of the graph
y, equals, f, left parenthesis, x, right parenthesis
between the points x, equals, a and x, equals, b?
\begin{aligned} \int \sqrt{(dx)^2 + (dy)^2} \end{aligned}