Welcome back! Over the past few lessons, we’ve seen that certain calculations like area and volume can be performed using integrals, which are limits of sums of easier area/volume calculations. In our last lesson today, we’ll follow this same thinking to use integrals as limits of sums to calculate lengths of curves and surface areas of solids.

Lesson 25: Arc Length of a Curve and Surface Area

Learning goals:

  • Determine the length of a curve given by the graph of a function between two points
  • Determine the surface area of a solid of revolution.

Topic:

WeBWorK:

  • Applications – Arc Length
  • Applications – Surface Area

Motivating question

How can we use integration to find the length of a curve in the plane or the surface area of a solid of revolution?

Warmup exercise 1

Evaluate the integral:

\[\int_1^2 \sqrt{1+16x^2}dx\]

Show answer 1

Use $u$-substitution and trigonometric substitution. The final answer is about 6.0859. See Video 1 below to see what the integral is used for and for a worked solution.

Main idea

For areas

We saw in Lesson 21 (link here) that the way we can calculate areas of irregularly shaped regions is by

  1. chopping them up into bits,
  2. finding the approximate areas of the bits,
  3. adding them up, and
  4. taking a limit.

The result is an integral where the integrand is the length of a segment (often this is the height of a function)

\[A = \int_a^b f(x) dx\]

For volumes

We saw in Lesson 23 (link here) that the way we can calculate volumes of solids is by

  1. chopping them up into bits,
  2. finding the approximate volumes of the bits,
  3. adding them up, and
  4. taking a limit.

The result is an integral where the integrand is the area of a cross section (often this is a flat disk or washer or a cylindrical shell)

\[A = \int_a^b A(x) dx\]

Now we apply this main idea to lengths of curves (arc length) and to surface areas of solids of revolution.

Main idea for arc length

For arc length, we follow the same steps from the main idea above:

  1. chop the curve into bits,
  2. approximate the length of each bit by the length of a line segment joining the two endpoints (using the distance formula, which is really just the Pythagorean theorem $d = \sqrt{(x_2 – x_1)^2 + (y_2-y_1)^2}$),
  3. add up the lengths of the line segments, and
  4. take a limit.

The result is an integral that looks like:

Theorem: Let $y= f(x)$ be the graph of a function $f(x)$ which is differentiable on the interval $[a,b]$. The length of the curve between $x=a$ and $x=b$ is given by

\[ {\rm arc \; length} = \int_a^b \sqrt{1 + (f'(x))^2}dx\]

Video 1 shows where this formula comes from and how to apply it in an example.

Note that the video and many other sources use the lower case letter $s$ to represent arc length. Don’t confuse this with the upper case $S$, which the next videos use to represent surface area.

Main idea for surface area

Again, for surface area of a solid of revolution we apply the main idea above:

  1. chop the solid into bits perpendicular to the axis of rotation,
  2. approximate the surface area of each bit by the surface area of a band or strip whose shape is the shape of the base of a cone,
  3. add up the surface areas of these bands, and
  4. take a limit.

The result is an integral that looks like:

Theorem: Let $y= f(x)$ be the graph of a non-negative function $f(x)$ which is differentiable on the interval $[a,b]$. Consider the solid obtained by rotating this graph around the $x$-axis. The surface area of this solid is given by

\[ {\rm surface \; area} = \int_a^b 2 \pi f(x) \sqrt{1 + (f'(x))^2}dx\]

Video 2 shows where this formula comes from and how to apply it in an example.

Video 2

Video 3 below shows how to apply this formula to find the surface area of a solid which is obtained by rotating a function of $y$ about the $y$-axis.

Video 3

Exit ticket

Find the length of the curve given by the graph of $y=-\frac{1}{2}x + 25$ on the interval $[1,4]$.