Arc Length of a Curve and Surface Area

In this section, we use definite integrals to find the arc length of a curve. We can think of arc length as the distance you would travel if you were walking along the path of the curve. Many real-world applications involve arc length. If a rocket is launched along a parabolic path, we might want to know how far the rocket travels. Or, if a curve on a map represents a road, we might want to know how far we have to drive to reach our destination.

reversed.) The techniques we use to find arc length can be extended to find the surface area of a surface of revolution, and we close the section with an examination of this concept.

Arc Length of the Curve y = f(x)

to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for

f (x) .

to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.)

We start by using line segments to approximate the length of the curve. For

i = 0, 1, 2,…, n,

Although it might seem logical to use either horizontal or vertical line segments, we want our line segments to approximate the curve as closely as possible. [link] depicts this construct for

n = 5 .

To help us find the length of each line segment, we look at the change in vertical distance as well as the change in horizontal distance over each interval. Because we have used a regular partition, the change in horizontal distance over each interval is given by

Δ x .

The change in vertical distance varies from interval to interval, though, so we use

Δ y_{i} = f (x_{i}) - f (x_{i - 1})

to represent the change in vertical distance over the interval

[x_{i - 1}, x_{i}],

By the Pythagorean theorem, the length of the line segment is

\sqrt{{(Δ x)}^{2} + {(Δ y_{i})}^{2}} .

Now, by the Mean Value Theorem, there is a point

x_{i}^{*} \in [x_{i - 1}, x_{i}]

Then the length of the line segment is given by

Δ x \sqrt{1 + {[f^{'} (x_{i}^{*})]}^{2}} .

Although it is nice to have a formula for calculating arc length, this particular theorem can generate expressions that are difficult to integrate. We study some techniques for integration in Introduction to Techniques of Integration. In some cases, we may have to use a computer or calculator to approximate the value of the integral.

Arc Length of the Curve x = g(y)

We have just seen how to approximate the length of a curve with line segments. If we want to find the arc length of the graph of a function of

y,

Then the length of the line segment is

\sqrt{{(Δ y)}^{2} + {(Δ x_{i})}^{2}},

If we now follow the same development we did earlier, we get a formula for arc length of a function

x = g (y) .

Area of a Surface of Revolution

The concepts we used to find the arc length of a curve can be extended to find the surface area of a surface of revolution. Surface area is the total area of the outer layer of an object. For objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces. For curved surfaces, the situation is a little more complex. Let

f (x)

We wish to find the surface area of the surface of revolution created by revolving the graph of

y = f (x)

As we have done many times before, we are going to partition the interval

[a, b]

and approximate the surface area by calculating the surface area of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For

i = 0, 1, 2,…, n,

to generate an approximation of the surface of revolution as shown in the following figure.

Notice that when each line segment is revolved around the axis, it produces a band. These bands are actually pieces of cones (think of an ice cream cone with the pointy end cut off). A piece of a cone like this is called a frustum of a cone.

To find the surface area of the band, we need to find the lateral surface area,

S,

of the frustum (the area of just the slanted outside surface of the frustum, not including the areas of the top or bottom faces). Let

r_{1}

be the radii of the wide end and the narrow end of the frustum, respectively, and let

l

Since a frustum can be thought of as a piece of a cone, the lateral surface area of the frustum is given by the lateral surface area of the whole cone less the lateral surface area of the smaller cone (the pointy tip) that was cut off (see the following figure).

The cross-sections of the small cone and the large cone are similar triangles, so we see that

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the

x -axis .

Note that the slant height of this frustum is just the length of the line segment used to generate it. So, applying the surface area formula, we have

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select

x_{i}^{*} \in [x_{i - 1}, x_{i}]

is continuous, by the Intermediate Value Theorem, there is a point

x_{i}^{* *} \in [x_{i - 1}, x_{i}]

Then the approximate surface area of the whole surface of revolution is given by

This almost looks like a Riemann sum, except we have functions evaluated at two different points,

x_{i}^{*}

the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both

x_{i}^{*}

Those of you who are interested in the details should consult an advanced calculus text.

to get a formula for the surface area of surfaces of revolution about the

y -axis .

Key Concepts

Key Equations

For the following exercises, find the length of the functions over the given interval.

y = 5 x from x = 0 to x = 2

2 \sqrt{26}

y = - \frac{1}{2} x + 25 from x = 1 to x = 4

x = 4 y from y = −1 to y = 1

2 \sqrt{17}

Pick an arbitrary linear function $x = g (y)$

over any interval of your choice $(y_{1}, y_{2}) .$

Determine the length of the function and then prove the length is correct by using geometry.

Find the surface area of the volume generated when the curve $y = \sqrt{x}$

revolves around the $x -axis$

from $(1, 1)$

to $(4, 2),$

as seen here.

This figure is a surface. It has been formed by rotating the curve y=squareroot(x) about the x-axis. The surface is inside of a cube to show 3-dimensions.

\frac{π}{6} (17 \sqrt{17} - 5 \sqrt{5})

Find the surface area of the volume generated when the curve $y = x^{2}$

revolves around the $y -axis$

from $(1, 1)$

to $(3, 9) .$

This figure is a surface. It has an elliptical shape to the top, forming a “bowl”.

For the following exercises, find the lengths of the functions of $x$

over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

y = x^{3 / 2}

from $(0, 0) to (1, 1)$

\frac{13 \sqrt{13} - 8}{27}

y = x^{2 / 3}

from $(1, 1) to (8, 4)$

y = \frac{1}{3} {(x^{2} + 2)}^{3 / 2}

from $x = 0 to x = 1$

\frac{4}{3}

y = \frac{1}{3} {(x^{2} - 2)}^{3 / 2}

from $x = 2$

to $x = 4$

[T] $y = e^{x}$

on $x = 0$

to $x = 1$

2.0035

y = \frac{x^{3}}{3} + \frac{1}{4 x}

from $x = 1 to x = 3$

y = \frac{x^{4}}{4} + \frac{1}{8 x^{2}}

from $x = 1 to x = 2$

\frac{123}{32}

y = \frac{2 x^{3 / 2}}{3} - \frac{x^{1 / 2}}{2}

from $x = 1 to x = 4$

y = \frac{1}{27} {(9 x^{2} + 6)}^{3 / 2}

from $x = 0 to x = 2$

10

[T] $y = sin x$

on $x = 0 to x = π$

For the following exercises, find the lengths of the functions of $y$

over the given interval. If you cannot evaluate the integral exactly, use technology to approximate it.

y = \frac{5 - 3 x}{4}

from $y = 0$

to $y = 4$

\frac{20}{3}

x = \frac{1}{2} (e^{y} + e^{− y})

from $y = −1 to y = 1$

x = 5 y^{3 / 2}

from $y = 0$

to $y = 1$

\frac{1}{675} (229 \sqrt{229} - 8)

[T] $x = y^{2}$

from $y = 0$

to $y = 1$

x = \sqrt{y}

from $y = 0 to y = 1$

\frac{1}{8} (4 \sqrt{5} + ln (9 + 4 \sqrt{5}))

x = \frac{2}{3} {(y^{2} + 1)}^{3 / 2}

from $y = 1$

to $y = 3$

[T] $x = tan y$

from $y = 0$

to $y = \frac{3}{4}$

1.201

[T] $x = {cos}^{2} y$

from $y = - \frac{π}{2}$

to $y = \frac{π}{2}$

[T] $x = 4^{y}$

from $y = 0 to y = 2$

15.2341

[T] $x = ln (y)$

on $y = \frac{1}{e}$

to $y = e$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $x -axis .$

If you cannot evaluate the integral exactly, use your calculator to approximate it.

y = \sqrt{x}

from $x = 2$

to $x = 6$

\frac{49 π}{3}

y = x^{3}

from $x = 0$

to $x = 1$

y = 7 x

from $x = −1 to x = 1$

70 π \sqrt{2}

[T] $y = \frac{1}{x^{2}}$

from $x = 1 to x = 3$

y = \sqrt{4 - x^{2}}

from $x = 0 to x = 2$

8 π

y = \sqrt{4 - x^{2}}

from $x = −1 to x = 1$

y = 5 x

from $x = 1 to x = 5$

120 π \sqrt{26}

[T] $y = tan x$

from $x = - \frac{π}{4} to x = \frac{π}{4}$

For the following exercises, find the surface area of the volume generated when the following curves revolve around the $y -axis .$

If you cannot evaluate the integral exactly, use your calculator to approximate it.

y = x^{2}

from $x = 0 to x = 2$

\frac{π}{6} (17 \sqrt{17} - 1)

y = \frac{1}{2} x^{2} + \frac{1}{2}

from $x = 0 to x = 1$

y = x + 1

from $x = 0 to x = 3$

9 \sqrt{2} π

[T] $y = \frac{1}{x}$

from $x = \frac{1}{2}$

to $x = 1$

y = \sqrt[3]{x}

from $x = 1 to x = 27$

\frac{10 \sqrt{10} π}{27} (73 \sqrt{73} - 1)

[T] $y = 3 x^{4}$

from $x = 0$

to $x = 1$

[T] $y = \frac{1}{\sqrt{x}}$

from $x = 1$

to $x = 3$

25.645

[T] $y = cos x$

from $x = 0$

to $x = \frac{π}{2}$

The base of a lamp is constructed by revolving a quarter circle $y = \sqrt{2 x - x^{2}}$

around the $y -axis$

from $x = 1$

to $x = 2,$

as seen here. Create an integral for the surface area of this curve and compute it.

This figure is a surface. It is half of a torus created by rotating the curve y=squareroot(2x-x^2) about the x-axis.

2 π

A light bulb is a sphere with radius $1 / 2$

in. with the bottom sliced off to fit exactly onto a cylinder of radius $1 / 4$

in. and length $1 / 3$

in., as seen here. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is $1 / 4$

in. Find the surface area (not including the top or bottom of the cylinder).

This figure has two images. The first is a sphere on top of a cylinder. The second is a lightbulb.

[T] A lampshade is constructed by rotating $y = 1 / x$

around the $x -axis$

from $y = 1$

to $y = 2,$

as seen here. Determine how much material you would need to construct this lampshade—that is, the surface area—accurate to four decimal places.

This figure has two images. The first is similar to a frustum of a cone with edges bending inwards. The second is a lamp shade.

10.5017

[T] An anchor drags behind a boat according to the function $y = 24 e^{− x / 2} - 24,$

where $y$

represents the depth beneath the boat and $x$

is the horizontal distance of the anchor from the back of the boat. If the anchor is $23$

ft below the boat, how much rope do you have to pull to reach the anchor? Round your answer to three decimal places.

[T] You are building a bridge that will span $10$

ft. You intend to add decorative rope in the shape of $y = 5 \| sin ((x π) / 5) \|,$

where $x$

is the distance in feet from one end of the bridge. Find out how much rope you need to buy, rounded to the nearest foot.

23

For the following exercises, find the exact arc length for the following problems over the given interval.

y = ln (sin x)

from $x = π / 4$

to $x = (3 π) / 4 .$

(Hint: Recall trigonometric identities.)

Draw graphs of $y = x^{2},$

y = x^{6},

and $y = x^{10} .$

For $y = x^{n},$

as $n$

increases, formulate a prediction on the arc length from $(0, 0)$

to $(1, 1) .$

Now, compute the lengths of these three functions and determine whether your prediction is correct.

2

Compare the lengths of the parabola $x = y^{2}$

and the line $x = b y$

from $(0, 0) to (b^{2}, b)$

as $b$

increases. What do you notice?

Solve for the length of $x = y^{2}$

from $(0, 0) to (1, 1) .$

Show that $x = (1 / 2) y^{2}$

from $(0, 0)$

to $(2, 2)$

is twice as long. Graph both functions and explain why this is so.

Answers may vary

[T] Which is longer between $(1, 1)$

and $(2, 1 / 2) :$

the hyperbola $y = 1 / x$

or the graph of $x + 2 y = 3 ?$

Explain why the surface area is infinite when $y = 1 / x$

is rotated around the $x -axis$

for $1 \leq x < \infty,$

but the volume is finite.

For more information, look up Gabriel’s Horn.

Arc Length of a Curve and Surface Area

Arc Length of the Curve y = f(x)

Arc Length of the Curve x = g(y)

Area of a Surface of Revolution

Key Concepts

Key Equations

Glossary