Arc Length of the Curve [latex]y[/latex] = [latex]f[/latex]([latex]x[/latex])

In previous applications of integration, we required the function [latex]f(x)[/latex] to be integrable, or at most continuous. However, for calculating arc length we have a more stringent requirement for [latex]f(x).[/latex] Here, we require [latex]f(x)[/latex] to be differentiable, and furthermore we require its derivative, [latex]{f}^{\prime }(x),[/latex] to be continuous. Functions like this, which have continuous derivatives, are called smooth. (This property comes up again in later chapters.)

Let [latex]f(x)[/latex] be a smooth function defined over [latex]\left[a,b\right].[/latex] We want to calculate the length of the curve from the point [latex](a,f(a))[/latex] to the point [latex](b,f(b)).[/latex] We start by using line segments to approximate the length of the curve. For [latex]i=0,1,2\text{,…},n,[/latex] let [latex]P=\left\{{x}_{i}\right\}[/latex] be a regular partition of [latex]\left[a,b\right].[/latex] Then, for [latex]i=1,2\text{,…},n,[/latex] construct a line segment from the point [latex]({x}_{i-1},f({x}_{i-1}))[/latex] to the point [latex]({x}_{i},f({x}_{i})).[/latex] 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. (Figure) depicts this construct for [latex]n=5.[/latex]

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 [latex]\text{Δ}x.[/latex] The change in vertical distance varies from interval to interval, though, so we use [latex]\text{Δ}{y}_{i}=f({x}_{i})-f({x}_{i-1})[/latex] to represent the change in vertical distance over the interval [latex]\left[{x}_{i-1},{x}_{i}\right],[/latex] as shown in (Figure). Note that some (or all) [latex]\text{Δ}{y}_{i}[/latex] may be negative.

By the Pythagorean theorem, the length of the line segment is [latex]\sqrt{{(\text{Δ}x)}^{2}+{(\text{Δ}{y}_{i})}^{2}}.[/latex] We can also write this as [latex]\text{Δ}x\sqrt{1+{((\text{Δ}{y}_{i})\text{/}(\text{Δ}x))}^{2}}.[/latex] Now, by the Mean Value Theorem, there is a point [latex]{x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right][/latex] such that [latex]{f}^{\prime }({x}_{i}^{*})=(\text{Δ}{y}_{i})\text{/}(\text{Δ}x).[/latex] Then the length of the line segment is given by [latex]\text{Δ}x\sqrt{1+{\left[{f}^{\prime }({x}_{i}^{*})\right]}^{2}}.[/latex] Adding up the lengths of all the line segments, we get

This is a Riemann sum. Taking the limit as [latex]n\to \infty ,[/latex] we have

We summarize these findings in the following theorem.

Note that we are integrating an expression involving [latex]{f}^{\prime }(x),[/latex] so we need to be sure [latex]{f}^{\prime }(x)[/latex] is integrable. This is why we require [latex]f(x)[/latex] to be smooth. The following example shows how to apply the theorem.

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 the second volume of this text. In some cases, we may have to use a computer or calculator to approximate the value of the integral.

Arc Length of the Curve [latex]x[/latex] = [latex]g[/latex]([latex]y[/latex])

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 [latex]y,[/latex] we can repeat the same process, except we partition the [latex]y\text{-axis}[/latex] instead of the [latex]x\text{-axis}.[/latex] (Figure) shows a representative line segment.

Then the length of the line segment is [latex]\sqrt{{(\text{Δ}y)}^{2}+{(\text{Δ}{x}_{i})}^{2}},[/latex] which can also be written as [latex]\text{Δ}y\sqrt{1+{((\text{Δ}{x}_{i})\text{/}(\text{Δ}y))}^{2}}.[/latex] If we now follow the same development we did earlier, we get a formula for arc length of a function [latex]x=g(y).[/latex]

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 [latex]f(x)[/latex] be a nonnegative smooth function over the interval [latex]\left[a,b\right].[/latex] We wish to find the surface area of the surface of revolution created by revolving the graph of [latex]y=f(x)[/latex] around the [latex]x\text{-axis}[/latex] as shown in the following figure.

As we have done many times before, we are going to partition the interval [latex]\left[a,b\right][/latex] 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 [latex]i=0,1,2\text{,…},n,[/latex] let [latex]P=\left\{{x}_{i}\right\}[/latex] be a regular partition of [latex]\left[a,b\right].[/latex] Then, for [latex]i=1,2\text{,…},n,[/latex] construct a line segment from the point [latex]({x}_{i-1},f({x}_{i-1}))[/latex] to the point [latex]({x}_{i},f({x}_{i})).[/latex] Now, revolve these line segments around the [latex]x\text{-axis}[/latex] 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, [latex]S,[/latex] 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 [latex]{r}_{1}[/latex] and [latex]{r}_{2}[/latex] be the radii of the wide end and the narrow end of the frustum, respectively, and let [latex]l[/latex] be the slant height of the frustum as shown in the following figure.

We know the lateral surface area of a cone is given by

where [latex]r[/latex] is the radius of the base of the cone and [latex]s[/latex] is the slant height (see the following figure).

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

Solving for [latex]s,[/latex] we get

Then the lateral surface area (SA) of the frustum is

Let’s now use this formula to calculate the surface area of each of the bands formed by revolving the line segments around the [latex]x\text{-axis}\text{.}[/latex] A representative band is shown in the following figure.

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 [latex]{x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right][/latex] such that [latex]{f}^{\prime }({x}_{i}^{*})=(\text{Δ}{y}_{i})\text{/}\text{Δ}x.[/latex] This gives us

Furthermore, since [latex]f(x)[/latex] is continuous, by the Intermediate Value Theorem, there is a point [latex]{x}_{i}^{**}\in \left[{x}_{i-1},{x}_{i}\right][/latex] such that [latex]f({x}_{i}^{**})=(1\text{/}2)\left[f({x}_{i-1})+f({x}_{i})\right],[/latex] so we get

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, [latex]{x}_{i}^{*}[/latex] and [latex]{x}_{i}^{**},[/latex] over the interval [latex]\left[{x}_{i-1},{x}_{i}\right].[/latex] Although we do not examine the details here, it turns out that because [latex]f(x)[/latex] is smooth, if we let [latex]n\to \infty ,[/latex] the limit works the same as a Riemann sum even with the two different evaluation points. This makes sense intuitively. Both [latex]{x}_{i}^{*}[/latex] and [latex]{x}_{i}^{**}[/latex] are in the interval [latex]\left[{x}_{i-1},{x}_{i}\right],[/latex] so it makes sense that as [latex]n\to \infty ,[/latex] both [latex]{x}_{i}^{*}[/latex] and [latex]{x}_{i}^{**}[/latex] approach [latex]x.[/latex] Those of you who are interested in the details should consult an advanced calculus text.

Taking the limit as [latex]n\to \infty ,[/latex] we get

As with arc length, we can conduct a similar development for functions of [latex]y[/latex] to get a formula for the surface area of surfaces of revolution about the [latex]y\text{-axis}.[/latex] These findings are summarized in the following theorem.

Calculating the Surface Area of a Surface of Revolution 1

Let [latex]f(x)=\sqrt{x}[/latex] over the interval [latex]\left[1,4\right].[/latex] Find the surface area of the surface generated by revolving the graph of [latex]f(x)[/latex] around the [latex]x\text{-axis}.[/latex] Round the answer to three decimal places.

Show Solution

The graph of [latex]f(x)[/latex] and the surface of rotation are shown in the following figure.

(a) The graph of [latex]f(x).[/latex] (b) The surface of revolution.

We have [latex]f(x)=\sqrt{x}.[/latex] Then, [latex]{f}^{\prime }(x)=1\text{/}(2\sqrt{x})[/latex] and [latex]{({f}^{\prime }(x))}^{2}=1\text{/}(4x).[/latex] Then,

[latex]\begin{array}{cc}\hfill \text{Surface Area}& ={\int }_{a}^{b}(2\pi f(x)\sqrt{1+{({f}^{\prime }(x))}^{2}})dx\hfill \\ & ={\int }_{1}^{4}(2\pi \sqrt{x}\sqrt{1+\frac{1}{4x}})dx\hfill \\ & ={\int }_{1}^{4}(2\pi \sqrt{x+\frac{1}{4}})dx.\hfill \end{array}[/latex]

Let [latex]u=x+1\text{/}4.[/latex] Then, [latex]du=dx.[/latex] When [latex]x=1,[/latex] [latex]u=5\text{/}4,[/latex] and when [latex]x=4,[/latex] [latex]u=17\text{/}4.[/latex] This gives us

[latex]\begin{array}{cc}\hfill {\int }_{0}^{1}(2\pi \sqrt{x+\frac{1}{4}})dx& ={\int }_{5\text{/}4}^{17\text{/}4}2\pi \sqrt{u}du\hfill \\ & =2\pi {\left[\frac{2}{3}{u}^{3\text{/}2}\right]|}_{5\text{/}4}^{17\text{/}4}=\frac{\pi }{6}\left[17\sqrt{17}-5\sqrt{5}\right]\approx 30.846.\hfill \end{array}[/latex]

Calculating the Surface Area of a Surface of Revolution 2

Let [latex]f(x)=y=\sqrt[3]{3x}.[/latex] Consider the portion of the curve where [latex]0\le y\le 2.[/latex] Find the surface area of the surface generated by revolving the graph of [latex]f(x)[/latex] around the [latex]y\text{-axis}.[/latex]

Show Solution

Notice that we are revolving the curve around the [latex]y\text{-axis},[/latex] and the interval is in terms of [latex]y,[/latex] so we want to rewrite the function as a function of [latex]y[/latex]. We get [latex]x=g(y)=(1\text{/}3){y}^{3}.[/latex] The graph of [latex]g(y)[/latex] and the surface of rotation are shown in the following figure.

(a) The graph of [latex]g(y).[/latex] (b) The surface of revolution.

We have [latex]g(y)=(1\text{/}3){y}^{3},[/latex] so [latex]{g}^{\prime }(y)={y}^{2}[/latex] and [latex]{({g}^{\prime }(y))}^{2}={y}^{4}.[/latex] Then

[latex]\begin{array}{cc}\hfill \text{Surface Area}& ={\int }_{c}^{d}(2\pi g(y)\sqrt{1+{({g}^{\prime }(y))}^{2}})dy\hfill \\ & ={\int }_{0}^{2}(2\pi (\frac{1}{3}{y}^{3})\sqrt{1+{y}^{4}})dy\hfill \\ & =\frac{2\pi }{3}{\int }_{0}^{2}({y}^{3}\sqrt{1+{y}^{4}})dy.\hfill \end{array}[/latex]

Let [latex]u={y}^{4}+1.[/latex] Then [latex]du=4{y}^{3}dy.[/latex] When [latex]y=0,[/latex] [latex]u=1,[/latex] and when [latex]y=2,[/latex] [latex]u=17.[/latex] Then

[latex]\begin{array}{cc}\hfill \frac{2\pi }{3}{\int }_{0}^{2}({y}^{3}\sqrt{1+{y}^{4}})dy& =\frac{2\pi }{3}{\int }_{1}^{17}\frac{1}{4}\sqrt{u}du\hfill \\ & =\frac{\pi }{6}{\left[\frac{2}{3}{u}^{3\text{/}2}\right]|}_{1}^{17}=\frac{\pi }{9}\left[{(17)}^{3\text{/}2}-1\right]\approx 24.118.\hfill \end{array}[/latex]