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2.4 Arc Length of a Curve and Surface Area

Learning Objectives

  • Determine the length of a curve with equation \ds y=f(x) between the two given points.
  • Determine the length of a curve with equation \ds x=g(y) between the two given points.
  • Find the surface area of a surface of revolution.

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.

We begin by calculating the arc length of curves defined as functions of \ds x, then we examine the same process for curves defined as functions of \ds y. (The process is identical, with the roles of \ds x and \ds y 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)

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

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

This figure is a graph in the first quadrant. The curve increases and decreases. It is divided into parts at the points a=xsub0, xsub1, xsub2, xsub3, xsub4, and xsub5=b. Also, there are line segments between the points on the curve.
Figure 1. We can approximate the length of a curve by adding line segments.

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 \ds \Delta x. The change in vertical distance varies from interval to interval, though, so we use \ds \Delta {y}_{i}=f({x}_{i})-f({x}_{i-1}) to represent the change in vertical distance over the interval \ds \left[{x}_{i-1},{x}_{i}\right], as shown in Figure 2 below. Note that some (or all) \ds \Delta {y}_{i} may be negative.

This figure is a graph. It is a curve above the x-axis beginning at the point f(xsubi-1). The curve ends in the first quadrant at the point f(xsubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.
Figure 2. A representative line segment approximates the curve over the interval \ds \left[{x}_{i-1},{x}_{i}\right].

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

\ds \text{Arc Length}\approx \underset{i=1}{\overset{n}{\sum}}\sqrt{1+{\left[{f}^{\prime }({x}_{i}^{*})\right]}^{2}}\Delta x.

This is a Riemann sum. Taking the limit as \ds n\to \infty , we have

\ds \text{Arc Length}=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\sum}}\sqrt{1+{\left[{f}^{\prime }({x}_{i}^{*})\right]}^{2}}\Delta x=\int\limits_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}\,dx .

We summarize these findings in the following theorem.

Arc Length for y=f(x)

Let \ds f(x) be a smooth function over the interval \ds \left[a,b\right]. Then the arc length of the portion of the graph of \ds f(x) from the point \ds (a,f(a)) to the point \ds (b,f(b)) is given by

\ds \text{Arc Length}=\int\limits_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}\,dx .

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

Calculating the Arc Length of a Function of x

Let \ds f(x)=2{x}^{3\text{/}2}. Calculate the arc length of the graph of \ds f(x) over the interval \ds \left[0,1\right].

Solution

We have \ds {f}^{\prime }(x)=3{x}^{1\text{/}2}, so \ds {\left[{f}^{\prime }(x)\right]}^{2}=9x. Then, the arc length is

\ds \begin{array}{cc}\ds \hfill \text{Arc Length}&\ds =\int\limits_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}\,dx \hfill \\[5mm]\ds &\ds =\int\limits_{0}^{1}\sqrt{1+9x}\,dx .\hfill \end{array}

Substitute \ds u=1+9x. Then, \ds du=9\,dx . When \ds x=0, then \ds u=1, and when \ds x=1, then \ds u=10. Thus,

\ds \begin{array}{cc}\ds \hfill \text{Arc Length}&\ds =\int\limits_{0}^{1}\sqrt{1+9x}\,dx \hfill \\[5mm]\ds &\ds =\frac{1}{9}\int\limits_{0}^{1}\sqrt{1+9x}\cdot 9\,\,dx =\frac{1}{9}\int\limits_{1}^{10}\sqrt{u}\,du\hfill \\[7mm]\ds &\ds ={\frac{1}{9}\cdot\frac{2}{3}{u}^{3\text{/}2}}\Big|_{1}^{10}=\frac{2}{27}\left[10\sqrt{10}-1\right]\text{units}.\hfill \end{array}

Let \ds f(x)=\frac43{x}^{3\text{/}2}. Calculate the arc length of the graph of \ds f(x) over the interval \ds \left[0,1\right].

Answer

\ds \frac{1}{6}(5\sqrt{5}-1)

Hint

Use the process from the previous example. Don’t forget to change the limits of integration.

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 the subsequent chapters of this text. Before we know them, we may have to stop after setting up the integral.

Setting Up the Integral for the Arc Length of a Function of x

Let \ds f(x)={x}^{2}. Set up the integral for the arc length of the graph of \ds f(x) over the interval \ds \left[1,3\right].

Solution

We have \ds {f}^{\prime }(x)=2x, so \ds {\left[{f}^{\prime }(x)\right]}^{2}=4{x}^{2}. Then the arc length is given by

\ds \text{Arc Length}=\int\limits_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}\,dx =\int\limits_{1}^{3}\sqrt{1+4{x}^{2}}\,dx .

Let \ds f(x)= \ln (x). Set up the integral for the arc length of the graph of \ds f(x) over the interval \ds \left[1,e \right].

Answer

\ds \text{Arc Length}=\int\limits_1^e\sqrt{1+\frac1{x^2}} \,dx

Hint

Use the process from the previous example.

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 \ds y, we can repeat the same process, except we partition the y-axis instead of the x-axis. Figure 3 below shows a representative line segment.

This figure is a graph. It is a curve to the right of the y-axis beginning at the point g(ysubi-1). The curve ends in the first quadrant at the point g(ysubi). Between the two points on the curve is a line segment. A right triangle is formed with this line segment as the hypotenuse, a horizontal segment with length delta x, and a vertical line segment with length delta y.
Figure 3. A representative line segment over the interval \ds \left[{y}_{i-1},{y}_{i}\right].

Then the length of the line segment is \ds \sqrt{{(\Delta y)}^{2}+{(\Delta {x}_{i})}^{2}}, which can also be written as \ds \sqrt{1+\left(\frac{\Delta {x}_{i}}{\Delta y}}\right)^{2}}\Delta y. If we now follow the analogous steps to what we have done before, we get a formula for arc length of a function \ds x=g(y).

Arc Length for x=g(y)

Let \ds g(y) be a smooth function over an interval \ds \left[c,d\right]. Then, the arc length of the graph of \ds g(y) from the point \ds (c,g(c)) to the point \ds (d,g(d)) is given by

\ds \text{Arc Length}=\int\limits_{c}^{d}\sqrt{1+{\left[{g}^{\prime }(y)\right]}^{2}}dy.

Setting up the Integral for the Arc Length of a Function of y

Let \ds g(y)=3{y}^{3}. Set up the integral for the arc length of the graph of \ds g(y) over the interval \ds \left[1,2\right].

Solution

We have \ds {g}^{\prime }(y)=9{y}^{2}, so \ds {\left[{g}^{\prime }(y)\right]}^{2}=81{y}^{4}. Then the arc length is

\ds \text{Arc Length}=\int\limits_{c}^{d}\sqrt{1+{\left[{g}^{\prime }(y)\right]}^{2}}dy=\int\limits_{1}^{2}\sqrt{1+81{y}^{4}}\,dy.

Let \ds g(y)=\frac1y. Set up the integral for the arc length of the graph of \ds g(y) over the interval \ds \left[1,4\right].

Answer

\ds \text{Arc Length}=\int\limits_1^4\sqrt{1+\frac1{y^4}}\,dy

Hint

Use the process from the previous example.

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

This figure has two graphs. The first graph is labeled “a” and is a curve in the first quadrant beginning at the y-axis. The curve is y=f(x). The second graph is labeled “b” and has the same curve y=f(x). There is also a solid surface formed by rotating the curve about the x-axis.
Figure 4. (a) A curve representing the function \ds f(x). (b) The surface of revolution formed by revolving the graph of \ds f(x) around the x-axis.

As we have done many times before, we are going to partition the interval \ds \left[a,b\right] and approximate the surface area by surface areas of simpler shapes. We start by using line segments to approximate the curve, as we did earlier in this section. For \ds i=0,1,2\text{,…},n, let \ds P=\left\{{x}_{i}\right\} be a regular partition of \ds \left[a,b\right]. Then, for \ds i=1,2\text{,…},n, construct a line segment from the point \ds ({x}_{i-1},f({x}_{i-1})) to the point \ds ({x}_{i},f({x}_{i})). Now, revolve these line segments around the x-axis to generate an approximation of the surface of revolution as shown in the following figure.

This figure has two graphs. The first graph is labeled “a” and is a curve in the first quadrant beginning at the y-axis. The curve is y=f(x). The area under the curve above the x-axis has been divided into regions with vertical lines. The second graph is labeled “b” and has the same curve y=f(x). There is also a solid surface formed by rotating the curve about the x-axis.
Figure 5. (a) Approximating \ds f(x) with line segments. (b) The surface of revolution formed by revolving the line segments around the x-axis.

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, \ds 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 \ds {r}_{1} and \ds {r}_{2} be the radii of the wide end and the narrow end of the frustum, respectively, and let \ds l be the slant height of the frustum as shown in the following figure.

This figure is a graph. It is a frustum of a cone above the x-axis with the y-axis in the center. The radius of the bottom of the frustum is rsub1 and the radius of the top is rsub2. The length of the side is labeled “l”.
Figure 6. A frustum of a cone can approximate a small part of surface area.

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

\ds \text{Lateral Surface Area}=\pi rs,

where \ds r is the radius of the base of the cone and \ds s is the slant height (see the following figure).

This figure is a cone. The cone has radius r, height h, and length of side s.
Figure 7. The lateral surface area of the cone is given by \ds \pi rs.

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).

This figure is a graph. It is a frustum of a cone. The radius of the bottom of the frustum is rsub1 and the radius of the top is rsub2. The length of the side is labeled “l”. There is also the top of a cone with broken lines above the frustum. It has length of side s.
Figure 8. Calculating the lateral surface area of a frustum of a cone.

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

\ds \frac{{r}_{2}}{{r}_{1}}=\frac{s-l}{s}.

Solving for \ds s, we get

\ds \begin{array}{ccc}\hfill \ds\frac{{r}_{2}}{{r}_{1}}&\ds =\hfill &\ds \frac{s-l}{s}\hfill \\[5mm]\ds \hfill {r}_{2}s&\ds =\hfill &\ds {r}_{1}(s-l)\hfill \\[5mm]\ds \hfill {r}_{2}s&\ds =\hfill &\ds {r}_{1}s-{r}_{1}l\hfill \\[5mm]\ds \hfill {r}_{1}l&\ds =\hfill &\ds {r}_{1}s-{r}_{2}s\hfill \\[5mm]\ds \hfill {r}_{1}l&\ds =\hfill &\ds ({r}_{1}-{r}_{2})s\hfill \\[5mm]\ds \hfill s&\ds =\hfill &\ds \frac{{r}_{1}l}{{r}_{1}-{r}_{2}}.\hfill \end{array}

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

\ds \begin{array}{cc}\ds \hfill S&\ds =\text{(Lateral SA of large cone)}-\text{(Lateral SA of small cone)}\hfill \\[5mm]\ds &\ds =\pi {r}_{1}s-\pi {r}_{2}(s-l)\hfill \\[5mm]\ds &\ds =\pi {r}_{1}\left(\frac{{r}_{1}l}{{r}_{1}-{r}_{2}}\right)-\pi {r}_{2}\left(\frac{{r}_{1}l}{{r}_{1}-{r}_{2}}-l\right)\hfill \\[5mm]\ds &\ds =\frac{\pi {r}_{1}^{2}l}{{r}_{1}-{r}_{2}}-\frac{\pi {r}_{1}{r}_{2}l}{{r}_{1}-{r}_{2}}+\pi {r}_{2}l\hfill \\[5mm]\ds &\ds =\frac{\pi {r}_{1}^{2}l}{{r}_{1}-{r}_{2}}-\frac{\pi {r}_{1}{r}_{2}l}{{r}_{1}-{r}_{2}}+\frac{\pi {r}_{2}l({r}_{1}-{r}_{2})}{{r}_{1}-{r}_{2}}\hfill \\[5mm]\ds &\ds =\frac{\pi {r}_{1}^{2}l}{{r}_{1}-{r}_{2}}-\frac{\pi {r}_{1}{r}_{2}l}{{r}_{1}-{r}_{2}}+\frac{\pi {r}_{1}{r}_{2}l}{{r}_{1}-{r}_{2}}-\frac{\pi {r}_{2}{}^{2}l}{{r}_{1}-{r}_{2}}\hfill \\[5mm]\ds &\ds =\frac{\pi ({r}_{1}^{2}-{r}_{2}^{2})l}{{r}_{1}-{r}_{2}}=\frac{\pi ({r}_{1}-{r}_{2})({r}_{1}+{r}_{2})l}{{r}_{1}-{r}_{2}}=\pi ({r}_{1}+{r}_{2})l.\hfill \end{array}

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. A representative band is shown in the following figure.

This figure has two graphics. The first is a curve in the first quadrant. Around the x-axis is a frustum of a cone. The edge of the frustum is against the curve. The edge begins at f(xsubi-1) and ends at f(xsubi). The second image is the same curve with the same frustum. the height of the frustum is delta x and the curve is labeled y=f(x).
Figure 9. A representative band used for determining surface area.

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

\ds \begin{array}{cc}\ds \hfill S&\ds =\pi ({r}_{1}+{r}_{2})l\hfill \\[5mm]\ds &\ds =\pi (f({x}_{i-1})+f({x}_{i}))\sqrt{(\Delta {x})^{2}+{(\Delta {y}_{i})}^{2}}\hfill \\[5mm]\ds &\ds =\pi (f({x}_{i-1})+f({x}_{i}))\sqrt{1+{\left(\frac{\Delta {y}_{i}}{\Delta x}\right)}^{2}}\Delta x.\hfill \end{array}

Now, as we did in the development of the arc length formula, we apply the Mean Value Theorem to select \ds {x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right] such that \ds {f}^{\prime }({x}_{i}^{*})=\frac{\Delta {y}_{i}}{\Delta x}. This gives us

\ds S=\pi (f({x}_{i-1})+f({x}_{i}))\sqrt{1+{({f}^{\prime }({x}_{i}^{*}))}^{2}}\Delta x.

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

\ds S=2\pi f({x}_{i}^{**})\sqrt{1+{({f}^{\prime }({x}_{i}^{*}))}^{2}}\Delta x.

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

\ds \text{Surface Area}\approx \underset{i=1}{\overset{n}{\sum}}2\pi f({x}_{i}^{**})\sqrt{1+{({f}^{\prime }({x}_{i}^{*}))}^{2}}\Delta x.

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

Taking the limit as \ds n\to \infty , we get

\ds \text{Surface Area}=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\sum}}2\pi f({x}_{i}^{**})\sqrt{1+{({f}^{\prime }({x}_{i}^{*}))}^{2}}\Delta x=\int\limits_{a}^{b}2\pi f(x)\sqrt{1+{({f}^{\prime }(x))}^{2}}\,dx .

Let us now consider the surface obtained by rotating the same curve y=f(x) about the y-axis assuming that a\ge0 to ensure that the curve is on one side of the axis of revolution. We can still use the approach as above, approximating the sought surface area with the sum of surface areas of the bands, it’s only that now the frustum bands are going to be formed by rotating the line segment connecting the points (x_{i-1}, f(x_{i-1})) and (x_i, f(x_i)) on the curve about the y-axis. This means that the parameters of the frustum are going to be r_1=x_{i-1}, r_2=x_i, while l=\sqrt{(\Delta x)^2+(\Delta y_i)^2} remains the same. Therefore, we get

\ds \begin{array}{cc}\ds \hfill S&\ds =\pi ({r}_{1}+{r}_{2})l\hfill \\[5mm]\ds &\ds =\pi ({x}_{i-1}+{x}_{i})\sqrt{(\Delta {x})^{2}+{(\Delta {y}_{i})}^{2}}\hfill \\[5mm]\ds &\ds =\pi ({x}_{i-1}+{x}_{i})\sqrt{1+{\left(\frac{\Delta {y}_{i}}{\Delta x}\right)}^{2}}\Delta x\hfill\\[5mm]\ds &\ds =2\pi \bar{x}_i\sqrt{1+{\left(f(x_i^*)\right)}^{2}}\Delta x,\hfill \end{array}

where \ds\bar{x}_i=\frac{x_{i-1}+x_i}{2} is the midpoint of [x_{i-1},x_i] and x_i^*\in[x_{i-1},x_i] comes from the application of the Mean Value Theorem, as before. Analogously, to what we’ve done above, we then obtain

\ds \text{Surface Area}=\underset{n\to \infty }{\text{lim}}\underset{i=1}{\overset{n}{\sum}}2\pi \bar{x}_i\sqrt{1+{\left(f(x_i^*)\right)}^{2}}\Delta x=\int\limits_{a}^{b}2\pi x\sqrt{1+{({f}^{\prime }(x))}^{2}}\,dx .

As with the arc length, we can now use a similar approach to derive formulas for the areas of surfaces obtained by revolving a curve with equation x=g(y) about the x and y-axes. These findings are summarized in the following theorem.

Surface Area of a Surface of Revolution

Let \ds f(x) be a smooth function over the interval \ds \left[a,b\right].

  • If f(x)\ge0, x\in[a,b], then the area of the surface obtained by revolving the graph of \ds f(x) around the x-axis is given by
    \ds \text{Surface Area}=\int\limits_{a}^{b}2\pi f(x)\sqrt{1+{({f}^{\prime }(x))}^{2}}\,dx .
  • If a\ge0, then the area of the surface obtained by revolving the graph of \ds f(x) around the y-axis is given by
    \ds \text{Surface Area}=\int\limits_{a}^{b}2\pi x\sqrt{1+{({f}^{\prime }(x))}^{2}}\,dx .

Similarly, let \ds g(y) be a smooth function over the interval \ds \left[c,d\right].

  • If g(y)\ge0, y\in[c,d], then the area of the surface obtained by revolving the curve \ds x=g(y) around the y-axis is given by
    \ds \text{Surface Area}=\int\limits_{c}^{d}2\pi g(y)\sqrt{1+{({g}^{\prime }(y))}^{2}}\,dy .
  • If c\ge0, then the area of the surface obtained by revolving the curve \ds x=g(y) around the x-axis is given by
    \ds \text{Surface Area}=\int\limits_{c}^{d}2\pi y\sqrt{1+{({g}^{\prime }(y))}^{2}}\,dy .

Calculating the Surface Area of a Surface of Revolution 1

Let \ds f(x)=\sqrt{x} over the interval \ds \left[1,4\right]. Find the surface area of the surface generated by revolving the graph of \ds f(x) around the x-axis.

Solution

The graph of \ds f(x) and the surface of rotation are shown in the following figure.

This figure has two graphs. The first is the curve f(x)=squareroot(x). The curve is increasing and begins at the origin. Also on the graph are the vertical lines x=1 and x=4. The second graph is the same function as the first graph. The region between f(x) and the x-axis, bounded by x=1 and x=4 has been rotated around the x-axis to form a surface.
Figure 10. (a) The graph of \ds f(x). (b) The surface of revolution.

We have \ds f(x)=\sqrt{x}. Then, \ds {f}^{\prime }(x)=\frac1{2\sqrt{x}} and \ds {({f}^{\prime }(x))}^{2}=\frac1{4x}. Then,

\ds \begin{array}{cc}\ds \hfill \text{Surface Area}&\ds =\int\limits_{a}^{b}2\pi f(x)\sqrt{1+{({f}^{\prime }(x))}^{2}}\,dx \hfill \\[5mm]\ds &\ds =\int\limits_{1}^{4}2\pi \sqrt{x}\sqrt{1+\frac{1}{4x}}\,dx \hfill \\[5mm]\ds &\ds =\int\limits_{1}^{4}2\pi \sqrt{x+\frac{1}{4}}\,dx .\hfill \end{array}

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

\ds \begin{array}{cc}\ds \hfill \int\limits_{0}^{1}2\pi \sqrt{x+\frac{1}{4}}\,dx &\ds =\int\limits_{5\text{/}4}^{17\text{/}4}2\pi \sqrt{u}du\hfill \\[5mm]\ds &\ds =2\pi {\left[\frac{2}{3}{u}^{3\text{/}2}\right]}\Big|_{5\text{/}4}^{17\text{/}4}=\frac{\pi }{6}\left[17\sqrt{17}-5\sqrt{5}\right].\hfill \end{array}

Let \ds f(x)=\sqrt{1-x} over the interval \ds \left[0,1\text{/}2\right]. Find the surface area of the surface generated by revolving the graph of \ds f(x) around the x-axis.

Answer

\ds \frac{\pi }{6}(5\sqrt{5}-3\sqrt{3})

Hint

Use the process from the previous example.

Calculating the Surface Area of a Surface of Revolution 2

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

Solution

Notice that we are revolving the curve around the y-axis, and the interval is in terms of \ds y, so we want to rewrite the function as a function of \ds y. We get \ds x=g(y)=\frac13{y}^{3}. The graph of \ds g(y) and the surface of rotation are shown in the following figure.

This figure has two graphs. The first is the curve g(y)=1/3y^3. The curve is increasing and begins at the origin. Also on the graph are the horizontal lines y=0 and y=2. The second graph is the same function as the first graph. The region between g(y) and the y-axis, bounded by y=0 and y=2 has been rotated around the y-axis to form a surface.
Figure 11. (a) The graph of \ds g(y). (b) The surface of revolution.

We have \ds g(y)=\frac13{y}^{3}, so \ds {g}^{\prime }(y)={y}^{2} and \ds {({g}^{\prime }(y))}^{2}={y}^{4}. Then

\ds \begin{array}{cc}\ds \hfill \text{Surface Area}&\ds =\int\limits_{c}^{d}2\pi g(y)\sqrt{1+{({g}^{\prime }(y))}^{2}}\,dy\hfill \\[5mm]\ds &\ds =\int\limits_{0}^{2}2\pi \left(\frac{1}{3}{y}^{3}\right)\sqrt{1+{y}^{4}}\,dy\hfill \\[5mm]\ds &\ds =\frac{2\pi }{3}\int\limits_{0}^{2}{y}^{3}\sqrt{1+{y}^{4}}\,dy.\hfill \end{array}

Let \ds u={y}^{4}+1. Then \ds du=4{y}^{3}dy. When \ds y=0, \ds u=1, and when \ds y=2, \ds u=17. Then

\ds \begin{array}{cc}\ds \hfill \frac{2\pi }{3}\int\limits_{0}^{2}{y}^{3}\sqrt{1+{y}^{4}}\,dy&\ds =\frac{2\pi }{3}\int\limits_{1}^{17}\frac{1}{4}\sqrt{u}\,du\hfill \\[5mm]\ds &\ds =\frac{\pi }{6}{\left[\frac{2}{3}{u}^{3\text{/}2}\right]}\Big|_{1}^{17}=\frac{\pi }{9}\left[{(17)}^{3\text{/}2}-1\right].\hfill \end{array}

Let \ds g(y)=\sqrt{9-{y}^{2}} over the interval \ds y\in \left[0,2\right]. Find the surface area of the surface generated by revolving the graph of \ds g(y) around the y-axis.

Answer

\ds 12\pi

Hint

Use the process from the previous example.

Key Concepts

  • The arc length of a curve can be calculated using a definite integral.
  • The arc length is first approximated using line segments, which generates a Riemann sum. Taking a limit then gives us the definite integral formula. The same process can be applied to functions of \ds y.
  • The concepts used to calculate the arc length can be generalized to find the surface area of a surface of revolution.
  • The integrals generated by both the arc length and surface area formulas are often difficult to evaluate.

Key Equations

  • Arc Length of a Function of x
    \ds \text{Arc Length}=\int\limits_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}\,dx
  • Arc Length of a Function of y
    \ds \text{Arc Length}=\int\limits_{c}^{d}\sqrt{1+{\left[{g}^{\prime }(y)\right]}^{2}}dy
  • Surface Area of a Function of x revolved about the x-axis
    \ds \text{Surface Area}=\int\limits_{a}^{b}2\pi f(x)\sqrt{1+{[{f}^{\prime }(x)]}^{2}}\,dx
  • Surface Area of a Function of x revolved about the y-axis
    \ds \text{Surface Area}=\int\limits_{a}^{b}2\pi x\sqrt{1+{[{f}^{\prime }(x)]}^{2}}\,dx
  • Surface Area of a Function of y revolved about the y-axis
    \ds \text{Surface Area}=\int\limits_{c}^{d}2\pi g(y)\sqrt{1+{[{g}^{\prime }(y)]}^{2}}\,dy
  • Surface Area of a Function of y revolved about the x-axis
    \ds \text{Surface Area}=\int\limits_{c}^{d}2\pi y\sqrt{1+{[{g}^{\prime }(y)]}^{2}}\,dy

Exercises

For the following exercises, find the length of the curve with the given equation over the specified interval.

1.  \ds y=5x\ \text{from}\ x=0\ \text{ to }x=2

Answer

\ds 2\sqrt{26}

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

3.  \ds x=4y\ \text{from}\ y=-1\text{ to }y=1

Answer

\ds 2\sqrt{17}

4.  Pick an arbitrary linear function \ds x=g(y) over any interval of your choice \ds [{y}_{1},{y}_{2}]. Determine the length of the corresponding curve with calculus and then verify the answer is correct by using geometry.

5.  Find the surface area of the surface generated when the curve \ds y=\sqrt{x} from \ds (1,1) to \ds (4,2) revolves around the x-axis, as shown below.

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.

Answer

\ds \frac{\pi }{6}(17\sqrt{17}-5\sqrt{5})

6. Find the surface area of the surface generated when the curve \ds y={x}^{2} from \ds (1,1) to \ds (3,9) revolves around the y-axis.

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 given curves. If you cannot evaluate the integral, leave your answer in the integral form (set up).

7. \ds y={x}^{3\text{/}2} from \ds (0,0)\text{ to }(1,1)

Answer

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

8.  \ds y={x}^{2\text{/}3} from \ds (1,1)\text{ to }(8,4)

(Hint: rewrite the equation of the curve in the form x=g(y).)

9.  \ds y=\frac{1}{3}{({x}^{2}+2)}^{3\text{/}2} from \ds x=0\text{ to }x=1

Answer

\ds \frac{4}{3}

10.  \ds y=\frac{1}{3}{({x}^{2}-2)}^{3\text{/}2} from \ds x=2 to \ds x=4

11. [Set Up] \ds y={e}^{x} on \ds x=0 to \ds x=1

Answer

\ds \int\limits_0^1 \sqrt{1+e^{2x}}\,dx

12.  \ds y=\frac{{x}^{3}}{3}+\frac{1}{4x} from \ds x=1\text{ to }x=3

13.  \ds y=\frac{{x}^{4}}{4}+\frac{1}{8{x}^{2}} from \ds x=1\text{ to }x=2

Answer

\ds \frac{123}{32}

14.  \ds y=\frac{2{x}^{3\text{/}2}}{3}-\frac{{x}^{1\text{/}2}}{2} from \ds x=1\text{ to }x=4

15.  \ds y=\frac{1}{27}{(9{x}^{2}+6)}^{3\text{/}2} from \ds x=0\text{ to }x=2

Answer

10

16. [Set Up] \ds y= \sin (x) from \ds (0,0)\text{ to }(\pi,0)

17.  \ds y=\frac{5-3x}{4} from \ds y=0 to \ds y=4

Answer

\ds \frac{20}{3}

18. \ds x=\frac{1}{2}({e}^{y}+{e}^{-y}) from \ds y=-1\text{ to }y=1

19.  \ds x=5{y}^{3\text{/}2} from \ds (0,0) to \ds (5,1)

Answer

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

20. [Set Up] \ds x={y}^{2} from \ds (1,1) to \ds (9,3)

21. [Set Up] \ds x=\sqrt{y} from \ds y=1\text{ to }y=4

Answer

\ds \int\limits_1^4 \sqrt{1+\frac1{4y}}\,dy

22. \ds x=\frac{2}{3}{({y}^{2}+1)}^{3\text{/}2} from \ds y=1 to \ds y=3

23. [Set Up] \ds x= \tan (y) from \ds (0,0) to \ds \left(1,\frac{\pi}{4}\right)

Answer

\ds\int\limits_0^{\pi\text{/}4} \sqrt{1+\sec^4(y)}\,dy

24. [Set Up] \ds x={ \cos }^{2}(y) from \ds y=-\frac{\pi }{2} to \ds y=\frac{\pi }{2}

25. [Set Up] \ds x={4}^{y} from \ds x=1\text{ to }x=16

Answer

\ds\int\limits_0^2 \sqrt{1+(\ln(4))^2\cdot4^{2y}}\,dy

26. [Set Up] \ds x=\text{ln}(y) from \ds \left(-1,\frac{1}{e}\right) to \ds (1,e)

For the following exercises, find the surface area of the surface generated when the given curve revolves around the x-axis. If you cannot evaluate the integral, leave your answer in the integral form (set up).

27.  \ds y=\sqrt{x} from \ds x=2 to \ds x=6

Answer

\ds \frac{49\pi }{3}

28.  \ds y={x}^{3} from \ds x=0 to \ds x=1

29.  \ds y=7x from \ds (-1,-7)\text{ to }(1,7)

Answer

\ds 70\pi \sqrt{2}

30. [Set Up] \ds y=\frac{1}{{x}^{2}} from \ds x=1\text{ to }x=3

31. \ds y=\sqrt{4-{x}^{2}} from \ds (0,2)\text{ to }(2,0)

Answer

\ds 8\pi

32.  \ds y=\sqrt{4-{x}^{2}} from \ds x=-1\text{ to }x=1

33.  \ds y=5x from \ds x=1\text{ to }x=5

Answer

\ds 120\pi \sqrt{26}

34. [Set Up] \ds y= \cos (x) from \ds x=-\frac{\pi }{3}\text{ to }x=\frac{\pi }{6}

For the following exercises, find the surface area of the surface generated when the given curve revolves around the y-axis. If you cannot evaluate the integral, leave your answer in the integral (set up) form.

35. \ds y={x}^{2} from \ds x=0\text{ to }x=2

Answer

\ds \frac{\pi }{6}(17\sqrt{17}-1)

36. \ds y=\frac{1}{2}{x}^{2}+\frac{1}{2} from \ds x=0\text{ to }x=1

37.  \ds y=x+1 from \ds (0,1)\text{ to }(3,4)

Answer

\ds 9\sqrt{2}\pi

38. [Set Up] \ds y=\frac{1}{x} from \ds x=\frac{1}{2} to \ds x=1

39.  \ds y=\sqrt[3]{x} from \ds x=1\text{ to }x=27

Answer

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

40. [Set Up] \ds y=3{x}^{4} from \ds x=0 to \ds x=1

41. [Set Up] \ds y=\frac{1}{\sqrt{x}} from \ds x=1 to \ds x=3

Answer

\ds\int\limits_{1\text{/}\sqrt3}^1 \frac{2\pi}{y^2}\sqrt{1+\frac4{y^6}}\,dy

42. [Set Up] \ds y= \cos (x) from \ds (0,1) to \ds \left(\frac{\pi }{2},0\right)

43.  The base of a lamp is constructed by revolving a quarter circle \ds y=\sqrt{2x-{x}^{2}} around the y-axis from \ds x=1 to \ds x=2, as shown below. Set up an integral for the surface area of the base of the lamp 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.

Answer

\ds 2\pi

44. A light bulb is a sphere with radius \ds 1\text{/}2 in. with the bottom sliced off to fit onto a cylinder of radius \ds 1\text{/}4 in. and hight \ds 1\text{/}3 in., as shown below. The sphere is cut off at the bottom to fit exactly onto the cylinder, so the radius of the cut is \ds 1\text{/}4 in. Find the surface area of the light bulb (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.

45. [Set Up] A lampshade is constructed by rotating a curve \ds y=1\text{/}x around the y-axis from \ds y=1 to \ds y=2, as shown below. Set up an integral for the amount of material you would need to construct this lampshade, that is, its surface area.

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.

Answer

\ds\int\limits_1^2\frac{2\pi}y\sqrt{1+\frac1{y^4}}\,dy

46. [Set Up] An anchor drags behind a boat according to the function \ds y=24{e}^{-x\text{/}2}-24, where \ds y represents the depth beneath the boat and \ds x is the horizontal distance to the anchor from the back of the boat. If the anchor is 23 ft below the boat, set up the integral to determine how much rope you have to pull to reach the anchor.

47. [Set Up] You are building a bridge that will span 10 ft. You intend to add decorative rope in the shape of \ds y=5| \sin ((\pi x)\text{/}5)|, where \ds x is the distance in feet from one end of the bridge. Set up the integral to determine how much rope you need.

Answer

\ds\int\limits_0^{10} \sqrt{1+\pi^2\cos^2\left(\frac{\pi x}5\right)}

48.  Find the arc length of the curve \ds y=\text{ln}( \sin (x)) from \ds x=\frac{\pi}4 to \ds x=\frac{3\pi}4.

(Hint: Recall trigonometric identities.)

Glossary

arc length
the arc length of a curve can be thought of as the distance a person would travel along the path of the curve
frustum
a portion of a cone; a frustum is constructed by cutting the cone with a plane parallel to the base
surface area
the surface area of a solid is the total area of the outer layer of the 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

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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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