Section 7.4 Arc Length and Surface Area
In previous sections we have used integration to answer the following questions:
- Given a region, what is its area?
- Given a solid, what is its volume?
Subsection 7.4.1 Arc Length
Consider the graph of on given in Figure 7.4.2.(a). How long is this curve? That is, if we were to use a piece of string to exactly match the shape of this curve, how long would the string be?
As we have done in the past, we start by approximating; later, we will refine our answer using limits to get an exact solution.
The length of straight-line segments is easy to compute using the Distance Formula. We can approximate the length of the given curve by approximating the curve with straight lines and measuring their lengths.
Graph of the function on The curve begins at the point from which it slopes upwards until reaching a peak at the point From the point, the curve slopes downwards until reaching the -axis at the point
Graph of the function on with four straight lines which will be used to approximate the length of this curve. The four lines are evenly spaced out in intervals of on the -axis. Each line begins at a point on the curve and ends at a point on the same curve after travelling a distance of on the -axis. The first line begins at the same point as the curve, from which it linearly increases until reaching the point which is also a point on the curve. The second line begins at the same point the first line ends, given by from which it linearly increases until reaching the point which is the peak of the curve. The third line begins at the same point the second line ends, given by from which it linearly decreases until reaching the point which is also a point on the curve. The fourth line begins at the same point the third line ends, given by from which it linearly decreases until reaching the point which is the end of the curve.
In Figure 7.4.2.(b), the curve has been approximated with 4 line segments (the interval has been divided into 4 subintervals of equal length). It is clear that these four line segments approximate very well on the first and last subinterval, though not so well in the middle. Regardless, the sum of the lengths of the line segments is so we approximate the arc length of on to be
In general, we can approximate the arc length of on in the following manner. Let be a partition of into subintervals. Let represent the length of the th subinterval
Graph of the th subinterval of the function which is graphed on the interval The graph contains a line between the start and endpoint of the curve, which will be used to approximate the length of the th subinterval curve. The th subinterval of the curve begins at the point from which it heads upwards in a concave arc until reaching the point The straight line then passes through the start and endpoints of the curve, and respectively. The line lies below the curve for the entire interval that the curve is plotted on. The graph also contains the measurements and giving the respective length of the change in and between the start and endpoint of the subinterval of the curve.
Figure 7.4.3 zooms in on the th subinterval where is approximated by a straight line segment. The dashed lines show that we can view this line segment as the hypotenuse of a right triangle whose sides have length and Using the Pythagorean Theorem, the length of this line segment is
Summing over all subintervals gives an arc length approximation
As shown here, this is not a Riemann Sum. While we could conclude that taking a limit as the subinterval length goes to zero gives the exact arc length, we would not be able to compute the answer with a definite integral. We need first to do a little algebra.
In the above expression factor out a term:
Now pull the term out of the square root:
This is nearly a Riemann Sum. Consider the term. The expression measures the “change in /change in ” that is, the “rise over run” of on the th subinterval. The Mean Value Theorem of Differentiation (Theorem 3.2.4) states that there is a in the th subinterval where Thus we can rewrite our above expression as:
As the integrand contains a square root, it is often difficult to use the formula in Theorem 7.4.4 to find the length exactly. When exact answers are difficult to come by, we resort to using numerical methods of approximating definite integrals. The following examples will demonstrate this.
Example 7.4.5. Finding arc length.
Solution 1.
We find note that on is differentiable and is also continuous. Using the formula, we find the arc length as
Graph of the function on the interval between and The curve begins at the point from which it heads upwards in a convex arc until reaching the point A straight line plotted between the start and endpoints of the curve would lie entirely above the curve on the interval between and and would showcase the shortest distance between the two points.
A graph of is given in Figure 7.4.6.
Solution 2. Video solution
Example 7.4.7. Finding arc length.
Solution 1.
This function was chosen specifically because the resulting integral can be evaluated exactly. We begin by finding The arc length is
Graph of the function The curve is highlighted on the interval between and The curve begins near the point from which it heads downwards in a convex arc until crossing the -axis at approximately The curve then continues in the convex arc, until it once again reaches the -axis at approximately
A graph of is given in Figure 7.4.8; the portion of the curve measured in this problem is in bold.
Solution 2. Video solution
The previous examples found the arc length exactly through careful choice of the functions. In general, exact answers are much more difficult to come by and numerical approximations are necessary.
Example 7.4.9. Approximating arc length numerically.
Solution.
This is somewhat of a mathematical curiosity; in Example 5.4.14 we found the area under one “hump” of the sine curve is 2 square units; now we are measuring its arc length.
The setup is straightforward: and Thus
This integral cannot be evaluated in terms of elementary functions so we will approximate it with Simpson’s Method with
Using a computer with the approximation is our approximation with is quite good.
Subsection 7.4.2 Surface Area of Solids of Revolution
We have already seen how a curve on can be revolved around an axis to form a solid. Instead of computing its volume, we now consider its surface area.
Graph of an arbitrary function on the interval The curve is a concave arc starting at at some arbitrary value from which it slopes upwards until ending at at some slightly higher value. The plot of the graph also contains a subinterval on the -axis, given by A line is drawn through the points and which approximates the length of the curve on the interval
We begin as we have in the previous sections: we partition the interval with subintervals, where the th subinterval is On each subinterval, we can approximate the curve with a straight line that connects and as shown in Figure 7.4.12.(a). Revolving this line segment about the -axis creates part of a cone (called a frustum of a cone) as shown in Figure 7.4.12.(b). The surface area of a frustum of a cone is
The length is given by we use the material just covered by arc length to state that
for some in the th subinterval. The radii are just the function evaluated at the endpoints of the interval. That is,
Since is a continuous function, the Intermediate Value Theorem states there is some in such that we can use this to rewrite the above equation as
Summing over all the subintervals we get the total surface area to be approximately
which is a Riemann Sum. Taking the limit as the subinterval lengths go to zero gives us the exact surface area, given in the following theorem.
Theorem 7.4.13. Surface Area of a Solid of Revolution.
- The surface area of the solid formed by revolving the graph of
where about the -axis is - The surface area of the solid formed by revolving the graph of
about the -axis, where is
(When revolving about the -axis, the radii of the resulting frustum are and their average value is simply the midpoint of the interval. In the limit, this midpoint is just This gives the second part of Theorem 7.4.13.)
Example 7.4.14. Finding surface area of a solid of revolution.
Find the surface area of the solid formed by revolving on around the -axis, as shown in Figure 7.4.15.
Solution 1.
The setup is relatively straightforward. Using Theorem 7.4.13, we have the surface area is:
The integration step above is nontrivial, utilizing the integration method of Trigonometric Substitution from Section 6.4.
It is interesting to see that the surface area of a solid, whose shape is defined by a trigonometric function, involves both a square root and an inverse hyperbolic trigonometric function.
Solution 2. Video solution
Example 7.4.16. Finding surface area of a solid of revolution.
- the
-axis - the
-axis.
Solution 1.
- The integral is straightforward to setup:
- Since we are revolving around the
-axis, the “radius” of the solid is not but rather Thus the integral to compute the surface area is:This integral can be solved using substitution. Set the new bounds are to We then have
Solution 2. Video solution
Our final example is a famous mathematical “paradox.”
Example 7.4.18. The surface area and volume of Gabriel’s Horn.
Consider the solid formed by revolving about the -axis on Find the volume and surface area of this solid. (This shape, as graphed in Figure 7.4.19, is known as “Gabriel’s Horn” since it looks like a very long horn that only a supernatural person, such as an angel, could play.)
Solution 1.
To compute the volume it is natural to use the Disk Method. We have:
Gabriel’s Horn has a finite volume of cubic units. Since we have already seen that regions with infinite length can have a finite area, this is not too difficult to accept.
We now consider its surface area. The integral is straightforward to setup:
Integrating this expression is not trivial. We can, however, compare it to other improper integrals. Since on we can state that
By Key Idea 6.8.17, the improper integral on the left diverges. Since the integral on the right is larger, we conclude it also diverges, meaning Gabriel’s Horn has infinite surface area.
Hence the “paradox”: we can fill Gabriel’s Horn with a finite amount of paint, but since it has infinite surface area, we can never paint it.
Somehow this paradox is striking when we think about it in terms of volume and area. However, we have seen a similar paradox before, as referenced above. We know that the area under the curve on is finite, yet the shape has an infinite perimeter. Strange things can occur when we deal with the infinite.
Solution 2. Video solution
A standard equation from physics is “Work = force × distance”, when the force applied is constant. In Section 7.5 we learn how to compute work when the force applied is variable.
Exercises 7.4.3 Exercises
Terms and Concepts
1.
T/F: The integral formula for computing Arc Length was found by first approximating arc length with straight line segments.
2.
T/F: The integral formula for computing Arc Length includes a square-root, meaning the integration is probably easy.
Problems
Exercise Group.
In the following exercises, find the arc length of the function on the given interval.
Exercise Group.
In the following exercises, set up the integral to compute the arc length of the function on the given interval. Do not evaluate the integral.
Exercise Group.
In the following exercises, use Simpson’s Rule, with to approximate the arc length of the function on the given interval. Note: these are the same problems as in Exercises 13–18.
Exercise Group.
In the following exercises, find the surface area of the described solid of revolution.
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Exercise Group.
The following arc length and surface area problems lead to improper integrals. Although the hypotheses of Theorem 7.4.4 and Theorem 7.4.13 are not satisfied, the improper integrals converge, and formulas for arc length and surface area still give the correct result.
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Find the length of the curve on (Note: this describes the top half of an ellipse with a major axis of length 6 and a minor axis of length 2.)
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