Arc Length and Surface Area
Calculus Techniques Meet History
David W. Stephens
The Bryn Mawr School
Baltimore, MD
NCTM – Baltimore 2004
15 October 2004
1
Contact Information
Email:
stephensd@brynmawrschool.org
The post office mailing address is:
David W. Stephens
109 W. Melrose Avenue
Baltimore, MD 21210
410-323-8800
The PowerPoint slides will be available on my school website:
http://207.239.98.140/UpperSchool/math/stephensd/StephensFirstPage.htm
2
Why is Arclength a Fascinating Topic?

This is a late topic in BC Calculus.
 The seniors are getting near the end of their
high school years … and the AP exam is on
the doorstep.
 Calculus is a great capstone course in high
school, because it brings together all of the
mathematics that the students have
previously learned.
3
How is Arclength a Fascinating Topic?

Calculus students already know about arclength on
a circle from their geometry class
 They understand radians (although perhaps they
still struggle with the importance of radians) … and
radians are crucial for calculus.
 It is valuable to tie in new methods to ones they
already know. Calculus topics often lend
themselves to doing this.
4
Calculus Strategies: Integration

The definite integral is
an accumulation of
products … that is the
sum of products of two
quantities, so definite
integrals can be
thought of as
measurements of areas.
f x = 0.1 25-x2
4
2
-5
5
5
Calculus Strategies: Integration

In any application of integration (such as areas
under a curve, volumes, arclength, work, distances,
or total costs), there is a three step strategy:



Cut the <area, volume, arclength, work, etc> into small
pieces.
Code the quantity to be measured on a representative
small piece, because we understand the geometry of the
small parts.
Recombine the parts (with sums / definite integrals).
6
Calculus Strategies: Integration

Step 1 (Cut the desired
result into small
pieces.)
f x = 0.1 25-x2
4
2
f x = 0.1 25-x2
2
-5
-5
5
5
-2
7
Calculus Strategies: Integration

Step 2 (Code the quantity to be measured on a
representative small piece)

It looks like this:
f x = 0.1 25-x2
4
2

dA = y dx
-5

5
The width (x) is cut into infinitesimally small
parts, and the height (y) depends on the
function under which the area is to be
measured.
8
Calculus Strategies: Integration

Step 3 (Recombine the parts with sums /
definite integrals )
f x = 0.1 25-x2
4

dA = y dx
2
-5

5
Adding up all of these simpler parts becomes
b
A   ydx
a
5

A   (25  x 2 )dx
0
9
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
300 BC Euclidean Geometry
Euclid (325 – 265 BC, probably at Alexandria, Egypt)
The subject of plane geometry was known as far back
as 2000 BC – 2500 BC. Perhaps the Chinese and
other Asian cultures knew this information
independently at about this same time as well.
Distance is measured with a straightedge.
History of mathematics available at:
http://www-gap.dcs.st-and.ac.uk/~history/BiogIndex.html
10
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
1629 to 1640’s Cartesian coordinates
Rene Descartes (France 1596 – 1650)
Points were located with numbers, “marrying”
geometry and algebra. Fermat knew these
results in about 1629 as well.
2
2
length  ( x1  x2 )  ( y1  y2 )
Length is now calculated, rather than measured.
11
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
1660-1670
Integral Calculus
Isaac Newton (1643 – 1727, England) and
Gottfried Leibniz (1646 – 1716, Germany)
Ideas of cutting a length into small pieces and
measuring the small pieces with plane geometry
methods and then recombining the pieces was a
new strategy.
2
2
ds  (dx)  (dy)
(Details to be shown later.)
12
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
1920 – 1945 Measurement of Coastlines
Lewis F. Richardson (England 1881 - 1953)
Richardson investigated to find out that the reported
length of coastlines in Europe (and he is known
especially for a discussion of the coastline of
England) varied by as much as 20%.
13
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
1975 Fractals
Benoit Mandelbrot (Poland 1924 - )
(His family was Lithuanian Jewish. He now resides in the USA.)
Methods were developed to look at the similarity of small
pieces of a line or surface to the whole line or surface.
Measurements (and the accumulation of parts of the
measurements) seemed to depend on the scale of the
measurement tool.
14
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
300 AD Theorems of Pappus
Pappus ( 290-350 AD, Alexandria, Egypt)
Pappus stated two useful theorems, long before the methods of
calculus were in existence, which help to calculate volume
and surface area. In uncanny ways, these ancient theorems
are verified by the much newer methods of the integral
calculus and the fractals.
15
A Whirlwind Histo-Mathematical Tour
How Do We Calculate Length?
1980’s Gaussian Quadrature
Texas Instruments Calculator Algorithm
The method for performing the numerical
integration fnInt is a fast, usually accurate, but
complicated and fascinating algorithm.
(This is a method used for any integration, not just for calculating
length, but it has a connection to the other methods.)
16
Arclength Meets History

Here is how a class might proceed, building
up the ideas for calculus in a historicalmathematical way.
 This discussion will proceed as if all of you
are not actually familiar with the calculus
topic of arclength.
17
What is an “arc”?
arc – Middle English word derived from Latin arcus meaning bow, as
in bow-and-arrow, and, later, arch or curve. In his 1551 Pathway to
Knowledge, Recorde used arche, arche lyne (also spelled archline),
and bowe lyne (also spelled bowline) for the arc of a circle. Billingsley
uses the word arke in his 1570 translation of Euclid’s Elements.
This is from Historical Modules for the teaching and Learning
of Secondary Mathematics (December 2002, Mathematical
Association of America). This definition comes from “Lengths,
Areas and Volumes” (page 193).
18
What do little pieces of most functions
look like?

Most functions have a curve to them, so the
question of the length of an arc amounts to
calculating the length of a piece of a
function.
 Calculus students have been well trained to
say that little pieces of most functions look
like ….
 line segments, because functions are usually
locally linear.
19
Setting up Arclength

So calculating the length of a curve comes
down to methods to measure the length of a
line segment.
Cut the <area, volume, arclength, work, etc> into
small pieces.
 Code the quantity to be measured on a
representative small piece, because we
understand the geometry of the small parts.
 Recombine the parts (with sums / definite
integrals).

20
Setting up Arclength :
Now we follow the history …

Pythagoras (569 – 475 BC, Samos, Ionia)

No coordinates available
c  a 2  b2
c
b
a
21
Setting up Arclength:
Use Pythagoras in a calculus class
We want to know the length of y = x2 on the
interval [0 , 4].
 We do this in four pieces to begin.

f x = x2
15
10
5
-20
-10
10
20
22
Setting up Arclength:
Use Pythagoras in a calculus class
f x = x2
15
10
5
7
5
-20
-10
10
20
3
1
1
1
1
1
Here are the four triangles whose
hypotenuses are straight, under the
assumption that curves are locally
linear.
s  2  10  26  50
So s = about 16.747
23
Setting up Arclength:
Use Pythagoras in a calculus class

Notice that we have
(1) cut the curve into small pieces even though
Pythagoras would not have understood the idea
of a function with coordinates,
 (2) used the geometry of Pythagoras to calculate
the lengths of the four pieces, and
 (3) recombined with addition. No calculus was
used, but the ideas of calculus were employed.

24
Setting up Arclength:
Add Descartes to the question

For each of the
triangles,
coordinates are
used to locate the
points on the
function, and the
distance formula is
that of Pythagoras
with adaptations
for the coordinates.
length  ( x1  x2 )2  ( y1  y2 )2
f x = x2
15
10
5
-20
-10
10
20
25
Setting up Arclength:
Add Descartes to the question
The coordinates of the
points are
(4, 16) ,
(3 , 9) ,
(2 , 4),
( 1, 1),
and (0, 0)
f x = x2
15
10
5
-20
-10
10
20
26
Setting up Arclength:
Add Descartes to the question

Using the distance
formula on each of the
triangles gives the
same results as before.
7
5
3
1
1
1
1
1
s  2  10  26  50
27
Setting up Arclength:
A Detour to the 20th Century

Calculus students accept the idea of local
linearity fairly easily, even though it is a
novel idea at first.
 To challenge their acceptance of this idea
(and recall it is late in the senior year at the
end of a long and challenging AP course),
let’s move to Richardson and Mandelbrot …
and the coastline of England (and other
places).
28
Length of a Coastline
1926 paper, "Does the Wind Possess a Velocity?" ("The question, at
first sight foolish, improves on acquaintance," he wrote.) Wondering about coastlines and wiggly national borders, Richardson
checked encyclopedias in Spain and Portugal, Belgium and the
Netherlands and discovered discrepancies of twenty percent in the
estimated lengths of their common frontiers.
Mandelbrot's analysis of this question struck listeners as either
painfully obvious or absurdly false. He found that most people
answered the question in one of two ways: "I don't know, it's not
my field," or "I don't know, but I'll look it up in the encyclopedia."
In fact, he argued, any coastline is—in a sense—infinitely
A Geometry of Nature 95
Richard F. Voss
prediction in the 1920s, studied fluid turbulence by throwing a
sack of white parsnips into the Cape Cod Canal, and asked in a
1926 paper, "Does the Wind Possess a Velocity?" ("The question, at
first sight foolish, improves on acquaintance," he wrote.) Wondering about coastlines and wiggly national borders, Richardson
checked encyclopedias in Spain and Portugal, Belgium and the
Netherlands and discovered discrepancies of twenty percent in the
estimated lengths of their common frontiers.
Mandelbrot's analysis of this question struck listeners as either
painfully obvious or absurdly false. He found that most people
answered the question in one of two ways: "I don't know, it's not
my field," or "I don't know, but I'll look it up in the encyclopedia."
In fact, he argued, any coastline is—in a sense—infinitely
A FRACTAL SHORE. A computer-generated coastline: the details are random,
but the fractal dimension is constant, so the degree of roughness or
irregularity looks the same no matter how much the image is magnified.
From Chaos by James Gleick
(Penguin Books 1987, page
95)
29
Length of a Coastline

Some of the stories told about the
measurement of coastlines include the
importance of knowing the length of the
coastlines of England and Norway during
World War II, so that the navies knew how
long a coastline they needed to defend.
 Later it became a fascinating mathematical
topic.
30
Length of a Coastline

We can actually do some measurements now
to see how this paradox of Lewis Richardson
goes.
We will simulate this with the maps of
Jaggedland and Smootherland
 We can measure with different “smallest” units
available.

31
Length of a Coastline

Use a 3 inch straightedge.
Start at some point on the map.
 Swing the 3 inch straightedge until it first hits
another point on the map.
 Move the end of the 3 inch straightedge until it
is at the last endpoint
 Count how many 3 inch measurements you can
make, continuing until you are back at the
starting point.

32
Length of a Coastline

Use a 1 inch straightedge.


Use a ½ inch straightedge.


Do the same process as above.
Do the same process as above.
Use the scale on the map to convert the total
number of inches to miles.
33
Length of a Coastline
Use actual maps of Florida, Norway, England, the Chesapeake Bay, and the
Mississippi River in classes.
A Student Worksheet
# of
3 inch
parts
Miles
# of
1 inch
parts
Miles
# of
1/2 inch
parts
Miles
Florida
England
Norway
Chesapeake Bay
Mississippi River
Observations:
34
Length of a Coastline

Actual mileages …whatever “actual” means
(since we are now skeptical about whether there is a real
answer ????)


Florida ….1,350 miles
England … 5,581miles (6261 including islands)
(11,072 miles for Great Britain, 19491 including islands)
Norway …
 Chesapeake Bay … 11,864 miles of shoreline
 Mississippi River … 2,350 to 2,552 miles


(depending on who you ask)
35
Length of a Coastline

What seems to be the results and
connections?
As the measuring tool gets shorter, the total
length gets longer, but not always!
 What measurement tool does a geological survey
use? Why?
 Actual length seems to be the result of practical
methods, but they are not definite answers.

36
Length of a Coastline

Small pieces on the maps are measured as
the Greeks would have done it (!!), and the
Pythagorean theorem could have been used
to calculate from the vertical and horizontal.
 Old meets new.
 Mathematics is still evolving and new
methods and ideas are still being added.
 It is OKAY to combine new and old ideas!
37
Length of a Coastline
Coastline Paradox
Determining the length of a country's coastline is not as simple as it first
appears, as first considered by L. F. Richardson (1881-1953). In fact,
the answer depends on the length of the ruler you use for the
measurements. A shorter ruler measures more of the sinuosity of bays
and inlets than a larger one, so the estimated length continues to
increase as the ruler length decreases.
In fact, a coastline is an example of a fractal, and plotting the length of the
ruler versus the measured length of the coastline on a log-log plot
gives a straight line, the slope of which is the fractal dimension of the
coastline (and will be a number between 1 and 2).
from http://mathworld.wolfram.htm
38
Length of a Coastline
How Long is the Coast of
Great Britain?
Figure 1: The coastline of Great Britain In 1967, Benoit
Mandelbrot published [7] ``How Long is the Coastline
of Great Britain'' in Nature. In it, he posed the simple
question of how one measures the length of a coastline.
As with any curve, the obvious answer for the
mathematician is to approximate the curve with a
polygonal path, each side of which is of length є . (See
Figure 2.)
Then by evaluating the length of these polygonal paths
as є0 , we expect to see the length estimate approach
a limit. Unfortunately, it appears that for coastlines, as
є0 , the approximated length L(є) infinity as well.
Figure 2: Approximating the
coastline of Great Britain
39
Length of a Coastline

In a later book, [10, pp 28-33,] Mandelbrot discusses the extensive
experimental work on this problem which was done by Lewis Fry
Richardson. Richardson discovered that for any given coastline, there
were constants F and D such that to approximate the coastline with a
polygonal path, one requires roughly Fe-D intervals of length . Thus,
the length estimate can be given as L(є) = Fe 1-D

The reason has to do with the inherent ``roughness'' of a coastline. In
general, a coastline is not the type of curve we are usually used to seeing
in mathematics. Although it is a continuous curve, it is not smooth at
any point. In fact, at any resolution, more inlets and peninsulas are
visible that were not visible before. (See Figure 3.) Thus as we look at
finer and finer resolutions, we reveal more and more lengths to be
approximated, and our total estimate of length appears to increase
without bound.
http://www.math.vt.edu/people/hoggard/FracGeomReport/node2.html
40
Length of a Coastline

Contrast this idea with the foundations of
calculus which assert that a limit is attained
when we cut the length into smaller and
smaller pieces.
 We make the assumption …and
conclusion… that there is a finite length and
that our methods of the integral calculus will
help calculate that length.
41
Length of a Coastline

What is the length of the coastline of Britain? Benoit
Mandelbrot proposed this question to demonstrate the
complexity of measurement and scale. There are a number
of almanacs that provide this information. However, if one
examines the measuring techniques used to determine the
length of Britain's coastline, it becomes obvious that this
measurement is only an estimate based on the accuracy of
the measuring device. Smaller units mean greater accuracy.
But we can continue that line of thinking indefinitely, just as
we do with fractions. There are always smaller fractions, an
infinite number. Therefore, the coastline of Britain is an
infinite length, however, it is confined within a finite space.
We can begin to understand then that perimeter can have an
infinite length confined within a finite area.
http://home.inreach.com/kfarrell/measure.html
42
Length of a Coastline

How Long Is Australia's Coastline? (an explanation)

At first blush the question seems eminently reasonable, but it is as open-ended
as the classical "how long is a piece of string?" The answer to both is the same:
it all depends.
 Dr Robert Galloway of the CSIRO Division of Land Use Research in Canberra
was recently confronted with the question when compiling an inventory of
Australia's coastal lands. Looking up the published figures he found the
following answers:
 The great disparity has to do with the precision with which the measurement is
made. The larger and more detailed the map, and the more finely the
measurement is made, the longer will be the coastline. Ultimately one could
walk around the coast itself with a measuring stick, but the answer still depends
on whether you use seven-league boots or a metre rule.
 (It's a philosophical point whether the coastline tends to any limit as precision
improves. Some say it does, others not.)
 To settle on a reliable, repeatable figure, Dr. Galloway got together 162 maps
covering the Australian coast and enlisted the help of Ms Margo Bahr of the
Division.
43
Length of a Coastline





A few points of methodology had to be agreed on before the exercise could
begin:
How far up estuaries should the coastline be taken? It was decided that all inlets
would be arbitrarily (but consistently) cut off whenever their mapped width was
less than 1 km. Within Sydney Harbour, for example, Kirribilli Point was joined
to Garden Island. Straits less than 1 km wide were ignored, treating the island as
though it were part of the mainland.
Islands less than 12 ha. were ignored. Measuring the coastline of the 2600
islands larger than that would be tedious in the extreme. Instead, a 16% sample
was taken and a graph of coast length against area drawn.
This plot gave a good correlation, allowing island coastlines to be derived
simply from their area. However, the ten largest islands (including Tasmania)
were, for accuracy, measured directly. (Macquarie Island and Lord Howe Island
were not included.)
Mangroves were regarded as part of the land, with the coastline following their
seaward fringe; channels between mangroves were treated as estuaries. All coral
reefs were excluded.
http://www.maths.mq.edu.au/numeracy/tutorial/cts2.htm
44
Length of a Coastline

Finally came the question, which tools to use: a pair of dividers, a map
measuring wheel, or a length of string or fine wire? On a test run, dividers gave
consistent results only if the same starting point was used; the wheel was rapid
but inaccurate. Fine wire (not string) laid on the drawn coastline proved
surprisingly consistent and accurate (as good as dividers set to a 0.7 km interval)
and so was chosen for the task. Down to work!
 When the 162nd map was put aside, the total length of the mainland coast plus
Tasmania worked out to be 30 270 km. Adding on the length of the coast of all
the islands greater than 12 ha, about 16 800 km, gave a grand total of 47 070
km.
 As a matter of interest and undeterred by their prior efforts, the two workers
examined the effect of different divider lengths on the measured coastline. As
expected, the apparent length of the coast of the mainland diminished steadily as
the divider length was increased; shrinking to 10 830 km at a 1000-km intercept.
 A simple formula was derived that linked coast length to the measuring
intercept. Using this formula to extrapolate a divider length of just 1 mm, gave a
length of about 132 000 km for the mainland of Australian rather more than
three times the circumference of the earth!
45
Length of a Coastline
How Long Is Australia's Coastline?
The correct answer is .... it depends!
It depends on which source you read, apparently.
Source
Year Book of Australia (1978)
Australian Encyclopedia
Australian Handbook
Length
36 735 km
19 658 km
19 320 km
So who is correct? The answer is: all of them! Each source used a ruler with
different sized increments on it. If you measure the coastline with a ruler
that is just 1 mm long, you would get a length of 132 000 km!
46
Length of a Coastline









4.2 Calculating coastline and population [in New Zealand for Maori tribes]
4.2.1 Coastline calculations
Under the proposed allocation method, inshore fishstocks and 60% of deepwater
fishstocks would be allocated according to the length of an Iwi’s coastline. Exactly
how would coastline lengths be worked out?
It is proposed that a 1:50,000 scale map of New Zealand would be used. Iwi would
have to reach agreement with neighbouring Iwi as to their respective coastline lengths.
The exact coastline length for a quota management area would then be calculated as
follows:
rivers would be cut off at the coast and the distance across the river mouth included in
the coastline measurement;
the coastline length of harbours and bays whose natural entrance points are greater
than 10 km apart would be included in the coastline measurement;
the juridical bay formula (see below) would be applied to harbours and bays whose
natural entrance points are less than 10 km apart in order to determine whether those
harbours and bays would be included in the coastline measurement; and
with the exception of the islands in the Chatham Islands group, coastline
measurements would not include the coastline of islands claimed by Iwi to be part of
their traditional takiwa.
47
Length of a Coastline







The juridical bay formula
The juridical bay formula is applied to bays where the natural entrance
points are less than 10 km apart in order to determine whether the distance
across the entrance of the bay or the actual coastline of the bay should be
added to the coastline measurement. The formula works as follows (see
Figure 3):
a straight line is drawn between the natural entrance points of the bay;
a semicircle is drawn on the straight line (using the straight line as the
diameter of the circle) and the surface area of the semicircle is calculated;
the surface area of the bay enclosed by the straight line is also calculated
using map information software;
if the surface area of the semicircle is smaller than the water surface area of
the bay, then the distance between the natural entrance points is included in
the coastline measurement, (see figure 3a, the shoreline of the bay is not
included) ; or
if the surface area of the semicircle is bigger than the water surface area of
the bay, then the shoreline of the bay is measured and included in the
coastline measurement.( see figure 3b)
48
Length of a Coastline
A drawing of
the New
Zealand
juridical bay
formula
http://www.tokm.co.nz/allocation/1997/implementing.htm
49
Length of a Coastline
The Florida Shoreline and its Measurement
The FLDEP is responsible for monitoring and managing approximately 680
miles of Florida’s coastline.
This includes the state’s entire coastline except for Monroe County (Florida
Keys) and Federal sites. These management efforts are the result of two
contributing factors. In addition, nature constantly changes the shoreline
through normal coastal processes and occasional storm events. Other
shoreline changes result from man’s engineering activities associated
with ports and harbors and shoreline stabilization. The FLDEP must
identify and quantify these changes and manage the coastline to
preserve Florida’s most important natural resource – its beaches.
50
Length of a Coastline
Traditional Method to Measure the Florida Shoreline
Measurements are taken at fifty feet intervals and at points of slope
change along these cross sections. The nearshore survey extends
from the waterline to either the 30 ft contour line or 2,400 ft from
the shoreline, whichever is closer. Survey monuments on the
baseline are maintained so that subsequent surveys may be taken
in the same locations providing a common reference point to
allow comparisons. Because of the time required to collect these
data, only three to four counties’ shorelines are surveyed each
year using this technique. Aerial photography is taken in
conjunction with the surveys to provide visual record and
supplement the coastline management process.
http://64.233.161.104/search?q=cache:77CGD4GeIfgJ:www.thsoa.org/pdf/hy9
9/9_3.pdf+coastline+measurement+Florida&hl=en
51
Length of a Coastline
Advancements in survey technology have provided new tools
for gathering data to manage the coastline. Airborne lidar is
one such advancements. Lidar is an acronym for LIght
Detection And Ranging. Lidar works similar to radar, but a
laser is used instead of radio waves for distance
measurements. Each laser pulse is transmitted from the
airborne platform to the surface below. Some of the light
energy is reflected from the water surface and detected by
onboard optical sensors. The remaining energy continues
through the water column, is reflected from the bottom and
is detected by the onboard sensors. The time difference
between the two energy returns indicates the water depth.
52
Length of a Coastline
The traditional methods would require about 2 years
time to complete a measurement of the Florida
coastline, where the new methods require about 1
month.
though the use of airborne lidar the areas were
surveyed in approximately one month producing
data on an 8m by 8m grid spacing as opposed to
the historic 1,000 ft cross sections. The speed and
cost effectiveness of SHOALS will enhance
FLDEP’s ability to “resurvey more beach areas on
a more frequent basis”
53
Length of a Coastline

The sophistication of methods, as well as the
extraordinary effort that is expended to
measure coastlines, indicates to students that
(1) the process is important,
 (2) the assumptions about how to accomplish
the measurement are heeded, and
 (3) mathematical methods and practical
considerations are both part of the process.

54
Arclength using calculus


Now we get to the
calculus methods to
measure arclength.
Step 1: Cut the length of the
curve into small pieces.
(“Small” is undefined)

ds
dy
dx
We get little triangles, and we
still believe that the curves are
usually locally linear, so that
the hypotenuse is very close
to linear.
55
Arclength using calculus

Step 2: Code the
quantity to be
measured on a
representative small
piece, because we
understand the
geometry of the small
parts
ds
dy
dx
ds  (dx)2  (dy)2
56
Arclength using calculus

Step 3: Recombine the
parts (with sums /
definite integrals)
ds  (dx)2  (dy)2
b
s   (dx) 2  (dy ) 2
a
57
Arclength using calculus
This is not quite in the form that we prefer, since
we need a dx or dy outside the square root. So
…
b
s   (dx) 2  (dy ) 2
a
b
s
a
b
dx
dy
dy 2


2
2
2
 ( )  ( )  * (dx)   1  ( ) dx
dx 
dx
 dx
a
58
Arclength using calculus
Similar derivations from the original
b
s   (dx) 2  (dy ) 2
a
can be done by dividing by either (dy)2 or (dt)2 to get
the companion arclength formulas:
b
s
a
dx
( ) 2  1dy
dy
b
s
a
dx 2 dy 2
( )  ( ) dt
dt
dt
59
Examples of arclength

Let’s look at an
example:
f(x) = x2 on [0 , 4 ]
S=
4

1  (2 x) 2 dx
0
=
60
Examples of arclength

Here’s another example:
For f(x) = sin -1 (x), suppose that
students have not learned how to
integrate the arctrig functions.
So y = sin -1(x) becomes x = sin y

S=
2

(cos( y ))2  1dy
=
0
61
Examples of arclength

Here’s an example
using parametric
functions:
x =cos(t)
y = sin(3t)
2
S=

( sin(t ))  (3cos(3t )) dt
2
2
0
=
62
Surface area: Back to the past

There is an interesting way to remember how
to calculate surfaces area in calculus that
relies on a very old geometry idea.

We look at the ideas of Pappus (300 AD, Egypt)
63
Theorems of Pappus:
Area and Volume … Arclength and Surface Area

Theorem 1:


If a region is rotated about an axis that does not
intersect with the region, then the volume generated
equals the product of the area of the cross section of
the region and the distance that the center of that
cross section travels.
V = 2 r * area
 So dV = 2r * xdy
or 2r * ydx

These are the shell methods of volumes of revolution.
64
Theorems of Pappus:
Area and Volume … Arclength and Surface Area
-5
5
-2
-4
Rotate the yellow region about x = -1.
Pappus says that the volume generated by the black strip is
dV = 2 ( x  1)*( y  y )dx
1
2
65
Theorems of Pappus:
Area and Volume … Arclength and Surface Area

Theorem 2:

If a arc is rotated about an axis that does not
intersect with the arc, then the surface area
generated equals the product of the length of the
arc and the distance that the center of that arc
travels.
dy 2
 SA = 2r * arclength =
2r * 1  ( ) dx
dx

The radius is determined by the axis about which the
arc is rotated (and there are the other two arclength
formulas mentioned earlier)
66
Theorems of Pappus:
Area and Volume … Arclength and Surface Area

It seems that this second theorem works
perfectly well as a geometry formula …
2
-5
5
-2
Since we get the lateral area of a
cylindrical prism.
67
Theorems of Pappus:
Area and Volume … Arclength and Surface Area

…but it seems a bit suspicious for an arc
that is not perpendicular to the axis of
rotation…
f x = 0.1 25-x2
A
2
-5
B
5
-2
But remember that arc AB is very small.
68
Surface Area (examples)
Calculate the surface area
when the curve f(x) = x2 on
[0 , 4] is rotated about the
y axis.
b
SA =

a
4
dy 2
2 r * 1  ( ) dx   2 x 1  (2 x) 2 dx
dx
0
whose value can be done with a u du substitution or a fnInt on a
calculator.
69
Surface Area (examples)
Calculate the surface area when
the curve f(x) = x2 on [0 , 4] is
rotated about the x axis.
SA =
b

a
4
dy 2
2 r * 1  ( ) dx   2 y 1  (2 x) 2 dx
dx
0
4
  2 x 2 1  (2 x) 2 dx
0
70
Surface Area (examples)
Calculate the surface area when the

curve f(y) = sin(y) on [0 , 2 ] is
rotated about the x axis.
b
SA =

a

dx 2
2 r * 1  ( ) dy 
dx
2
 2 y
1  (cos( y )) dy
2
0
The numerical integral is needed to calculate this value.

fnInt( 2 y 1  (cos( y)) , y, 0 ,
)
2
2
71
Surface Area (examples)
Calculate the surface area when the curve
defined by x = cos(t) and y = sin(t) is rotated
about the x-axis. Use 0  t  
SA =
b

a

dx 2 dy 2
2 r * ( )  ( ) dt   2 y (  sin(t ) 2  (cos(t )) 2 dt
dt
dt
0

=
 (2 (sin(t )) *1)dt  4
0
72
Surface Area (examples)
Note that we have just
calculated the surface
area of a sphere.
4 3
V = r
3
SA =
4 r 2
73
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals
Now, we are in the mid to late 1980’s with
technology.
 The Texas Instruments family of calculators
has a fancy method to evaluate the definite
integrals which is fast and usually quite
accurate.


Their website describes it as a version of
Gaussian quadrature.
74
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals
TI-S2 FAQ Tl-83 f AO TI-8S FAQ TI-86 FAQ
Numeric integration on the TI-82, TI-83, TI-85, TI-86 - how does it work?
The method used in the TI-82, 83, 85, and 86 is known as Gauss-Kronrod integration. It is a good deal more
sophisticated than commonly known methods like Simpson's rule or even Romberg integration. In fact, the
widely used mainframe library program QUADPACK uses this technique. A detailed discussion is much
better left to the literature on this subject, which is voluminous. A few comments, however, may help you in
utilizing this function.
The concept begins with Gaussian quadrature rules, which have the property that by sampling a function at "n"
points, an integral can be estimated that would be exact if the function were a polynomial of degree "2n-l".
For comparison, the trapezoid rule is of polynomial degree one and Simpson's rule is of polynomial degree 3.
But while Simpson's rule gives an exact integral for cubic polynomials based on three points, a Gauss
quadrature rule with three points will give the exact integral of polynomials up to order 5. The next concept is
to use a pair of Gauss quadrature rules of different order to provide both an integral result and a corresponding
error estimate. The development of optimal extension pairs (pairs for which the lower order sample points are
a subset of the higher order ones so that extra function evaluations are avoided) was done by A.S. Kronrod in
1965. The use of these pairs allow the method to be "adaptive". The integral is developed by beginning with
the whole interval, computing the integral and the error estimate. If the error is too large, the interval is
bisected and the process repeated on each half. Anytime a segment passes with regard to its "share" of the
total error budget, that integral is added to the total integral. Other segments that don't pass are further
divided. Note that this is quite different than some "adaptive" methods that iterate with smaller intervals until
the result changes by less than some tolerance. Our method computes error from two integral estimates and
only takes new function samples in rapidly changing regions that require it.
Formerly on http://www.ti.com/
75
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals




From a user standpoint, this means that you need to specify an error "tolerance" that you
want to achieve for the integral. In our implementation this is an "absolute" error bound.
This means the algorithm will quit when the absolute value of the difference in its two
integrals is less than the tolerance you specify. Hence, you may not want to ask for a
tolerance of .00001 if you compute an integral that has a value of order 10,000 because
you are asking for 10 accurate digits. Another thing to watch out for is the fact the
function is sampled a finite number of times on the first integration attempt. These points
are not equally spaced and are in fact clustered toward the end points. However, it is not
uncommon to have a function that is virtually constant or perhaps zero throughout much
of the range for integration. In this case, the integrator may quit early and fail to detect the
existence of function behavior in other regions. An example of this is:
fn!nt(eA(-t/le-6),t,0,l). This may seem appropriate to an electrical engineer who wants to
compute all the "energy" in this waveform with a microsecond order fall time. However,
just over a value of 0.002 in the interval 0 to 1, the value of this expression becomes less
than le-999 and underflows to zero. So the algorithm returns a value of zero for this
integral. A range of (0 , 0001) would be more appropriate for the problem and also
because the integral is small, a tol [tolerance] of le-10 is not excessive.
References:
"Numerical Methods and Software", David Kahaner, Cleve Moler and Stephen Nash,
Prentice Hall, 1989, Chapter 5.
76
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals

It works something like this:
The interval along the x-axis is divided into two
equal pieces.
 The area under each half is calculated.
 The half-intervals are each cut in half again, and
the area is calculated again.
 If the area originally calculated does not change
when the interval is subdivided, then it is
assumed that the area is close to correct.

77
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals

If the area is significantly different (whatever
“significantly different” is), then it is further
subdivided again and again until one
calculation is not significantly different than
the preceding one.

Thus, parts of a function are divided more than
other parts. (Which parts?)
78
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals

However, in exchange for the accuracy (and often
the speed) of the calculations, there are
occasional examples for which the calculators
make BIG errors, or for which answers are not
available.



Ex: fnInt(1/x , x , 1 , 1 x 1020)
Ex: fnInt(sin(x) , x , 0 , 1000) is really slow.
Ex: fnInt(sin(x) , x , 0 , 500 )

gives 2.718 x 10-9 for an answer, which is only slightly
different from the exact answer of 0.
79
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals
Ex: fnInt(e^(-10x), x, 0, 10000)
gives 0, when the actual answer is 0.1
10000

0
e
10 x
1 10 x
dx   e
10
10000
0
1 100000 0
  (e
 e )  0.1
10
There is a similar strategy similar to the coastline
measurement being used, but the entire function
is not sampled if changes are not detected earlier
enough, so major errors in calculation can occur.
80
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals
f(x) =
{
1
x<2
1001 2 < x < 3
1
x>3
Calculators can accept these piecewise defined functions:
81
Gaussian Quadrature …
How the TI calculators evaluate numerical definite integrals
The area under the curve on [0, 1000] ought
to be 2000, but fnInt (Y1 , x , 0 , 1000) gives
1000.
f(x) =
{
1
x<2
1001 -2 < x < 3
1
x>3
82
In conclusion …
Arclength and surface area may not be the
most crucial topics in an AP Calculus class.
…but there are some connections with ideas
and the historical development that make the
topic fascinating and valuable.
83
In conclusion …

We followed the ideas of measuring a
straight line from the methods of Pythagoras,
which the students learned in algebra and
geometry classes. So the new calculus
methods have pedagogical ties to previous
knowledge.
 We connected the geometry of the
Pythagorean theorem to the algebra of the
coordinate plane.
84
In conclusion …

We help students to make the leap to the
measurement of an arc as a series of line
segments, because of the local linearity of
most functions.
 This develops the “official” methods of
measuring arclength.
85
In conclusion …

But….the anecdotes from history bring some
authentic relevance to why these lengths are
valuable:
Geography in the 20th century use calculus ideas
to measure coastlines.
 Calculator technology adapt issues that plague
coastline measurement, developing highly
effective ways to evaluate definite integral.
 Calculus improves the methods of algebra and
geometry to do effective calculations.

86
In conclusion …
Students often cannot help but be eager to
learn, especially when the intellectual
twists and turns are authentic and
unexpected.
87