Volume of Solid
Revolution, Arc Length,
and Surface Area of
Revolution
By: Pragya Singh and Arielle
Berman
*All cartoon images and references used in this project are property of Nickelodeon.
1
.Arc Length. Surface Area. Disk. Washer.
Long ago, these four calculus topics lived in
harmony.
Then everything changed when the AP Exam
attacked. Only the Calcatar, master of all
four topics, could stop it, but when the
class needed him the most, he vanished.
A school year has passed and my friend and
I have discovered the new Calcatar, you.
And although your calc-bending skills are
great, you have a lot to learn before you
can ace the test. But we believe you can
get the 5….
2
Table of Contents
The History of Archimedes…………………….…….4
Real World Application………………………..……..4
Volume of Solid Revolution…………………….....5-6
Washer Method………………………….…….……...7
Arc Length………………………………..…………..7
Surface Area………………………………..………...7
Analytical Example: Disk Method………………..….9
Analytical Example: Washer Method………….…...10
Analytical Example: Arc Length……………..……..11
Analytical Example: Surface Area…………….…...12
AP level Multiple Choice………………….....….13-14
Conceptual Example………………………….…….15
AP level free response…………………………..…..16
Analytical Exercises………………………...............17
AP Multiple Choice Exercises………………..….18-19
Full AP Free Response………………………..…….20
Uncle Iroh’s Wisdom…………………………..……21
Works Cited…………………………………..……..22
3
Archimedes
(287 BC – 212 BC)
Eureka! Born in Syracuse, Sicily, then a Greek citystate, Archimedes was not only an inventor of many
important devices including the water screw for raising
water to irrigate the fields, but he was also an important
figure in the world of mathematics throughout his life.
Probably most famous for discovering mass
displacement with water in a bathtub, Archimedes has
had many other great accomplishments in mathematics.
Although his method to find the volume of a curve
revolved around an axis- originally called “Method of
Exhaustion”- is now outdated, its great effects on
integral calculus are still evident today.
Real World Application
The measurement of volume is necessary in every aspect of life, but
standard volume equations only encompass standard shapes- spheres, cubes,
pyramids, etc. Volume of solid revolution allows mathematicians to find the
volume of obscure objects using equations and actual measurements. This
technique is extended with the “washer method”- an equation used to find
the volume of a solid with a portion cut out of the center. Moreover, the arc
length equation allows for the ability to find the length, and ultimately the
surface area- of a complex curve and shape within certain bounds which is
useful in real life to find the length and surface area of an object too big for
physical measurement- such as a planet.
4
Volume of Solid Revolution
Steps:
1.Area Between Two Curves
In order to calculate the volume of a curve revolved around an axis, it is necessary to
generate the area between the curve and the x-axis or- in the case of the washer
method- the area between two different curves. The way to do this is create a
representative rectangle, a rectangle with the height ∆x and the height of the top curve
f(x) and the bottom curve g(x), and take the integral of the f(x)-g(x) on the interval
[a,b].
𝒃
𝒚=
𝒇 𝒙 − 𝒈 𝒙 𝒅𝒙
𝒂
2.Axis of Revolution
One of the most important parts of this calculation is knowing where the axis of
revolution, or the line the area is revolved around, is. This is because the
location of the axis revolution determines the equation needed to solve the
problem which will be shown later. The axis of revolution can either be a
vertical or horizontal line anywhere on the coordinate plane.
3.
The Bounds
The bounds of the problem are determined by the equation and axis of
revolution. If the axis of revolution is horizontal the bounds will be xcoordinates. Imagine placing CDs on a self next to one another from left to right,
eventually filling the entire shelf with volume; this is essentially what you are
doing with disks of volume with radius f(x)-g(x) and width ∆x. If the axis of
revolution is vertical the bounds will be y-coordinates. In this case, finding the
volume is like stacking CDs on top of one another from the ground, or a, to a
certain height, b. The bounds could also be determined if two separate curves
intersect at two different points and that area is then revolved around an axis.
5
3.
Equations
𝒃
𝒚=𝝅
[𝑹 𝒙 ]𝟐
𝒂
1.Horizontal Axis of Revolution
This equation makes sense because when finding the area of a disk
with the miniscule width of ∆x, the equation is 𝜋𝑟 2 , and in this
case r is equal to the length between the curve and the axis of
revolution.
2.Vertical Axis of Revolution
When finding the volume of solid revolved around a vertical axis,
you must convert the equation into terms of y because now you are
stacking disks and integrating in terms of y.
*Remember*: The representative rectangles should be perpendicular to the
axis of revolution.
For example: Find the volume of 𝑦 = 𝑥 2 rotated about the y-axis bounded by
the x-axis and y=5.
𝑦 = 𝑥 2 must be rewritten as 𝑥 = 𝑦
Rep Rec Magnified
𝒙=𝟎
𝒙= 𝒚
This figure with form the shape of a vase.
R(x)= 𝑦 and disks are “stacked” from y=0 to y=5.
𝟓
𝑽𝒐𝒍𝒖𝒎𝒆 = 𝝅 𝟎 [ 𝒚]𝟐dx but can be inserted into
your calculator in terms of x.
Toph may be blind and can’t read our
explanations, but even she can agree that
Solids of Revolution are a blast!
6
Washer Method
Washer Method is a technique used to find the volume of solids with holes formed
when the axis of revolution is not touching the area that is being revolved. The
general form is:
𝒃
𝒚=𝝅
([𝑹 𝒙 ]𝟐 −[𝒓 𝒙 ]𝟐 )𝒅𝒙
𝒂
Where R(x) is the outer radius from the axis of revolution to the top or left-most
curve, and r(x) is the inner radius from the axis of revolution to the bottom or rightmost curve of the bounded area. These radii can be represented as two
representative rectangles labeled with their respective curves.
Arc Length
𝒃
𝟏 + [𝒇′ 𝒙 ]𝟐 𝒅𝒙
𝒔=
𝒂
This equation for arc length can only be used if f(x) is a smooth, continuous
function. It essentially calculates the length of tiny secant lines with a width of dx
from a to b.
Surface Area
Surface area is the arc length of a curve revolved around an axis, so, like arc length,
the curve must be smooth, continuous, and therefore differentiable everywhere
within the bounds. When finding the surface area of a curve revolved around an
axis, you are actually finding the surface area of tiny frustums with the equation
𝜋 𝑟1 + 𝑟2 𝑙 which becomes 2𝜋𝑟𝑙 where l is the length of the secant line to the
curve.
Revolved Around a Horizontal Axis:
𝒃
𝑺 = 𝟐𝝅
𝒓(𝒙) 𝟏 + [𝒇′ 𝒙 ]𝟐 𝒅𝒙
𝒂
where r(x)=f(x)
Revolved Around a Vertical Axis:
𝒃
𝐒 = 𝟐𝛑
𝒙 𝟏 + [𝒇′ 𝒙 ]𝟐 𝒅𝒙
𝒂
When finding surface area of a curve revolved around a vertical axis, “x” must always be
included in the integral because this is the changing radius between the axis of revolution and the
curve. It can also be manipulated and written in terms of y where “x” would be r(y) and dx would
be dy. Since arc length can be written in terms of y or x, we translate it into terms of x because
this is a more convenient way to think of equations.
7
All of this may seem a little confusing
right now… Even Sokka doesn’t seem
to understand it all.
But don’t worry! Ahead you will find
examples of all these types of problems
as well an AP level multiple choice and
free response!
8
Analytical Example: Disk Method
*Requires a Graphing Calculator
Find the volume of the solid of revolution formed by revolving the
2
region bounded by y  4 x  16
and from x=0 to x=2 and the x-axis.
y  4 x 2  16
𝑦=0
1. Recognize that the equation represents the radius of the volume,
which is the top curve minus the bottom curve, or in this case
[−4𝑥 2 + 16 − 0]. Since the axis of revolution is horizontal, the
rep recs must be vertical, as shown above. The bounds of 0 to 2
are given in the problem. Plug the information into the general
𝒃
form 𝒚 = 𝝅 𝒂 [𝑹 𝒙 ]𝟐 to get:
 0  4 x  16  dx
2
2
2
2. Solve: To solve this problem, plug it in to your calculator. The
answer should be: 857.864
Aang uses disc-like airbending!
9
Analytical Example: Washer Method
The region enclosed by the graph of y  x 2 and the line x=4,
and the x-axis is revolved about the x=0. The volume generated is
The radiuses in this case would be
R(x)=
0
4
r(x)=
x2
0
Since the region is being revolved about the y-axis, the equation
2
must be in terms of y. So, y  x  x  y
At x=4, y=16 since y  4 2.Therefore the bounds are from 0 to 16.
The integral is
16
  R( x) 2  r ( x) 2 dy
General form
0
16
16
  (4  ( y ) )dy    (16  y )dy
2
0
 16 y 
2
2
Plug in for radius
0
y
) ]  (256  128)  
2 0
16
Integrate
10
Analytical Example: Arc Length
*Requires a Graphing Calculator
Find the arc length of y  4 x 2  16 from x=0 to x=4.
1. General Form of Arc length:
b
S   1   f ' ( x)
2
a
2. Since y=f(x), then y’= f’(x). Find the derivative of y.
y  4x 2 16  y'  8x
3. Plug into general form. The lower and upper bounds are 0
and 4.
4
S   1   8 x 
2
0
4. Solve.
4
4
S   1   8 x  S   1  64x 2  64.291
2
0
0
Still don’t understand?
Remain calm, unlike
Katara!
“I’m completely calm!”
11
Analytical Example: Surface Area
*Requires a Graphing Calculator
Find the surface area of y  4 x  16 from x=0 to x=4
revolved about the x-axis.
2
1. General Form of Surface Area:
b
SA  2  f ( x) 1   f ' ( x) dx
2
a
2. Since y=f(x), then y’= f’(x). Find the derivative of y.
y  4 x 2  16  y '  8 x
3. Plug into general form. The left and right bounds are 0 and 4.
2
2
SA  2  (4 x 2  16) 1   8 x  =1652.093
2
Now Katara starts to get it!
12
AP Level Multiple Choice
1
What is the length of the arc of y  x 4 from
4
x=0 to x=3?
(A)13.550
(B)7.341
(C)20.25 (D) 21.273
(E) 5.666
b
S   1   f ' ( x)
2
General Form
a
1 4
x , then f’(x) is the derivative of y.
4
1
1
y  x 4  4  x 41  x 3
Take the Derivative
4
4
Since f(x)=y 
b
3
 
S   1   f ' ( x)   1  x
2
a
0
3 2
Plug into the general form.
The upper bound is 3 and
the lower bound is 0
You must solve the problem in order to defeat Azula. The
answers are on the next slide but you better solve it right or feel
13
her wrath!
AP Level Multiple Choice
1
What is the length of the arc of y  x 4 from
4
x=0 to x=3?
(A)13.550
(B)7.341
(C)20.25 (D) 21.273
(E) 5.666
b
S   1   f ' ( x) dx
2
General Form
a
1 4
x , then f’(x) is the derivative of y.
4
1
1
y  x 4  4  x 41  x 3
Take the Derivative
4
4
Since f(x)=y 
b
3
  dx
S   1   f ' ( x) dx   1  x
2
a
0
3 2
Plug into the general form.
The upper bound is 3 and
the lower bound is 0
Since this is the calculator section, you can plug the equation
into your graphing calculate and you will get the answer
21.273. Therefore the correct answer is D.
•
•
•
If you forgot to square x3 you will get 7.341
1 4
If you didn’t take the derivative and plugged in y  4 x for f’(x)
then you will get 13.550
If you forgot to use add one under the square root you get 20.25
14
AP Conceptual
1.
While trying to escape from Azula, Ty Lee, and Mai who are chasing them
with a tank through the Earth Kingdom, Team Calcatar has been riding Appa
nonstop for days. Appa is tired, so Aang lets him walk part of the way. He
walks from x=0 to x=21.715 ft, but with Azula closing in quickly, they are
forced to fly starting from x=21.715 ft modeled by the equation
𝑦 = −.25 𝑥 − 50 2 + 200 but has to make a crash-landing at x=78.294 ft. He
walks once again until he reaches a large river at x=200. Set up but do not
integrate the equation to find the distance of Appa’s path.
•
Before arriving at an answer, you must gather all of the information necessary
to plug into the arc length general form. This includes the bounds- which are
given in this instance- and the derivative of the function of the curve you are
trying to find the arc length of.
If you have correctly taken the derivative it should be a form of:
𝑦 ′ = −.5 𝑥 − 50
Then, plug this information into the general form
𝒃
𝒔=
𝟏 + [𝒇′ 𝒙 ]𝟐 𝒅𝒙
𝒂
to get:
𝟕𝟖.𝟐𝟖𝟒
𝒚=
𝟏 + (−. 𝟓 𝒙 − 𝟓𝟎 )𝟐 𝒅𝒙
𝟐𝟏.𝟕𝟏𝟓
•
Once you have the distance of Appa’s flight path, you must then add the
distance he travels on land which is his distance from the origin to his takeoff
and from his landing to the large river.
𝟕𝟖.𝟐𝟖𝟒
𝟐𝟏.𝟕𝟏𝟓
𝟏 + (−. 𝟓 𝒙 − 𝟓𝟎 )𝟐 𝒅𝒙+21.715+(200-78.284)
15
AP Free-Response Example
R(x)
R
Momo says:
When attacking
these problems,
always try to
picture the radius
and how it is
being revolved!
r(x)
1.
Let R be the shaded region in the second quadrant enclosed by the graphs of
𝑦 = 1 + 4𝑒 𝑥 , 𝑦 = 1, 𝑥 = −1 and 𝑥 = −4.
a) While on the Fire Nation prison rig to save Heru, Aang creates a windfunnel so Katara and Sokka can shoot large chunks of coal through it at
the Fire Nation Army. Find the volume of the funnel generated when R is
revolved around the x-axis.
Explanation:
1) First determine the axis of revolution. We know that the representative
rectangles must be perpendicular to this, and since it is the x-axis the rep recs
must be vertical.
2) Then, calculate the bounds of the integral. This is determined by either a
boundary line given in the problem or where the two curves intersect. In this
case the bounds are 𝑥 = −1 and 𝑥 = −4.
Rep Recs Magnified
r(x)
y=1
R(x)
𝒚 = 𝟏 + 𝟒𝒆𝒙
y=0
y=0
Lastly, plug the information you have into the general form to get:
𝑽= 𝝅
−𝟏
[( 𝟏
−𝟒
+ 𝟒𝒆𝒙 − 𝟎)𝟐 − 𝟏 − 𝟎 𝟐 ]𝒅𝒙 =12.178
16
Analytical Exercises
*Some May Require a Graphing Calculator
Calculate using the given function and bounds.
1. Volume of the area bounded by y = xex , 𝑥 = 2, and the x-axis
Revolved around 𝑦 = 0.
2. Volume of the area bounded by y = ln 2𝑥 , 𝑥 = 3, and the x-axis
Revolved around the x-axis
3. Volume of area bounded by y = x 3 − 4x, the x-axis, and in the second
quadrant
a.
b.
4.
5.
Volume of the area bounded by y = 3 x + 1, 𝑦 = 1, 𝑥 = 6
Revolved around the x-axis
Volume of the area bounded by y = 3 − 7x 2 and the x-axis
Revolved around y=-2
1.
6.
7.
Revolved around the x-axis
Arc Length of the function from x=-2 to x=2
Surface area of the function bounded by the first quadrant revolved around the y-axis
Arc length of y = x 16 − x 2 from x=-1 to x=3
Arc length of y = ex/3 from x=-2 to x=7
a.
Surface area of the function with the same bounds revolved around the x-axis
y = 2 − .3x 2 bounded by the x-axis
Revolved around y=0
9. y = e−x/2 , x=-1, x=2, and the x-axis
Revolved around y=-1
10. Arc length of y = x 2x from x=.5 to x=2
11. Surface area of y = x from x=0 to x=2.5
Revolved around the y-axis
12. Surface area of 𝑦 = .5𝑥 2 + 2 from x=0 to x=7
Revolved around x axis
Calculate the volume formed when the area between these two intersecting
curves is revolved around the given axis.
1. y = x 3 /2, y = 3x, in the first quadrant
Revolved around the y-axis
8.
2.
3.
x2
y=5−
,𝑦=3
2
Revolved around the x-axis
y = x 2 + 2x + 6, y = 4x 2
Revolved around y=15
17
Exercise Section: AP
Multiple Choice
1. What is the arc length of 𝑦 = 3𝑥 2 − 2𝑥 − 5 from x= -1
and x=5/3?
(a) 10.004 (b)11.212 (c) 43.120 (d) 59.555 (e) 34.267
2.
What is surface area of the graph of 𝑦 = 𝑥 from the
interval [0,1] around the –axis?
(a) 2.000 (b)6.660 (c) 0.714
3.
(d)4.490
(e) 2.245
What is the volume of the region of 𝑦 = 𝑒 2𝑥 bounded by
x-axis, y-axis, and x=2 revolved about the x-axis?
(a) 2340.453 (b)744.989 (c)84.191 (d) 42.095 (e)100.000
The Kyoshi Warriors kick butt with Calculus
Exercise Section: AP
Multiple Choice (Cont.)
4. What is the volume of the region bounded by
𝑦 = −𝑥 2 + 𝑥 + 6 and y=4 revolved about the x-axis?
(a) 44.1
(b)34.678
(c)14.137 (d)25.446 (e)138.544
5. The graph of 𝑦 = 1 − cos2 x bounded by the x and y-axis from
the x=0 to x=𝜋 revolved about the x-axis. Its volume is found by
evaluating which of the following integrals?
𝜋
(a) 𝜋 0 (1 − cos2 x)dx
𝜋
(b) 𝜋 0 (1 − 𝑐𝑜𝑠 2 x)2 dx
𝜋
(c) 0 (1 − 𝑐𝑜𝑠 2 x)2 dx
𝜋
(d) 𝜋 0 (1 − 𝑐𝑜𝑠 4 𝑥)dx
(e) 2𝜋
𝜋
(1
0
− 𝑠𝑖𝑛2 𝑥)dx
Mai and Ty Lee are
intensely studying
calculus…
AP Free Response
(k,𝒆𝒌/𝟐 )
(0,1)
R
k
1. On the day of the Black Sun, Team Avatar fights Azula and her Dai Li Agents in the Fire
Nation. In order to protect her friends, Toph begins firing boulders in the shape of
frustums at Azula.
a. Write but do not evaluate the volume, in terms of k, of Toph’s earthbending boulder
represented as R revolved around the x-axis.
b. Write, but do not integrate, an expression involving an integral to find the entire surface
area of the boulder including the bases.
c. The path of this boulder once Toph shoots it is represented by the equation
𝑦 = −.25𝑥 2 + 2𝑥
How far does of Toph’s boulder travel after she shoots it in yards?
Get out of the
way twinkletoes!
20
Remember Uncle Iroh’s
Wisdom
You are stronger and wiser and
freer than you have ever been.
And now you have come to the
crossroads of your destiny. It’s
time for you to choose. It’s time
for you to choose calculus.
21
Works Cited
Our pictures were found courtesy of:
http://25.media.tumblr.com/tumblr_l7796mbglP1qa2ss4o1_500.png
http://www.racebending.com/v3/wpcontent/uploads/2009/10/aangface.jpg
www.fanpop.com
www.avatar.wikia.com
We also used the help of the Larson textbook and AP Central
Zuko says thanks you!
22