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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.

.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….

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Table of ContentsThe 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

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Archimedes (287 BC – 212 BC)

Real World ApplicationThe 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.

Eureka! Born in Syracuse, Sicily, then a Greek city-state, 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.

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Volume of Solid Revolution

Steps:1.Area Between Two CurvesIn 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].

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 BoundsThe bounds of the problem are determined by the equation and axis of revolution. If the axis of revolution is horizontal the bounds will be x-coordinates. 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.

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3. Equations

4. Horizontal Axis of RevolutionThis equation makes sense because when finding the area of a disk with the miniscule width of ∆x, the equation is , and in this case r is equal to the length between the curve and the axis of revolution.

5. Vertical Axis of RevolutionWhen 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 rotated about the y-axis bounded by the x-axis and y=5. must be rewritten as

Rep Rec Magnified

This figure with form the shape of a vase.R(x)= and disks are “stacked” from y=0 to y=5.dx but can be inserted into

your calculator in terms of x.

𝒙=𝟎 𝒙=√𝒚

Toph may be blind and can’t read our explanations, but even she can agree that Solids of Revolution are a blast!

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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:

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 right-most curve of the bounded area. These radii can be represented as two representative rectangles labeled with their respective curves.

Arc Length

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 which becomes where l is the length of the secant line to the curve. Revolved Around a Horizontal Axis:

where r(x)=f(x)

Revolved Around a Vertical Axis:

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.

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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!

Analytical Example: Disk Method*Requires a Graphing Calculator

Find the volume of the solid of revolution formed by revolving the region bounded by and from x=0 to x=2 and the x-axis.

1. Recognize that the equation represents the radius of the volume,

which is the top curve minus the bottom curve, or in this case . 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 form to get:

2. Solve: To solve this problem, plug it in to your calculator. The answer should be: 857.864

164 2 xy

164 2 xy

dxx 2

0

22 164

Aang uses disc-like airbending! 9

The region enclosed by the graph of and the line x=4, and the x-axis is revolved about the x=0. The volume generated is

2xy

The radiuses in this case would be

R(x)= 0 4

r(x)=0

Since the region is being revolved about the y-axis, the equation must be in terms of y. So,

At x=4, y=16 since .Therefore the bounds are from 0 to 16.The integral is

yxxy 2

24y

2x

dyxrxR 16

0

22 )()( General form

16

0

16

0

22 )16())(4( dyydyy

Integrate )128256(])2

1616

0

2yy

Plug in for radius

Analytical Example: Washer Method

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Analytical Example: Arc Length*Requires a Graphing Calculator

Find the arc length of from x=0 to x=4.

1. General Form of Arc length:

2. Since y=f(x), then y’= f’(x). Find the derivative of y.

3. Plug into general form. The lower and upper bounds are 0 and 4.

4. Solve.

164 2 xy

b

a

xfS 2)('1

xyxy 8'164 2

4

0

281 xS

291.64641814

0

24

0

2 xSxS

Still don’t understand? Remain calm, unlike Katara!

“I’m completely calm!”

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Analytical Example: Surface Area*Requires a Graphing Calculator

Find the surface area of from x=0 to x=4 revolved about the x-axis.

1. General Form of Surface Area:

2. Since y=f(x), then y’= f’(x). Find the derivative of y.

3. Plug into general form. The left and right bounds are 0 and 4.

164 2 xy

dxxfxfSAb

a 2)('1)(2

xyxy 8'164 2

2

2

22 81)164(2 xxSA

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Now Katara starts to get it!

=1652.093

AP Level Multiple ChoiceWhat is the length of the arc of from x=0 to x=3?

4

4

1xy

(A)13.550 (B)7.341 (C)20.25 (D) 21.273 (E) 5.666

b

a

xfS 2)('1 General Form

Since f(x)= , then f’(x) is the derivative of y.4

4

1xy

3144

4

14

4

1xxxy Take the Derivative

3

0

232 1)('1 xxfSb

a

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 her wrath! 13

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AP Level Multiple Choice

What is the length of the arc of from x=0 to x=3?

4

4

1xy

(A)13.550 (B)7.341 (C)20.25 (D) 21.273 (E) 5.666

dxxfSb

a 2)('1 General Form

Since f(x)= , then f’(x) is the derivative of y.4

4

1xy

3144

4

14

4

1xxxy Take the Derivative

dxxdxxfSb

a 3

0

232 1)('1Plug 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• If you didn’t take the derivative and plugged in for

f’(x) then you will get 13.550• If you forgot to use add one under the square root you get 20.25

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4

4

1xy

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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 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:

Then, plug this information into the general form

to get:

• 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.

+21.715+(200-78.284)

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AP Free-Response Example

1. Let R be the shaded region in the second quadrant enclosed by the graphs of , and .

a) While on the Fire Nation prison rig to save Heru, Aang creates a wind-funnel 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 and .

Rep Recs Magnified

r(x) R(x)

y=1

y=0

y=0

Lastly, plug the information you have into the general form to get:12.178

R

r(x)

R(x)

Momo says: When attacking these problems, always try to picture the radius and how it is being revolved!

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Analytical Exercises*Some May Require a Graphing Calculator

Calculate using the given function and bounds.

1. Volume of the area bounded by , , and the x-axisRevolved around .

2. Volume of the area bounded by , , and the x-axisRevolved around the x-axis

3. Volume of area bounded by , the x-axis, and in the second quadrant a. Revolved around the x-axis

b. Arc Length of the function from x=-2 to x=2

4. Volume of the area bounded by , , Revolved around the x-axis

5. Volume of the area bounded by and the x-axisRevolved around y=-2

1. Surface area of the function bounded by the first quadrant revolved around the y-axis

6. Arc length of from x=-1 to x=37. Arc length of from x=-2 to x=7

a. Surface area of the function with the same bounds revolved around the x-axis

8. bounded by the x-axis Revolved around y=0

9. , x=-1, x=2, and the x-axisRevolved around y=-1

10. Arc length of from x=.5 to x=2

11. Surface area of from x=0 to x=2.5Revolved around the y-axis

12. Surface area of 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.

13. , , in the first quadrantRevolved around the y-axis

14. , Revolved around the x-axis

15. , Revolved around y=15

Exercise Section: AP Multiple Choice

1. What is the arc length of 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 from the interval [0,1] around the –axis?

(a) 2.000 (b)6.660 (c) 0.714 (d)4.490 (e) 2.245

3. What is the volume of the region of 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

4. What is the volume of the region bounded by 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 x bounded by the x and y-axis from the x=0 to x= revolved about the x-axis. Its volume is found by evaluating which of the following integrals?

(a) )dx(b) dx(c) dx(d) dx(e) 2dx

Exercise Section: AP Multiple Choice (Cont.)

Mai and Ty Lee are intensely studying calculus…

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AP Free Response

(k,)

k

(0,1)

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

How far does of Toph’s boulder travel after she shoots it in yards?

R

Get out of the way twinkle-

toes!

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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.

Works Cited

Our pictures were found courtesy of:

http://25.media.tumblr.com/tumblr_l7796mbglP1qa2ss4o1_500.png

http://www.racebending.com/v3/wp-content/uploads/2009/10/aangface.jpg

www.fanpop.com

www.avatar.wikia.com

We also used the help of the Larson textbook and AP Central

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Zuko says thanks you!