volumes of solids of revolution calculus cm

7/31/2019 Volumes of Solids of Revolution Calculus CM

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AP CalculusVolumes of Solids of Revolution

Cu cu um M du

Professional DeveloPMent


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ContentsVolumes of Solids of Revolution: Rotating About a LineOther than the x- or y -Axis .................................................................1Catherine Want

About the Contributor ...................................................................... 15


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Volumes of Solids of Revolution:Rotating About a Line Other thanthe x - or y -AxisCatherine WantUniversity High SchoolIrvine, California

Questions involving the area o a region between curves, and the volume o the solidormed when this region is rotated about a horizontal or vertical line, appear regularly on both the AP Calculus AB and BC exams. Student per ormance on this problem isgenerally quite strong except when the solid is ormed using a line o rotation otherthan the x - or y -axis. Students have di culty nding volumes o solids with a line o rotation other than the x - or y -axis. My visual approach to these problems develops anunderstanding o how each line o rotation a ects the radius o the solid. Te approach Iuse (equivalent to the washer method) in the ollowing our examples is what I have oundto be most e ective in my own classroom.

Te our examples I have chosen include a line o rotation below the region to be rotated,above the region to be rotated, to the le o the region to be rotated, and to the right o theregion to be rotated.

Te disk method is used in all our examples. All rotations result in a solid with a hole.Te washer method uses one integral to nd the volume o the solid. I use two integrals,nding the answer as the volume o a solid minus the volume o the hole. I have oundthat when they set up these problems using two integrals, my students understandbetter what each part o the integral, especially the integrand, represents. Tus, they aresuccess ul with this type o problem.


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2

A Curriculum Module for AP Calculus

Example 1. Line of Rotation Below the Region toBe Rotated

Lets start by looking at the solid (with a hole) we get when we rotate the region boundedby y x= + 2 and y e x== about the line y = 2= .

Te two curves intersect at the points ( 1.980974, 0.13793483) and(0.44754216, 1.5644623).

Color will be used to help visualize the problem situation. Using the same color or thesame eature consistently rom problem to problem helps students ollow the processas they are learning to solve these problems. Orange will be used or the line o rotation.Purple is used or the top curve and blue or the bottom curve.

Drawing the disks used to solve the problem helps students ollow the steps in the setup

and ties the calculus language to what we are doing. I verbalize or my students what Ihave written here or you while I draw the diagrams on the board. You may want to try drawing the disks as I describe the procedure or you. All graphs were created using theree program Winplot. My avorite Winplot tutorial is athttp://spot.pcc.edu/~ssimonds/winplot/.

I nd that refecting the original shaded region across the line o rotation helps me make adrawing o the solid.

We will think o our solid as being made up o red disks minus green disks. Te ollowing

graph is a representation.First draw a typical red disk. Start by drawing a radius rom the orange line o rotationthrough the original shaded region until you get to the outside o the region. Stop at thispurple curve. Note: You can draw the disk in other ways. However, i the radius is notdrawn in this way, students o en have di culty getting the expression or the radius.

1

-1

-2

-3

-4

-6

y

x

2

-5

-1-2


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answer may not be accurate to three decimal places as needed. Te easiest thing to do is tostore these values in the calculator and then retrieve them when evaluating the integral.

We now will nd the volume o the green diskscreating the hole. o draw a typical green disk, rst

draw the radius rom the orange line o rotationuntil you hit the edge o the original shaded region.Stop at the blue curve. Starting at the orange line o rotation, we move up (vertically) to the edge o theshaded region. Vertical is the y direction, so thegreen radius involves y . Note that y blue goes only rom the x -axis to the blue curve. From the orangeline o rotation, we have to move two units up toget to the x -axis. Ten we move rom the x -axisto the curve to complete the radius. Tere ore, theradius o each green disk is r = 2 + y blue.

Te volume o one disk is ! r thickness 2 " . Once again, since the radius o the green disk is vertical, the thickness (height) is horizontal. And since horizontal is thex direction, thethickness is ! x . Since y blue = e

x , we have:

Volume o each green disk = ! r x2 ! = ! (2 + y blue)2 ! x = ! 22

+( )e x x " .We now stack up n green disks rom the point o intersection x = 1.980974 on the le tothe other point o intersection x = 0.44754216 on the right.

Te sum o the volumes o these n green disks is given by ! 22

1

+( )=

" e x xk

n

k # .

o obtain a per ectly smooth green solid, we let n ! " and ! x " 0 . Te volume o thisgreen solid is the limit o the sum o the volumes o n green disks as n ! " (and ! x " 0 ).

Volume o green solid = limn

x

k

n

e xk!"

=

+( )#$%&'() *

22

1

+ = ! 22

1980 974

0447542 16

+( )"

# e dx x.

.

= 52.258610.

1

-1

-2

-3

-4

-6

y

x

2

-5

-1-2

1

-1

-2

-3

-4

-6

y

x

2

-5

-1-2


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Volumes of Solids of Revolution

So, the volume o the solid we get when we rotate the region bounded by y x= + 2 ande

x= about the line y = ! 2 is:

VOLUME of red solid VOLUME of green solid

= ! 2 22

1980 974

044754 216

+ +( )"

# x dx.

.

! 22

1980 974

0447 542 16

+( )"

# e dx x.

.

= 19.724 or 19.725.

I have ound that students are better able to ocus on the idea o the solid o revolution asa di erence o two solids i I do not emphasize the role o limit notation in this process.Tere ore, in each example that ollows, I will show how to write the volume o each reddisk and each green disk, and then will re er to the de nite integral as the sum o in nitely many disks.

Example 2. Line of Rotation Above the Region toBe Rotated

Now lets rotate the same region about y = !2= . Were rotating the region bounded by y x= + 2 and about the line y = !2= .

Te solid will be made up o red disks minus green disks.

o draw the red disks that make up the solid, rstdraw the radius rom the orange line o rotationthrough the original shaded region until you getto the outside o the region. Stop at the blue curve.From the orange line o rotation through theshaded region, we move down (vertically). Verticalis the y direction, so the red radius involves y .Now y blue goes rom the x -axis to the blue curve.Tis is not the red radius.

Notice that y blue + radius = 2. Tere ore, the radius r o each red disk is r = 2 y blue. Te radius o the reddisk is vertical; the thickness (height) is horizontal.Since horizontal is thex direction, the thickness is! x , and the volume is ! r thickness 2 ! .Since y blue = e x , we have:

y

5

4

3

2

1

-1

6

x

1-1-2


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Volume o each red disk = ! r x2 ! = ! (2 y blue)2 ! x = ! 22

"( )e x x # .As in Example 1, the red disks are stacked up romx = 1.980974 on the le to x =0.44754216 on the right. We use a de nite integral to add up an in nite number o thesered disks. Tis gives us:

Volume o red solid = ! 22

1980 974

04475 4216

"( )"

# e dx x.

.

= 16.406065.

We will now nd the volume o the green disks creating the hole.

o draw the green disks that create the hole, rstdraw the radius rom the orange line o rotationuntil you hit the edge o the original shaded region.Stop at the purple curve. From the orange line o

rotation, we move down (vertically) to the edgeo the shaded region. Vertical is the y direction,so the green radius involves y . Note that y purple goes rom the x axis to the purple curve. Tis isnot the green radius. Notice that y purple + radius =2. Tere ore, the radius r o each green disk is r =2 y purple.

Te volume o one disk is ! r thickness 2 ! . Onceagain, since the radius o the green disk is vertical,

the thickness (height) is horizontal. And since horizontal is the x direction, the thickness

is ! x . Since y purple = x + 2 , we have:

Volume o each green disk = ! r x2 " = ! (2 y purple)2! x = ! 2 2

2

" +( ) x x# .Again, the green disks are stacked up romx = 1.980974 on the le to x = 0.44754216 onthe right, and we use a de nite integral to add up an in nite number o these green disks.Tis gives us:

Volume o green solid = ! 2 22

1980 974

04475 4216

" +( )"

# x dx.

.

= 7.870360.

y

5

4

3

2

1

-1

6

x

1-1-2


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So, the volume o the solid we get when we rotate the region bounded by y x= + 2 and

y e x

= about the line y = !2= is:


= ! 22

1980 974

04475 4216

"( )"

# e dx x.

.

! 2 22

1980 974

04475 4216

" +( )"

# x dx.

.

= 8.5357 = 8.535 or 8.536.

y

5

4

3

2

1

-1

6

x

1-1-2


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Example 3. Line of Rotation to the Left of theRegion to Be Rotated

Now lets rotate around the vertical line x = ! 3 . We will rotate the region bounded by y x= + 2 and e x= about the line x = ! 3 .

Our solid will be made up o red disksminus green disks. o draw the reddisks that make up the solid, rst draw the radius rom the orange line o rotation through the original shadedregion until you get to the outside o the region. Stop at the blue curve. Fromthe orange line o rotation, we moveright (horizontally) through the shadedregion. Horizontal is the x direction, sothe red radius involves x . Now x blue isthe x -coordinate o a point on the bluecurve, and can be pictured as runninghorizontally rom the y -axis to the bluecurve. Tis is not the radius.

Notice that, in the rst quadrant, radius = 3 + x blue where x blue is positive. In the secondquadrant,

radius = 3 the distance rom the y -axis to the blue curve.But x blue is negative, so this distance is |x blue| = x blue and we have radius = 3 (x blue) =3 + x blue. Tere ore, the radius r o each red disk is r = 3 + x blue.

Te volume o one disk is ! r thickness 2 ! . Since the radius o the red disk is horizontal,the thickness (height) is vertical. And since vertical is the y direction, the thickness is ! y .Since y blue = e x , then x blue = ln y and we have:

Volume o each red disk = ! r y2 ! = ! (3 + x blue)2! y = ! 32

+( )ln y y" .

Once again, we use a de nite integral to add up an in nite number o these red disks. Tered disks are stacked up rom y = 0.1379348256 on the bottom to y = 1.564462259 on thetop. Tis gives us:

Volume o red solid = ! 32

013793 4825 6

156446 2259

+( ) " ln.

.

y dy .

y

2

1

-1

x

1-2 -1-3-4-5-6-7


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We will now nd the volume o the green disks creating the hole. o draw the green disksthat create the hole, rst draw the radius rom the orange line o rotation until you hitthe edge o the original shaded region. Stop at the purple curve. From the orange lineo rotation, we move right (horizontally) through the shaded region. Horizontal is the x direction, so the green radius involvesx . Now, x purple is the x -coordinate o a point on thepurple curve and can be pictured as running horizontally rom the y -axis to the purplecurve. Tis is not the radius. Notice again that in the rst quadrant, radius = 3 + x purple where x purple is positive. In the second quadrant,

radius = 3 the distance rom the y -axis to the purple curve.

But x purple is negative, so this distance is |x purple| = x purple and we haveradius = 3 (x purple) = 3 + x purple. Tere ore, the radius r o each green disk isr = 3 + x purple.Since y purple = x + 2 , then x purple = y

2 2! and we have:

Volume o each green disk = ! ! r y2 " = (3 + x purple)2 ! y = + !( )( )"# 3 222

y y .

Te volume o one disk is ! r thickness 2 ! . Since the radius o the green disk is horizontal,the thickness (height) is vertical. And since vertical is the y direction, the thickness is ! y .

Again, we use a de nite integral to add up an in nite number o these green disks stackedup rom y = 0.1379348256 on the bottom to y = 1.564462259 on the top. Tis gives us:

Volume o green solid = ! 3 22

2

+ "( )( ) # y dy0. 1379 3482 56

1. 5644 6225 9

.

So, the volume o the solid we get when we rotate the region bounded by y x= + 2 and y e x== about the line x = ! 3= is:


= ! 32

013793 4825 6

156446 2259

+( ) " ln.

.

y dy ! 3 2 22

+ "( )( ) # y dy0. 1379 3482 56

1. 5644 6225 9

=15.5387 = 15.538 or 15.539.


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In the second quadrant,

radius = 1 + the distance rom the y -axis to the blue curve.

But x blue is negative, so this distance is |x blue| = x blue and we have

radius = 1 + (x blue) = 1 x blue. Tis is the same expression or the radius as in the rstquadrant. Tere ore, radius = 1 x blue.

Te volume o one disk is ! r thickness 2 ! . Since the radius o the green disk is horizontal,the thickness (height) is vertical. And since vertical is the y direction, the thickness is ! y .

We use a de nite integral to add up an in nite number o these green disks stacked uprom y = 0.1379348256 on the bottom to y = 1.564462259 on the top. Since y blue = e x , then

x blue = ln y and we have:

Volume o green solid = ! 12

"( ) # ln y dy0. 1379 3482 561. 5644 6225 9

.

So, the volume o the solid we get when we rotate the region bounded by y x= + 2 ande

x= about the line x = 1 is:


= ! 1 2 22

" "( )( ) # y dy0. 1379 3482 56

1. 5644 6225 9

! 12

013793 4825 6

156446 2259

"( ) # ln.

.

y dy = 12.72067 = 12.720 or 12.721.

I the x - or y -axis is the line o rotation, the problem can be done in the same way as theour examples shown here. Te radius will just be the appropriate x or y value. Yes, thereare patterns in the integrand depending on the line o rotation. But there is no reasonto memorize patterns and take the chance o making a mistake. A drawing is all that isneeded to get a problem o this type set up correctly.


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AP Examination Questions

Problems rom any calculus book can be used to practice this concept. Be care ul, however,that you dont assign problems where the line o rotation goes through the interior o the

shaded region. Tis situation can be interesting but also can involve some tedious algebra.You will notice when looking at the problems in the list below that there has never been anAP question where the line o rotation goes through the interior o the shaded region.

Since 1997 the ollowing ree-response questions rom the AP Calculus examinations haveinvolved revolving a region about a horizontal or vertical line.

1997 AB2d line of rotation x -axis

1998 AB/BC1c line of rotation x -axis

AB/BC1d line of rotation x -axis

1999 AB/BC2b line of rotation x -axis

AB/BC2c line of rotation y = k


2001 AB1c line of rotation x -axis

2002 AB/BC1b line of rotation y = 4

2002B AB1b line of rotation x -axis

BC3b line of rotation x -axis


2003B AB1c line of rotation x -axis

2004 AB/BC2 line of rotation y = 2

2004B AB1c line of rotation y = 3

AB1d line of rotation x = 4


2005 AB/BC1c line of rotation y = 1

2005B AB1b line of rotation x -axis



AB/BC1c line of rotation y -axis

2006B AB/BC1b line of rotation y = 2


2007B AB/BC1c line of rotation y = 1

2008B AB1b line of rotation x = 1

2008B BC4b line of rotation x -axis

For a complete listing o ree-response questions, you can access

apcentral.collegeboard.com/apc/public/exam/exam_questions/index.html.

Practice Problems

1. Find the volume o the solid resulting rom rotating the region enclosed by y x= 2 and y x= 3 about the vertical line x = 1 .

2. Find the volume o the solid resulting rom rotating the region enclosed by y x= sin and y x= 2 about the horizontal line y = 1 .

3. Find the volume o the solid resulting rom rotating the region enclosed by y x= 2 and y x=

3

about the vertical line x = ! 1= .

4. Find the volume o the solid resulting rom rotating the region enclosed by y x= sin and y x=

2

about the horizontal line y = ! 1= .


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Solutions to Practice Problems

1. Te radius o each large (red) disk is horizontal.

Red radius = 1 x purple (outside) = 1 y .Volume o the red solid = ! 1

2

0

1

"( ) # y dy .Te radius o each small (green) disk is also horizontal.Green radius = 1 x blue (inside) = 1 y3 .

Volume o the green solid = ! 1 32

0

1

"( ) # y dy .Volume o the solid o revolution = ! 1

2

0

1

"( ) # y dy ! 1 32

0

1

"( ) # y dy = 0.209.2. Te radius o each large (red) disk is vertical.

Red radius = 1 y blue (outside) = 1 x2 .

Volume o the red solid = ! 1 22

0

087 67

"( ) # x dx.

.

Te radius o each small (green) disk is also vertical.Green radius = 1 y purple (inside) = 1 sin x .

Volume o the green solid = ! 12

0

087 67

"( ) # sin.

x dx .

Volume o the solid o revolution =! 1 2

2

0

087 67

"( ) # x dx

.

! 1

2

0

087 67

"( ) # sin

.

x dx=0.573.

3. Te radius o each large (red) disk is horizontal.Red radius = 1 +x blue (outside) = 1 + y3 .

Volume o the red solid = ! 1 32

0

087 67

+( ) " y dy.

.

Te radius o each small (green) disk is also horizontal.Green radius = 1 + x purple (inside) = 1 + y .

Volume o the green solid = ! 12

0

087 67

+( ) " y dy.

.

Volume o the solid o revolution = ! 1 32

0

087 67

+( ) " y dy.

! 12

0

087 67

+( ) " y dy.

=0.822.


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4. Te radius o each large (red) disk is vertical.Red radius = 1 + y purple (top) = 1 + sin x .

Volume o the red solid = ! 12

0

087 67

+( ) " sin.

x dx .

Te radius o each small (green) disk is also vertical.Green radius = 1 + y blue (bottom) = 1 + x 2 .

Volume o the green solid = ! 1 22

0

087 67

+( ) " x dx.

.

Volume o the solid o revolution = ! 12

0

087 67

+( ) " sin.

x dx ! 1 22

0

087 67

+( ) " x dx.

= 1.131or 1.132.


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About the Contributor

Catherine Want was a high school math teacher or 33 years and has taught every level o mathematics. She was with Grossmont Union High School District in San Diego County,Cali ornia, or 15 years and Irvine Uni ed School District in Irvine, Cali ornia, or 18years. Catherine also spent a year as a Fulbright Exchange eacher in England.

Catherine taught both AP Calculus AB and AP Calculus BC or 25 years and AP Statisticsor nine years. She was an AP Calculus Exam Reader or seven years and a able Leaderor an additional six years. As a College Board consultant, she has presented at workshopsin Cali ornia, Oregon, Washington, Utah, Arizona, exas, Kentucky, Florida, New York,Alaska and Hawaii, and also has conducted workshops overseas in Saipan, Tailand,Indonesia, Germany and India. Catherine received commendation rom the University o Cali ornia, Irvine; was a Siemens eacher Excellence Award winner in 2003; and wasnamed a eacher o Merit by IN EL in 2005


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volumes of solids of revolution calculus cm

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