Right minus left

• y + 1 = 0.5 y2 – 3 0 = 0.5 y2 – y – 4 • 0 = y2 – 2y – 8 =(y – 4)(y + 2)

4

2 right leftx x dy

4 2

2( 1) (0.5 3)y y dy

4 2

20.5 4y y dy

3

6

y

2

2

y 4

24 |y 64

6

8 16 ( ) 8

62 8

32 824 6

3 6

18

What is the volumeif the yellow area is rotated about the

x-axis?

An easier question would be an easier

graph of f(x).

Start like we did for area. Take a narrow

red strip and then rotate it about the

x-axis.

The volume of a nickel is r2 times the width.

Find the volume of a stool with radius 10 ft. and height ft.

A. 10/cubic feet

B. 100 cubic feet

C. 10 cubic feet

D. 100 cubic feet

Find the volume of a stool with radius 10 ft. and height ft.

A. 10/cubic feet

B. 100 cubic feet

C. 10 cubic feet

D. 100 cubic feet

Set up n rectangles of width x

And the height

of each is f(x) so . . .

* 2

1

lim ( )n

i in

i

Volume f x x

What is the volume ifthe area between f(x) and y=0 is

revolved about the x-axis?

* 2

1

lim ( )n

i in

i

Volume f x x

2( )

b

a

f x dx

Example 1

• Find the volume when the area under y = x2 and over the x-axis is revolved about the x-axis.

• Between x=0 and x=2• Just add up all of the red nickels• As they slide from x=0 to x=2• The top function is . . . • Y= x2

By the definition of the definite integral

Volume = * 2 2

1

lim ( ( ) ( )bn

i in

i a

f x x f x dx

Example 1• Find the volume when the area under

y=x2

• Between x=0 and x=2• Is revolved about the x-axis

= x5/5

= 32/5

22 2

0

( )Volume x dx 20|

Example 2

• Find the volume when the area under y=the square root of x is revolved about the x-axis between x=0 and x=4.

• Volume =

• Volume =

4* 2 2

1 0

lim ( ) ( )n

i in

i

f x x x dx

Volume = =

A. 2B. 4C. 6D. 8

42

0

( )x dx

Volume = =

A. 2B. 4C. 6D. 8

42

0

( )x dx

• Volume =

4* 2 2

1 0

lim ( ) ( )n

i in

i

f x x x dx

240|2

xVolume

240 8

2

Revolve the shownarea about the x-axis.

A. ]

B. ]

C. ]

32

2

.C V x dx

3

2

.B V xdx

34

2

.A V x dx

Revolve the shownarea about the x-axis.

A. ]

B. ]

C. ]

32

2

.C V x dx

3

2

.B V xdx

34

2

.A V x dx

[3

2

2

' ?V x dx how many s

[

6.333

0.1

32

2

' ?V x dx how many s

Example 3 Washer Method

• Spin the shown region about the x-axis

• Show red strip perpendicular to the axis of revolution

Example 3 Washer Method

• Use the disc method for the top function

• Use it again for the bottom one

• Subtract the two answers

Example 3 Washer Method

.

2 2 2 2 b b b

top bottom top bottom

a a a

V y y dx y dx y dx

Speaker Washer Method

• Set the two functons equal to each other• Solve for x x2 = x3 or 0 = x3 - x2

• By factoring 0 = x2 ( x – 1 ) • so x2 =0 or x–1=0• Next we add up all of the red washers• From 0 to 1• Volume =

Volume =

= [(7-5)/35] = 2 /35

1* 2 * 2 2 2 3 2

1 0

lim ( ( ) ( ) ) ( ) ( )n

i i in

i

f x g x x x x dx

5 710

1 1[ ] | [ ]5 7 5 7

x x

Find the volume

A. []

B. []

C. []

/ 32

/3

. sec ( ) 0.16A x dx

/ 3

2

0

. 2 sec ( ) 1.6B x dx

/ 3

/3

. sec( ) 0.4C x dx

Find the volume

A. []

B. []

C. []

/ 32

/3

. sec ( ) 0.16A x dx

/ 3

2

0

. 2 sec ( ) 1.6B x dx

/ 3

/3

. sec( ) 0.4C x dx

.

A. []

B. []

C. []

2 /3/3. [tan ( ) 0.16 ]|A x x

/ 3/3. [tan( ) 0.16 ]|B x x

/ 3/3. [tan( )]|C x

/ 32

/3

sec ( ) 0.16x dx

.

A. []

B. []

C. []

2 /3/3. [tan ( ) 0.16 ]|A x x

/ 3/3. [tan( ) 0.16 ]|B x x

/ 3/3. [tan( )]|C x

/ 32

/3

sec ( ) 0.16x dx

.

A. []

B. []

C. []

2. 2 [tan ( ) 0.16 ]3 3

A

. 2 [tan( ) 0.16]3

B

. 2 [tan( ) 0.16 ]3 3

C

/ 3/3[tan( ) 0.16 ]|x x

.

A. []

B. []

C. []

2. 2 [tan ( ) 0.16 ]3 3

A

. 2 [tan( ) 0.16]3

B

. 2 [tan( ) 0.16 ]3 3

C

/ 3/3[tan( ) 0.16 ]|x x

]2 [tan( ) 0.16 ]3 3

]

9.83

0.2

2 [tan( ) 0.16 ]3 3

24

4

secv ydy

23

0 4

yv dy

Example 4

• Take the area bounded by x = y2 and y = x/2.

• Revolve that area about the y-axis• Red strip is perpendicular to axis of rev.

x = y2 and y = x/2

• Solve for x and set them equal

• y2 = 2y 2 2 2

0

y

right leftyV x x dy

x = y2 and y = x/2

• Solve for x and set them equal

• y2 = 2y

• y2 - 2y = 0

• y(y – 2) = 0 so y = 0 or y = 2

2 2 2

0

y

right leftyV x x dy

x = y2 and y = x/2

• Solve for x and set them equal

• y2 = 2y

• y2 - 2y = 0

• y(y – 2) = 0 so y = 0 or y = 2

2 2 2

0

y

right leftyV x x dy

2 22 2

02

y

yV y y dy

3 54 2 2 32 3( ) ( )

2 5(32) 3(32)

3 5 3 5 3 5

2 22 2

02

y

yV y y dy

23 5

0

4 |

3)|

(5

y y

What is the volumeif the grey area is revolved about the

x-axis?

What are the limits of integration.

2 2

0(1 )

2

xV dx

What is the volumeif the yellow area is revolved about the

y-axis?

Red strip must be perpendicular to the axis of revolution.

y

yV dy

What is the volumeif the yellow area is revolved about the

y-axis?

Red strip must be perpendicular to the axis of revolution.

2

0V dy

What is the volumeif the yellow area is revolved about the

y-axis?

Red strip must be perpendicular to the axis of revolution.

22

0

3

2

yV dy

=-/3[0-1]

22

0sin cosV x xdx

2tan ( )4

ydy

22

0cos [ sin ]V x x dx

3cos

3

xV

/ 20|

sin cosV x xdx 1 2tan ( )

4

ydy

2tan ( )4

ydy

2

0sin cosV x xdx

.2

0sin cosV x xdx

.

0.5

0.1

2

0sin cosV x xdx

Region bounded by y=x2+1 and y=x+3 is revolved about the x-axis.

Region bounded by y=x2+1 and y=x+3 is revolved about the x-axis.

23.4

0.2

cos(A+B)=cosAcosB-sinAsinB

• cos(x+x) = cosx cos x – sinx sinx

• =cos2x-sin2x=2 cos2x -1

• Thus cos2x = (1+cos(2x))/2

• Similarly sin2x = (1-cos(2x))/2

• sin(x+x) = sin x cos x + cos x sin x

• sin(2x) = 2 sin x cos x

First two places y=cosxand y=2sinxcosx cross?

]A. x= /3, B. x=/3, /2

C. x=/6, /2

What is the area between them?

A. .

B. .

C. .

2

6

. sin(2 ) cos( )A x x dx

2

6

. cos( ) sin(2 )B x x dx

6

2

. cos( ) sin(2 )C x x dx

]

A. ]

B. ]

C. ]

2

6

sin(2 ) cos( )x x dx

/ 2

/ 6

1 |. cos(2 ) sin( )2 |

C x x

/ 2

/ 6

1 |. cos(2 ) sin( )2 |

B x x

/ 2

/ 6

1 |. cos(2 ) sin( )2 |

A x x

]

0.25

0.1

/ 2

/ 6

1 |cos(2 ) sin( )2 |

x x