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