Calculus Chapter 7
7.1 Integrals as Net Change
100
A honey bee makes several trips from the hive to a flower garden. What is the total distance traveled by the bee?
ft min
50 200ft 200ft 200 200 200 100
700
0
-50
-100
2 4
200ft
6 minutes
8
100ft
10
700 feet
7.1 Integrals as Net Change
What is the displacement of the bee?
100 feet towards the hive
100 ft min
50
200ft 200ft
0
-50
-100
2 4
-200ft
6 minutes
8 10
-100ft
7.1 Integrals as Net Change
To find the displacement (position shift) from the velocity function, we just integrate the function. The negative areas below the x -axis subtract from the total displacement
.
Displacement
a b
7.1 Integrals as Net Change
To find distance traveled we have to use absolute value.
Distance Traveled
b a
Find the roots of the velocity equation and integrate in pieces, just like when we found the area between a curve and the x -axis. (Take the absolute value of each integral.)
Or you can use your calculator to integrate the absolute value of the velocity function.
-2
2
1
0
-1
1
7.1 Integrals as Net Change
1
1
2
2
1
3
2
Displacement:
Distance Traveled:
1
1
1 1
2 2
1 1
2 2
4
2
5
2
2 1
2 4 velocity graph
1
0
-1
-2
1 2 3 4 5 position graph
7.1 Integrals as Net Change
In the linear motion equation : dS
dt
V ( t ) is a function of time .
dS
For a very small change in time, V ( t ) can be considered a constant.
S
t We add up all the small changes in S to get the total distance.
S
1 2 3 t S
V
1
V
2
V
3
t
7.1 Integrals as Net Change
S
t We add up all the small changes in S to get the total distance.
S
1 2 3 t
S
S
S
V
1
V
2
V
3
t
n k
1
V n
t n
1
V n
t
As the number of subintervals becomes infinitely large (and the width becomes infinitely small), we have integration.
S
7.1 Integrals as Net Change
Let v (t) = 2 t 3 – 14 t 2 + 20 t , Determine when the particle is moving to the right, left, and stopped.
Find the particle’s distance and displacement after
3 seconds.
0 ++0 ------- 0 +++++
• v ( t ) = 2 t 3 – 14 t 2 + 20 t
• v ( t ) = 2 t ( t 2 – 7 t + 10)
• v ( t ) = 2 t ( t
– 2)( t
– 5)
•
•
0 2 5
Right (0,2) U (5,
)
Left (2,5)
•
Stopped t = 0, 2s, 5s
7.1 Integrals as Net Change
Let v ( t ) = 2 t 3 – 14 t 2 + 20 t , Determine when the particle is moving to the right, left, and stopped.
Find the particle’s distance and displacement after
0
2
3 seconds.
( 2 t
3
14 t
2
20 t ) dt
3
2
( 2 t
3
14 t
2
20 t ) dt
32
3
37
6
101 u
6
3
0
( 2 t
3
14 t
2
20 t
9 u
2
) dt
7.1 Integrals as Net Change
This same technique is used in many different reallife problems.
7.1 Integrals as Net Change
National Potato Consumption
The rate of potato consumption for a particular country was:
2.2 1.1
t where t is the number of years since 1970 and C is in millions of bushels per year.
t
7.1 Integrals as Net Change
t
2.2 1.1
t
We add up all these small amounts to get the total consumption: total consumption
From the beginning of 1972 to the end of 1973:
2
4 t dt
2.2
t
1 ln1.1
1.1
t
4
2
7.066
million bushels
7.1 Integrals as Net Change
Work: work
Calculating the work is easy when the force and distance are constant.
When the amount of force varies, we get to use calculus!
7.1 Integrals as Net Change
Hooke’s law for springs: F
kx k = spring constant x = distance that the spring is extended beyond its natural length
x =2 M
7.1 Integrals as Net Change
F =10 N
Hooke’s law for springs: F
kx
It takes 10 Newtons to stretch a spring 2 meters beyond its natural length.
10
2
5
k F
x
How much work is done stretching the spring to 4 meters beyond its natural length?
x =4
M
7.1 Integrals as Net Change
F ( x )
How much work is done stretching the spring to 4 meters beyond its natural length?
For a very small change in x , the force is constant.
dw
dw
5
5
W x
2
5
2 x
4
0
dw
5
W
0
4
W
40 newton-meters
W
40 joules
7.2 Areas in the Plane y
1 x 2
How can we find the area between these two curves?
y
2
x
We could and figure it out, but there is split the area into several sections, use subtraction an easier way.
7.2 Areas in the Plane y
1 x
2
Consider a very thin vertical strip.
The length of the strip is: y
1
y
2 or
2
x
2
y
2
x
Since the width of the strip is a very small change in x , we could call it dx .
7.2 Areas in the Plane y
1 y
1 x
2 y
1
y
2 dx y
2
Since the strip is a long thin rectangle, the area of the strip is:
2
x
2
y
2
x
If we add all the strips, we get:
1
2
2
x
2 x dx
7.2 Areas in the Plane y
2
x y
1 x
2
1
2
2
x
2 x dx 2 x
1 x
3
1
3 2 x
2
2
1
6
4
8 1 1
2 2
3 3 2
3
2
1 1
3 2
6
27
6
9
2
7.2 Areas in the Plane
The formula for the area between curves is:
Area
a b
1
f
2
dx
We will use this so much, that you won’t need to “memorize” the formula!
y
x
7.2 Areas in the Plane dx y
x dy dx y
x y x 2 y x 2
If we try vertical strips, we have
to integrate in two parts:
0
2 x dx
2
4 x
2
dx
We can find the same area using a horizontal strip.
Since the width of the strip is dy , we find the length of the strip by solving for x in terms of y .
y x 2 y
2 x y x
7.2 Areas in the Plane y
x
We can find the same area using a horizontal strip.
0 y
x y
2 x
2 y
dy y y x 2
2 y dy length of strip x y x 2
Since the width of the strip is dy , we find the length of the strip by solving for x in terms
1
2 of y .
2
2
1 y y y
3
3
2
0 width of strip
2
10
3
4
8
3
7.2 Areas in the Plane
1 Sketch the curves.
2 Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.)
3
Write an expression for the area of the strip.(If the width is dx , the length must be in terms of x .If the width is dy , the length must be in terms of y .
4 Find the limits of integration. (If using dx , the limits are x values; if using dy , the limits are y values.)
5 Integrate to find area.
3
3
7.3 Volumes
Find the volume of the pyramid:
Consider a horizontal slice through the pyramid.
The volume of the slice is s 2 dh .
If we put zero at the top of the pyramid and make down the positive direction, then s=h .
3
0
3 h
7.3 Volumes s dh
V slice
2 h dh
3
2
V h dh
0
1
3 h
3
3
9
0
This correlates with the formula:
1
V Bh
3
1
3
9
7.3 Volumes
Definition:
Volume of a Solid
The
volume
of a solid with known integrable cross section area A ( x ) from x = a to x = b is the integral of A from a to b ,
V
a b
A ( x ) dx
7.3 Volumes
1
Method of Slicing (p384):
Sketch the solid and a typical cross section.
2 Find a formula for V ( x ) .
(Note that V ( x ) is used instead of A ( x ) .)
3
Find the limits of integration.
4 Integrate V ( x ) to find volume.
7.3 Volumes
A 45 o wedge is cut from a cylinder of radius 3 as shown.
Find the volume of the wedge.
y x
You could slice this wedge shape several ways, but the simplest cross section is a rectangle.
7.3 Volumes y x
If we let h equal the height of the slice then the volume of the slice is:
2 y h dx h
Since the wedge is cut at a 45 o angle:
45 o h
x x
Since x
2 y
2
9 y
9
x
2
7.3 Volumes
V y x
Even though we started with a cylinder, p does not
2 9
x
2
0
3
2 x 9
2
2 x dx y h dx du
x
2
2 enter the calculation!
u u
9
0 h
x
V
9
0
1
2
3 u 2
3
9
0 y
2
9
3
27 x
2
18
7.3 Volumes
Cavalieri’s Theorem: Two solids with equal altitudes and identical parallel cross sections have the same volume.
Identical Cross Sections
2
1
0
7.3 Volumes
1 y
x
2 3 4
Suppose I start with this curve.
My boss at the ACME Rocket
Company has assigned me to build a nose cone in this shape.
So I put a piece of wood in a lathe and turn it to a shape to match the curve.
2
1
0 1
7.3 Volumes y
x
2 3 4
How could we find the volume of the cone?
One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes.
The volume of each flat cylinder
(disk) is:
r
2
the thickness
2 dx r= the y value of the function
2
1
0
7.3 Volumes y
x The volume of each flat cylinder (disk) is:
r
2
the thickness
2 dx
1 2 3 4
If we add the volumes, we get:
0
4
2 x dx
0
4 x dx
2 x
2
4
8
0
7.3 Volumes
This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk.
A shape rotated about the x -axis would be:
A shape rotated about the y -axis would be: v v
b a
c
d f ( x )
2 dx f ( y )
2 dy
7.3 Volumes x
1 y
y y -axis is revolved about the y -axis. Find the volume.
4 y x
1 1
1
2
2
.707
3
1
3
.577
1
2
ln y
1
4
4
3
1
0
2 dy
We use a horizontal disk.
The thickness is dy .
1
0
The radius is the x value of the
1 function .
y
V
1
4
1 y
2 ln 2
2 dy
1
4
1 y dy
7.3 Volumes
4
3 y
2 x
The region bounded by y
x
2 and is
2 revolved about the y-axis.
y
x
2
1
Find the volume.
If we use a horizontal slice:
0 y
x
2 y
2 x
1 2 y
x y
2
x
The “disk” now has a hole in it, making it a “washer”.
The volume of the washer is:
R
2 r
2
thickness
7.3 Volumes y
x
2 y
x y
2 x y
2
x
V
0
4
1
2 y y dy
4
V
0
4
1
2 y y dy
4
2
1
0
4
3 y
2 x
1 y
x
2
2
The volume of the washer is:
R
2 r
2
thickness
R
2 r
2
dy
V
0
4
2
1
2 y
2
1
12 y
3
4
0
2
dy
8
16
3
outer radius
8
3 inner radius
7.3 Volumes
This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.
The washer method formula is: V
a b
2 2
R r dx
7.3 Volumes y
x
2 y
x y
2 x y
2
x
0
2
1
4
3 y
2 x
1
R r
2 y
x
2
If the same region is rotated about the line x = 2 :
The outer radius is: y
R 2
2
The inner radius is: r y
7.3 Volumes
2
1
0 y
x
2 y
2 x y
x y
2
x
V
0
4
2
0
4
R
2 r dy y
2
2
2 y
2
dy
4
3 y
2 x
1
R r
2 y
x
2
0
4
y
y
4
2
4 4
0
4 y
y
4
2
4 4
0
4
3 y
1
4 y
2
4
1
2 y
y
y dy
3
2 y
2
1
12 y
3
8
3 y 2
3
0
4
24
16
64
3 3
8
3
7.3 Volumes y x
2 x
y
1
2
3
4
5
1
0 1 2
y
x
2
1
1
5
4
y
1
dy 4
y
x
5
5
y dy
4
x
2
5 y axis.
y 1
2
y
0
5
1
4
y -
25
25
5
1
2 2
4
25
9
2 2
4
16
2
4
1
5
2
2
y
1
2
dy inner radius outer radius thickness of slice
cylinder 8
4
12
7.3 Volumes
5
4
3
2
1 y
x
2
1
Here is another way we could approach this problem:
0 1 2 cross section
If we take a vertical slice and revolve it about the y -axis we get a cylinder.
If we add all of the cylinders together, we can reconstruct the original object.
7.3 Volumes
5
4 r is the x value of the function.
3
2 y
x
2
1 h is the y value of the function .
1 thickness is dx .
0 1 2 cross section
The volume of a thin, hollow cylinder is given by:
7.3 Volumes
5 y
x
2
1
4
3
This is called the shell method because we use cylindrical shells.
2
1
0 1 2
=2
=2
thickness
2
dx r circumference h thickness
0
2
2
2
1
dx
2
1
4 x
4
1
2 x
2
2
0
If we add all the cylinders from the smallest to the largest:
2
0
2 x
3
x dx
2
12
7.3 Volumes
Definition:
Volume of a Solid
Shell Method
The volume of a solid rotated about the y-axis is
V
2
a
b x f ( x ) dx
7.3 Volumes
Definition:
Volume of a Solid
Shell Method
The volume of a solid rotated about the x -axis is
V
2
b a y f ( y ) dy
7.3 Volumes
Find the volume generated when this shape is revolved about the y -axis.
4
3
2
1
0 1 2 3 4 5 6 7 y
4 x
2
10 x
16
9
We can’t solve for x , so
8 we can’t use a horizontal slice directly.
7.3 Volumes
If we take a vertical slice and revolve it about the y-axis we get a cylinder.
Shell method:
Lateral surface area of cylinder
0
4
3
2
1
1 2 3 4 5 6 y
4
9
x
2
10 x
16
7 8
=2
Volume of thin cylinder
2
r h dx
7.3 Volumes
4
3
2
1
0
Volume of thin cylinder
2
r h dx
2
8
2
r x
9
4 x
2
10 x
16
dx
circumference
h
thickness
1 2 3 4 5 6 y
4
9
x
2
10 x
16
7
160
502.655 cm
3
8
7.3 Volumes
When the strip is parallel to the axis of rotation, use the shell method.
When the strip is perpendicular to the axis of rotation, use the washer method.
7.4 Lengths of Curves ds dy ds
2 dx dx
2 dy
2
By the Pythagorean Theorem: ds
dx
2 dy
2
S
ds
dx
2 dy
2
dx
2
dy
2 dx dx
2
2
dx
2
L
1
dy
2 dx
2
dx
Length of Curve (Cartesian)
We need to get dx out from under the radical.
L
a b
1
dy
2
dx dx
7.4 Lengths of Curves
Definition Arc Length: Length of a Smooth Curve
If a smooth curve begins at ( a , c ) and ends at ( b , d ), a < c and b < d , then the length (arc length) of the curve is
L
a b
1
dy
2
dx dx
L
c
d
1
dx dy
2 dy if y is a smooth function on [ a , b ] if x is a smooth function on [ c , d ]
7.4 Lengths of Curves
6
5
8
7
2
1
4
3 y x
0 x 3
9
9 y x dy dx
2 x
9
L
0
3
L
L
ln
37
6
4
3 37
2
L
0
3
0
3
1
dy
2 dx
dx
1
2 x
2 dx
2 x dx
9.74708875861
0 1 2 3
9
8
7
6
5
2
1
4
3
0 1 2 3
7.4 Lengths of Curves
If we check the length of a straight line:
9
2
3
2
C
2
81 9
90
C
C
2
C
9.49
2 The curve should be a little longer than the straight line, so our answer seems reasonable .
L
ln
37
6
4
3 37
2
9.74708875861
7.4 Lengths of Curves
1
L
1
1
1
dy
2 dx
dx
-1 0 x
2 y
2
1 y
2 x
2 y
1
x
2
1
3.1415926536
7.4 Lengths of Curves
0
1
2
3
If you have an equation that is easier to solve for x than for y, the length of the curve can be found the same way.
x
y
2
0 3
1 2 3 4 5 6 7 8 9
L
0
3
1
dx
2 dy
dy
Notice that x and y are reversed.
9.74708875861
7.4 Lengths of Curves
Surface Area: ds
Consider a curve rotated about the x -axis:
The surface area of this band is: 2
r
The radius is the y -value of the function, so the whole area is given by:
a b
2
y ds
This is the same ds that we had in the “length of curve” formula, so the formula becomes:
7.4 Lengths of Curves ds
Surface Area: r
To rotate about the y -axis, just reverse x and y in the formula!
Surface Area about x -axis (Cartesian):
S
a b
2
y 1
dy
2 dx
dx
7.4 Lengths of Curves
4
Rotate about the y -axis.
3
2
1
0 1 2 y
4
3 x 4
3 y
4
3 x x
3
4 y 3
3
3
4 y dx dy
3
4 x
4
3
2
1
0
7.4 Lengths of Curves
1 2 3 x
3
4 y 3 dx dy
3
4
S
0
4
2
3
4 y 3
1
3
2 dy
0
4
2
2
0
4
3
4 y
5
3
4 dy
3
4 y 3
25 dy
16
5
2
3
8 y
2
3 y
4
0
5
2
5
2
6
15
7.4 Lengths of Curves
4
Rotate about the y -axis.
3
2
1 y
4
3 x 4
0 1 s
2 3
From geometry:
. .
rs
15
dx dy
3
4
15
7.4 Lengths of Curves
Example: y
x rotated about x -axis.
3
2
1
0 1 2 3 4 5 6 7 8 9
S
2
0
9 y 1
dy
2
dx dx
117.319
-1 y
2
7.4 Lengths of Curves
Example: x
2 y
2
1 rotated about x -axis.
1
S
2
1
1 y 1
dy
2
dx dx x
2
0 1 y
1
x
2
12.5663706144
Check:
. .
4
r
3
4
12.5663706144
7.5 Applications a Find k : F
kx
24
k
30
k
A spring has a natural length of 1 m .
A force of 24 N stretches the spring to
1.8 m .
F
30 x b How much work would be needed to stretch the spring 3m beyond its natural length?
W
a b
W
15 x
2
3
0
W
0
3
W
135 newton-meters
7.5 Applications
Over a very short distance, even a non-constant force doesn’t change much, so work becomes:
If we add up all these small bits of work we get:
W
a b
7.5 Applications
20
A leaky 5 lb bucket is raised 20 feet
The rope weights 0.08 lb/ft.
The bucket starts with 2 gal (16 lb) of water and is empty when it just reaches the top.
0
Check: At x
0, F
16
At x
20, F
0
7.5 Applications
20
0
Work:
Bucket:
Water: The force is proportional to remaining rope.
20
5
20 x
16
4
16
4
5 x
W
a
At x x
0
20
0, F
20, F
4
5 x dx
7.5 Applications
20
0
Work:
Bucket:
Water:
W
0
20
16
4
5 x dx
W
16 20
16
2
W x x
5
2
20
0
2
160 ft-lb
5
7.5 Applications
20
0
Work:
Bucket:
Water:
Rope:
W
160 ft-lb
20
x
0.08
W
W
Total:
20
0
1.6
x
At x
.04
x
2
0, F
16 ft-lb
At x
20, F
0
100 160 16
276 ft-lb
10 ft
7.5 Applications
5 ft
4 ft
I want to pump the water out of this tank. How much work is done?
w
Fd
The force is the weight of the water.
The water at the bottom of the tank must be moved further than the water at the top.
10 ft
7.5 Applications
5 ft
4 ft
0 dx
10
Consider the work to move one “slab” of water: weight of slab
2
62.5
5 dx
dx distance 4
10 ft
7.5 Applications w
Fd weight of slab
62.5
5
2 dx
dx
4 ft
0 work distance
x
4
dx dx
10 distance
W
0
10 x
force
dx
5 ft
7.5 Applications
5 ft
4 ft work
x
dx
10 ft
A 1 horsepower pump, rated at
550 ft-lb/sec, could empty the tank in just under 14 minutes!
distance force
W
W
W
0
10 x
dx
1562.5
1562.5
1
2 x
2
50 40
4 x
10
0
W
441, 786 ft-lb
10 ft
7.5 Applications
10 ft
2 ft
A conical tank is filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3 .
How much work is required to pump the oil to the rim?
10
y y
7.5 Applications x
Consider one slice (slab) first:
5,10
W F d
W
y
2 x
1 x y
2
W
57
1
2
y dy
2
10
y
W
0
8
10
y
57
1
4
2 y dy
7.5 Applications
A conical tank if filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3 .
How much work is required to pump the oil to the rim?
10
y y x
5,10 y
2 x
1 x y
2
W
57
W
W
W
0
8
57
4
57
4
10
0
8
3
1
2
y dy
2
10 y
10
y
2 3 y y dy
3
57
y
4
4
1
4
8
0
10
y
2 y dy
10 ft
7.5 Applications
10 ft
2 ft
A conical tank if filled to within 2 ft of the top with salad oil weighing 57 lb/ft 3 .
How much work is required to pump the oil to the rim?
W
W
57
4
0
8
10
57 10
4 3 y
2 3 y y dy
3 y
4
4
8
0
W
57
4
5120
3
4096
4
W
30, 561 ft-lb
1 ft
7.5 Applications
3 ft
What is the force on the bottom of the aquarium?
2 ft
Force
weight of water
lb
62.5 2 ft 3 ft 1 ft
3 ft
375 lb
7.5 Applications
If we had a 1 ft x 3 ft plate on the bottom of a 2 ft deep wading pool, the force on the plate is equal to the weight of the water above the plate.
lb
62.5 ft
3
2 ft
3 ft 1 ft
375 lb density depth area
All the other water in the pool doesn’t affect the answer!
pressure
1 ft
7.5 Applications
What is the force on the front face of the aquarium?
2 ft
Depth (and pressure) are not constant.
If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure.
3 ft
1 ft
2 ft
2
0 y
7.5 Applications
3 ft
3 ft
Depth (and pressure) are not constant.
dy
2 ft
If we consider a very thin horizontal strip, the depth doesn’t change much, and neither does the pressure.
F y
62.5
3 dy depth density area
F
F
0
2
62.5
3 dy
2
187.5
375 lb
2 y
2
0
2 ft
3 ft
7.5 Applications
A flat plate is submerged vertically
We could have put the origin at the as shown. (It is a window in the surface, but the math shark pool at the city aquarium.) was easier this way.
Find the force on one side of the plate.
y
x x
y
Depth of strip:
5
y
Length of strip: 2 x
2 y
Area of strip: 2
6 ft
2 ft
3 ft
F y
7.5 Applications y
x x
y
6 ft
y
2 density depth area
Depth of strip:
5
y
Length of strip: 2 x
2 y
Area of strip:
2
F
0
3 y
2
F
3
125 5
0
2 y y dy
F
125
5
2 y
2
1
3 y
3
3
0
F
1687.5 lb
7.5 Applications
Normal Distribution:
34%
13.5%
2.35%
3
2
68%
2
3
95%
99.7%
“68, 95, 99.7 rule”
For many real-life events, a frequency distribution plot appears in the shape of a
“normal curve”.
x heights of 18 yr. old men shape of the curve is
lengths of pregnancies time for corn to pop
7.5 Applications
Normal Distribution:
34%
13.5%
2.35%
3
2
2
3
“68, 95, 99.7 rule”
Normal Probability
Density Function:
(Gaussian curve)
The area under the curve from a to b represents the probability of an event occurring within that range.
1
2
e
2
/ 2
2
