Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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5.6 Definite Integral Substitutions and the Area Between Curves
Recall that we initially developed the definite integral to compute the area under a curve.
Z b
Area = A =
f (x)dx
a
Suppose that we have two continuous functions, f and g, and f (x) ≥ g(x) for all x on [a, b].
We would like to find the area bounded by the graphs of y = f (x) and y = g(x) on the
interval [a, b].
Z b
Formula: Area= [f (x) − g(x)]dx
a
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Find the area bounded by the graphs of y = −x2 and y = x + 1 for 0 ≤ x ≤ 2.
Sketch the region.
Example Find the area bounded by the graphs of y = 3 − x and y = x2 − 9. Sketch the
region.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Finding the Area of a Region Using Several Integrals
Finding the area of some regions may require breaking the region up into several pieces, each
having different upper and/or lower boundaries.
Example Find the area bounded by the graphs of y = x2 and y = 2 − x2 for 0 ≤ x ≤ 2.
Sketch the region.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Integrating with respect to y
Z d
f (y) − g(y)dy
Formula: Area = A =
c
Example Find the area of the region bounded by the graphs of x = y 2 and x = 2 − y 2 .
Sketch the region.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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6.1 Volumes Using Cross-Sections
If a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and
the line is called the axis of revolution. The simplest such solid is a right circular cylinder
or disk, with is formed by revolving a rectangle about an axis adjacent to one side of the
rectangle. The volume of a disk is
V = πr2 w = πr2 dx or V = πr2 dy
If cross section is a disk, then
Z b
Z b
A(x)dx =
Volume of a solid = V =
a
a
Z d
Z d
π[r(x)]2 dx (axis of revolution is x-axis)
or
A(y)dy =
Volume of a solid = V =
c
c
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π[r(y)]2 dy (axis of revolution is y-axis)
Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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x
Example Suppose that the line segment y = + 1, 0 ≤ x ≤ 12, is revolved about the
3
x-axis. Compute the volume of this solid.
√
Example Revolve the region under the curve y = x on the interval [0, 4] about the x-axis
and find the volume of the resulting solid of revolution.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Find the volume of the solid resulting from revolving the portion of the curve
x2
from x = 0 to x = 2 about the y-axis.
y =2−
2
The Method of Washers
The disk method can be extended to cover solids of revolution with holes by replacing
the representative disk with a representative washer. The washer is formed by revolving a
rectangle about an axis.
Let R=outer radius.
Let r=inner radius.
Then Volume of washer= π(R2 − r2 )w.
Z b
π([R(x)]2 − [r(x)]2 )dx (axis revolution is x-axis)
Volume of solid of revolution=
a
Z d
Volume of solid of revolution=
π([R(y)]2 − [r(y)]2 )dy (axis revolution is y-axis)
c
Graphs:
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Let R be the region bounded by the graphs of y = x2 , x = 0, and y = 1. Compute
the volume of the solid formed by revolving R about the x-axis.
Example Let R be the region bounded by the graphs of y = x2 , x = 0, and y = 1. Compute
the volume of the solid formed by revolving R about the y-axis.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
Revolving the region about the line x = l or the line y = k
1. Line is above the region
2. Line is below the region
3. Line is to the right of the region
4. Line is to the left of the region
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Let R be the region bounded by the graphs of y = x2 , x = 0, and y = 1. Compute
the volume of the solid formed by revolving R about the line y = 2.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example
Let R be the region bounded by the graphs of y = 4 − x2 and y = 0. Compute the volume
of the solids formed by revolving R about
1. the line x = 4
2. the line y = −3
3. the line y = 7
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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6.2 Volumes Using Cylindrical Shells
Cylindrical shells look like a soda can.
Volume of cylindrical shell=circumference · height · thickness=2πrhw
Z b
Volume of solid of revolution using shells=
2πr(x)h(x)dx (vertical axis of revoa
lution)
where r=the distance between a representative rectangle and the axis of revolution
and
h =height of the representative rectangle
Z d
Volume of solid of revolution using shells=
2πr(y)h(y)dy (horizontal axis of revc
olution)
where r=the distance between a representative rectangle and the axis of revolution
and
h =height of the representative rectangle
Note that the representative rectangle is parallel to the axis of revolution
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Revolve the region bounded by the graphs of y = x and y = x2 in the first
quadrant about the y-axis. Find the volume of the resulting solid. Sketch the region. Use
shell method.
13
Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
Revolving the region about the line x = l or the line y = k
1. Line is above the region
2. Line is below the region
3. Line is to the right of the region
4. Line is to the left of the region
14
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Let R be the region bounded by the graphs of y = x, y = 2 − x and y = 0.
Compute the volume of the solid formed by revolving R about the lines
1. y = 2
2. y = −1
3. x = 3
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
6.3 Arc Length
Arc Length
Z bp
Formula: s=
1 + [f ′ (x)]2 dx
a
Graph:
Proof:
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example Use a graphing calculator to find the arc length of the curve f (x) = sin x over
[0, π].
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example A cable is to be hung between two poles of equal height that are 20 feet apart.
It can be shown that such a hanging cable assumes the shape
of a catenary. In this case,
x
x suppose that the cable takes the shape of y = 5 e 10 + e− 10 , −10 ≤ x ≤ 10, as seen in the
figure below. How long is the cable?
x3
1
1
+
on the interval
,2 .
Example Find the arc length of the graph of y =
6
2x
2
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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6.4 Areas of Surfaces of Revolution
Revolve the graph of y = f (x) about the x-axis on the interval [a, b] and obtain the surface
of revolution as seen below
The area S of the surface of revolution formed by revolving the graph of y = f (x) about a
horizontal axis is
Z b
Z b
Z b
p
p
′
2
rds = 2π
f (x) 1 + [f ′ (x)]2 dx
S = 2π
r(x) 1 + [f (x)] dx = 2π
a
a
or
a
Z d p
S = 2π
y 1 + [g ′ (y)]2 dy
c
where r(x) is the distance between the graph of f (x) and the axis of revolution.
The area S of the surface of revolution formed by revolving the graph of x = g(y) about a
vertical axis is
Z d
Z d
p
rds
r(y) 1 + [g ′ (y)]2 dy = 2π
S = 2π
c
c
or
Z b p
x 1 + [f ′ (x)]2 dx
S = 2π
a
where r(y) is the distance between the graph of g(y) and the axis of revolution.
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
Example [if time permits] Prove the formula for the surface area.
20
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
Example Find the surface area of the surface generated by revolving y =
about the x-axis.
21
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√
x, for 1 ≤ x ≤ 4,
Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example
Find the surface area of the surface generated by revolving y =
the y-axis.
22
√
x, for 1 ≤ x ≤ 4, about
Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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6.5 Work and Fluid Forces
The term work is used in everyday language to mean the total amount of effort required to
perform a task. In physics it has a technical meaning that depends on the idea of a force.
If an object moves along a straight line with position function s(t), then the force F on the
object in the same direction is given by Newton’s Second Law of Motion:
F = ma = ms′′ (t).
In the metric system, the mass is measured in kg, the displacement in meters, the time in
seconds, and the force in newtons (N = kg × m/s2 ). Thus, a force of 1 N acting on a mass
of 1 kg produces an acceleration of 1 m/s2 . In the American system, the fundamental unit
of force is the pound (lb).
In the case of constant acceleration, the force F is also constant and the work done is given
by
W = F d.
If F is measured in newtons and d in meters, then the unit for W is a newton-meter, which
is called a joule (J). If F is measured in pounds and d in feet, then the unit for W is a
foot-pound (ft-lb). 1 ft-lb ≈ 1.36 J.
Example How much work is done in lifting a 1.2-kg book off the floor to put it on a desk
that is 0.7 m high? Use the fact that the acceleration due to gravity is g = 9.8m/s2 .
Solution: Force=
Work=
Example How much work is done in lifting a 50-lb weight 10 ft off the ground?
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Math 152, Jagodina
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What happens if the force is not constant? Let’s suppose that the object moves along the
x-axis in the positive direction, from x = a to x = b, and at each point x between a and
b a force f (x) acts on the object, where f is a continuous function. We divide the interval
[a, b] into n subintervals with endpoints a, x1 , . . . , xn−1 , b and equal width △x. We choose
a sample point x∗i in the ith subinterval [xi−1 , xi ]. Then the force at that point is f (x∗i ). If
n is large, then △x is small, and since f is continuous, the values of f do not change very
much over the interval [xi−1 , xi ]. In other words, f is almost constant on the interval and so
the work Wi that is done in moving the object from xi−1 to xi is approximately
Wi ≈ f (x∗i )△x.
If we increase n, then the work becomes
W = lim [f (x∗1 )△x + f (x∗2 )△x + . . . + f (x∗n )△x] = lim
n→∞
n→∞
n
X
i=1
f (x∗i )△x =
Z b
f (x)dx.
a
Example When a particle is located x feet from the origin, a force of 3x3 + 5 pounds acts
on it. How much work is done in moving it from x = 2 to x = 5?
Hooke’s Law Hooke’s Law states that the force required to maintain a spring stretched x
units beyond its natural length is proportional to x:
f (x) = kx
where k is a positive constant called the spring constant. Hooke’s Law holds provided that
x is not too large.
The work is then
Z b−natural length
W =
f (x)dx.
a−natural length
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example A force of 40 N is required to hold a spring that has been stretched from its
natural length of 10 cm to a length of 15 cm. How much work is done in stretching the
spring from 15 cm to 18 cm?
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Pumping Liquids out of a Tank
Metric System
The force is F = m kg ·g m/s2 = mg N= V m3 · ρ kg/m3 ·g m/s2 =V ρg N.
Rb
The formula for work is W = a V ρg(h − y)dy where y is the distance between a random
slice/slab and the bottom of a tank and h is the distance between the slice/slab and the
spout/opening of the tank.
American System
The force is F = V ft3 · w lb/ft3 = V w ft where V is the volume of a random slice/slab and
w is the weight of liquid.
Rb
The formula for work is W = a F (h − y)dy where y is the distance between a random
slice/slab and the bottom of a tank and h is the distance between the slice/slab and the
spout/opening of the tank.
Example A cone monument is to be constructed with stones weighing 75 kg/m3 . Its radius
will be 25 m and its height will be 150 m. How much work is required?
(2.6 × 109 )π J
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Math 152, Jagodina
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Example Water in a cylinder of height 25 ft and radius 8 ft is to be pumped out. Water
weighs 62.4 lb/ft3 . Find the work required if
1. The tank is full of water and the water is to be pumped over the top of the tank.
3920707.632 ft-lb
2. The tank is full of water and the water must be pumped to a height of 5 ft above the
top of the tank.
5488990.684 ft-lb
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Math 152, Jagodina
Section 5.6 and Chapter 6 Packet
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Example A rectangular tank has length 10 ft, width 8 ft, and depth 7 ft. If the tank is
full, how much work does it take to pump all the water out through the top of the tank?
Water weighs 62.4 lb/ft3 .
122304 ft-lb
Example A water tank is in the form of a right circular cylinder with height 15 ft and
radius 7 ft. If the tank is full, find the work required to pump all of the water to a point 8
ft above the top of the tank. Water weighs 62.4 lb/ft3 .
2233333.085 ft-lb
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