Chapter 6 Some Applications of the Integral

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Section 6.1 More on Area

a. Representative Rectangle

b. Vertical Separation

c. Example

d. Integration with Respect to y

e. Example

Section 6.2 Volume by Parallel Cross Section; Disks and Washers

a. Volume of a Solid

b. Volume by Integration

c. Example

d. Disk Method About the x-axis

e. Example

f. Disk Method About the y-axis

g. Example

h. Washer Method About the x-axis

i. Example

j. Washer Method About the y-axis

k. Example

Section 6.3 Volume by the Shell Method

a. Shell Method Formulas

b. Average Value Point of View Shell Method About the y-axis

Chapter 6 Some Applications of the Integral

Section 6.4 The Centroid of a Region, Pappus’s Theorem

a. Centroid of a Region

b. Principle 1 and Principle 2

c. Calculating the Centroid

d. Example

e. Centroids Between Two Functions

f. Pappus’s Theorem on Volumes

g. Example

Section 6.5 The Notion of Work

a. Work = Force x Displacement

b. Work for Variable Force

c. Hooke's Law

d. Example

e. Counteracting the Force of Gravity

f. Example

Section 6.6 Fluid Force

a. Downward Fluid Force

b. Fluid Force Against a Wall

c. Example

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More on Area

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More on Area

Integrating the vertical separation gives Riemann Sums of the form

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More on Area

Example

Find the area A of the set shaded in Figure 6.1.4.

Solution

From x = −1 to x = 2 the vertical separation is the difference 2x2 − (x4 − 2x2). Therefore

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More on Area

Areas Obtained by Integration with Respect to y

In this case we are integrating with respect to y the horizontal separation F(y) − G( y) from y = c to y = d.

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More on Area

Example

Find the area of the region bounded on the left by the curve x = y2 and bounded on the right by the curve x = 3 − 2y2.

Solution

The region is sketched in Figure 6.1.6. The points of intersection can be found by solving the two equations simultaneously:

x = y2 and x = 3 − 2y2 together imply that

y = ±1.

The points of intersection are (1, 1) and (1,−1). The easiest way to calculate the area is to set our representative rectangles

horizontally and integrate with respect to y. We then find the area of the region by integrating the horizontal separation

(3 − 2y2) − y2 = 3 − 3y2 from y = −1 to y = 1:

( )

1 2 3 1

1 3 3 3 1 4

A y dy y y

 

=

− =  −  =

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Volume by Parallel Cross Section; Disks and Washers

Figure 6.2.1 shows a plane region Ω and a solid formed by translating Ω along a line perpendicular to the plane of Ω. Such a solid is called a right cylinder with cross section Ω.

If Ω has area A and the solid has height h, then the volume of the solid is a

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Volume by Parallel Cross Section; Disks and Washers

If the cross-sectional area A(x) varies continuously with x, then we can find the volume V of the solid by integrating A(x) from x = a to x = b:

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Volume by Parallel Cross Section; Disks and Washers

Example

Find the volume of the pyramid of height h given that the base of the pyramid is a square with sides of length r and the apex of the pyramid lies directly above the center of the base.

Solution

Set the x-axis as in Figure 6.2.5. The cross section at coordinate x is a square. Let s denote the length of the side of that square. By similar triangles

The area A(x) of the square at coordinate x is s2 = (r2 /h2)(h − x)2. Thus

1 1

2s 2r h x = h

− and therefore s r

(

h x

)

= h

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Volume by Parallel Cross Section; Disks and Washers

Solids of Revolution: Disk Method

The volume of this solid is given by the formula

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Volume by Parallel Cross Section; Disks and Washers

Example

We can generate a circular cone of base radius r and height h by revolving about the x-axis the region below the graph of

By (6.2.3),

volume of cone =

( )

r , 0

f x x x h

= h ≤ ≤

2 2 2 3

2 2

2 2

0 0

0

1

3 3

h

h r r h r x

x dx x dx r h

h h h

π π

π = =     = π

∫ ∫

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Volume by Parallel Cross Section; Disks and Washers

We can interchange the roles played by x and y. By revolving about the y-axis the region of Figure 6.2.10, we obtain a solid of cross-sectional area A(y) = π[g(y)]2 and volume

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Volume by Parallel Cross Section; Disks and Washers

Example

Let Ω be the region bounded below by the curve y = x2/3 + 1, bounded to the left by the y-axis, and bounded above by the line y = 5. Find the volume of the solid generated by revolving Ω about the y-axis.

Solution We need to express the right boundary of Ω as a function of y:

y = x2/3 +1 gives x2/3 = y − 1 and thus x = (y − 1)3/2.

The volume of the solid obtained by revolving Ω about the y-axis is given by the integral

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Volume by Parallel Cross Section; Disks and Washers

Solids of Revolution: Washer Method

The washer method is a slight generalization of the disk method. Suppose that f and g are nonnegative continuous functions with g(x) ≤ f (x) for all x in [a, b]. If we revolve the region Ω about the x-axis, we obtain a solid. The volume of this solid is given by the formula

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Volume by Parallel Cross Section; Disks and Washers

Example

Find the volume of the solid generated by revolving the region between y = x2 and y = 2x

about the x-axis.

Solution

The curves intersect at the points (0, 0) and (2, 4).

For each x from 0 to 2, the x cross section is a washer of outer radius 2x and inner radius x2. By (6.2.5),

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Volume by Parallel Cross Section; Disks and Washers

As before, we can interchange the roles played by x and y.

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Volume by Parallel Cross Section; Disks and Washers

Example

Find the volume of the solid generated by revolving the region between y = x2 and y = 2x

about the y-axis.

Solution

The curves intersect at the points (0, 0) and (2, 4).

For each y from 0 to 4, the y cross section is a washer of outer radius and inner radius ½y. By (6.2.6),

y

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Volume by the Shell Method

To describe the shell method of calculating volumes, we begin with a solid cylinder of radius R and height h, and from it we cut out a cylindrical core of radius r.

Since the original cylinder has volume π R2h and the piece removed has volume πr2h, the cylindrical shell that remains has volume

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Volume by the Shell Method

Now let [a, b] be an interval with a ≥ 0 and let f be a nonnegative function continuous on [a, b]. If the region bounded by the graph of f and the x-axis is revolved about the y-axis, then a solid is generated. The volume of this solid can be obtained from the formula

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Volume by the Shell Method

The volume generated by revolving Ω about the y-axis is given by the formula

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Volume by the Shell Method

Example

Find the volume of the solid generated by revolving the region between y = x2 and y = 2x

about the y-axis.

Solution

The curves intersect at the points (0, 0) and (2, 4).

For each x from 0 to 2 the line segment at a distance x from the y-axis generates a cylindrical surface of radius x, height (2x − x2), and lateral area 2πx(2x − x2). By (6.3.4),

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Volume by the Shell Method

The volume generated by revolving Ω about the x-axis is given by the formula

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Solution

The curves intersect at the points (0, 0) and (2, 4).

We begin by expressing the bounding curves as functions of y.

We write x = for the right boundary and x = ½y for the left boundary. For each y from 0 to 4 the line segment at a distance y from the x-axis generates a cylindrical surface of radius y, height ( − ½y), and lateral area 2πy( − ½y).

By (6.3.5),

Volume by the Shell Method

y

y

Example

Find the volume of the solid generated by revolving the region between y = x2 and y = 2x

about the x-axis.

y

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The Centroid of a Region

The Centroid of a Region

Previously you saw how to locate the center of mass of a thin rod. Suppose now

that we have a thin distribution of matter, a plate, laid out in the xy-plane in the shape of some region Ω. If the mass density of the plate varies from point to point, then the determination of the center of mass of the plate requires the evaluation of a double integral. If, however, the mass density of the plate is constant throughout Ω, then the center of mass depends only on the shape of Ω and falls on a point that we call the centroid. Unless Ω has a very complicated shape, we can locate the centroid by ordinary one-variable integration. We will use two guiding principles to locate the centroid of a plane region.

(

x y,

)

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The Centroid of a Region

Principle 1: Symmetry

If the region has an axis of symmetry, then the centroid lies somewhere along that axis. In particular, if the region has a center, then the center is the centroid.

Principle 2: Additivity

If the region, having area A, consists of a finite number of pieces with areas A1, . . . , An and centroids , then

(

x y,

)

(

x y1, 1

)

,,

(

x yn, n

)

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The Centroid of a Region

Denote the area of Ω by A.

The centroid of can be obtained from the following formulas:

(

x y,

)

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The Centroid of a Region

Example

Locate the centroid of the quarter-disk shown in Figure 6.4.4.

Solution

The quarter-disk is symmetric about the line y = x. Therefore we know that . Here

x = y

Since A = ¼πr2,

The centroid of the quarter-disk is the point

3

2

1 3 1 4

4 3

r r

y

r π

= π =

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The Centroid of a Region

Figure 6.4.6 shows the region Ω between the graphs of two continuous functions f and g. In this case, if Ω has area A and centroid , then

(

x y,

)

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Pappus’s Theorem on Volumes

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Pappus’s Theorem on Volumes

Applying Pappus’s Theorem

Example

Find the volume of the torus generated by revolving the circular disk (x − h)2 + (y − k)2 ≤ r2, h, k ≥ r

(a) about the x-axis, (b) about the y-axis.

Solution The centroid of the disk is the center (h, k). This lies k units from the x-axis and h units from the y-axis. The area of the disk is πr2. Therefore

(a) Vx = 2π(k)(πr2) = 2π2kr2 (b) Vy = 2π(h)(πr2) = 2π2hr2

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The Notion of Work

Suppose now that an object moves along the x-axis from x = a to x = b subject to constant force F. The work done by F during the displacement is by definition the force times the displacement:

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The Notion of Work

If an object moves from x = a to x = b subject to a constant force F, then the work done by F is the constant value of F times b − a.

What is the work done by F if F does not remain constant but instead varies continuously as a function of x?

As you would expect, we then define the work done by F as the average value of F times b − a:

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The Notion of Work

Hooke’s Law

You can sense a variable force in the action of a steel spring. Stretch a spring within its elastic limit and you feel a pull in the opposite direction. The greater the stretching, the harder the pull of the spring. Compress a spring within its elastic limit and you feel a push against you. The greater the compression, the harder the push. According to Hooke’s law (Robert Hooke, 1635–1703), the force exerted by the spring can be written

F(x ) = −kx

where k is a positive number, called the spring constant, and x is the

displacement from the equilibrium position. The minus sign indicates that the spring force always acts in the direction opposite to the direction in which the spring has been deformed (the force always acts so as to restore the spring to its equilibrium state).

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The Notion of Work

Hooke’s Law Example

Stretched ⅓ meter beyond its natural length, a certain spring exerts a

restoring force with a magnitude of 10 newtons. What work must be done to stretch the spring an additional ⅓ meter?

Solution

Place the spring on the x-axis so that the equilibrium point falls at the origin.

View stretching as a move to the right and assume Hooke’s law: F(x) = −kx.

When the spring is stretched ⅓ meter, it exerts a force of −10 newtons (10 newtons to the left). Therefore, −10 = −k( ⅓) and k = 30.

To find the work necessary to stretch the spring an additional ⅓ meter, we integrate the opposite force −F(x) = 30x from x = ⅓ to x = ²∕3:

2 / 3 2 / 3

2

1/ 3 1/ 3

30 30 1 5 joules

W =

x dx = 2 x  =

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The Notion of Work

Counteracting the Force of Gravity

To lift an object we must counteract the force of gravity. Therefore, the work required to lift an object is given by the equation

If an object is lifted from level x = a to level x = b and the weight of the object varies continuously with x—say the weight is w(x)—then the work done by the lifting force is given by the integral

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The Notion of Work

Example

A 150-pound bag of sand is hoisted from the ground to the top of a 50-foot

building by a cable of negligible weight. Given that sand leaks out of the bag at the rate of 0.75 pounds for each foot that the bag is raised, find the work required to hoist the bag to the top of the building.

Solution

Once the bag has been raised x feet, the weight of the bag has been reduced to 150 − 0.75x pounds. Therefore

( ) ( )

( ) ( )( )

50 2 50

0 0

2

150 0.75 150 1 0.75 2

150 50 1 0.75 50 6562.5 foot-pounds 2

W = x dx = x x

= =

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Fluid Force

For any fluid, the weight per unit volume is called the weight density of the fluid.

We’ll denote this by the Greek letter σ.

An object submerged in a fluid experiences a compressive force that acts at right angles to the surface of the body exposed to the fluid.

Fluid in a container exerts a downward force on the base of the container. What is the magnitude of this force? It is the weight of the column of fluid directly above it.

If a container with base area A is filled to a depth h by a fluid of weight density σ, the downward force on the base of the container is given by the product

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Fluid Force

Fluid force acts not only on the base of the container but also on the walls of the container. In Figure 6.6.2, we have depicted a vertical wall standing against a body of liquid. (Think of it as the wall of a container or as a dam at the end of a lake.) We want to calculate the force exerted by the liquid on this wall.

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Fluid Force

Example

A cylindrical water main 6 feet in diameter is laid out horizontally. Given that the main is capped half-full, calculate the fluid force on the cap.

Solution

Here σ = 62.5 pounds per cubic foot. From the figure we see that

The fluid force on the cap can be calculated as follows:

( )

2 9 2

w x = x

( ) ( )

3 2

0

3 2

0

62.5 2 9 62.5 2 9 1125 pounds.

F x x dx

x x dx

=

=

=

Figure

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