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In this section, we will learn about: Using integration to find out the volume of a solid.

1.
6

APPLICATIONS OF INTEGRATION

APPLICATIONS OF INTEGRATION

2.
APPLICATIONS OF INTEGRATION

6.2

Volumes

In this section, we will learn about:

Using integration to find out

the volume of a solid.

6.2

Volumes

In this section, we will learn about:

Using integration to find out

the volume of a solid.

3.
In trying to find the volume of a solid,

we face the same type of problem as

in finding areas.

we face the same type of problem as

in finding areas.

4.
We have an intuitive idea of what volume

However, we must make this idea precise

by using calculus to give an exact definition

of volume.

However, we must make this idea precise

by using calculus to give an exact definition

of volume.

5.
We start with a simple type of solid

called a cylinder or, more precisely,

a right cylinder.

called a cylinder or, more precisely,

a right cylinder.

6.
As illustrated, a cylinder is bounded by

a plane region B1, called the base, and

a congruent region B2 in a parallel plane.

The cylinder consists of

all points on line segments

perpendicular to the base

and join B1 to B2.

a plane region B1, called the base, and

a congruent region B2 in a parallel plane.

The cylinder consists of

all points on line segments

perpendicular to the base

and join B1 to B2.

7.
If the area of the base is A and the height of

the cylinder (the distance from B1 to B2) is h,

then the volume V of the cylinder is defined

V = Ah

the cylinder (the distance from B1 to B2) is h,

then the volume V of the cylinder is defined

V = Ah

8.
In particular, if the base is a circle with

radius r, then the cylinder is a circular

cylinder with volume V = πr2h.

radius r, then the cylinder is a circular

cylinder with volume V = πr2h.

9.
RECTANGULAR PARALLELEPIPEDS

If the base is a rectangle with length l and

width w, then the cylinder is a rectangular box

(also called a rectangular parallelepiped) with

volume V = lwh.

If the base is a rectangle with length l and

width w, then the cylinder is a rectangular box

(also called a rectangular parallelepiped) with

volume V = lwh.

10.
IRREGULAR SOLIDS

For a solid S that isn’t a cylinder, we first

‘cut’ S into pieces and approximate each

piece by a cylinder.

We estimate the volume of S by adding the volumes

of the cylinders.

We arrive at the exact volume of S through a limiting

process in which the number of pieces becomes large.

For a solid S that isn’t a cylinder, we first

‘cut’ S into pieces and approximate each

piece by a cylinder.

We estimate the volume of S by adding the volumes

of the cylinders.

We arrive at the exact volume of S through a limiting

process in which the number of pieces becomes large.

11.
IRREGULAR SOLIDS

We start by intersecting S with a plane

and obtaining a plane region that is called

a cross-section of S.

We start by intersecting S with a plane

and obtaining a plane region that is called

a cross-section of S.

12.
IRREGULAR SOLIDS

Let A(x) be the area of the cross-section of S

in a plane Px perpendicular to the x-axis and

passing through the point x, where a ≤ x ≤ b.

Think of slicing S

with a knife

through x and

computing the

area of this slice.

Let A(x) be the area of the cross-section of S

in a plane Px perpendicular to the x-axis and

passing through the point x, where a ≤ x ≤ b.

Think of slicing S

with a knife

through x and

computing the

area of this slice.

13.
IRREGULAR SOLIDS

The cross-sectional area A(x) will vary

as x increases from a to b.

The cross-sectional area A(x) will vary

as x increases from a to b.

14.
IRREGULAR SOLIDS

We divide S into n ‘slabs’ of equal width ∆x

using the planes Px1, Px2, . . . to slice the solid.

Think of slicing a loaf of bread.

We divide S into n ‘slabs’ of equal width ∆x

using the planes Px1, Px2, . . . to slice the solid.

Think of slicing a loaf of bread.

15.
IRREGULAR SOLIDS

If we choose sample points xi* in [xi - 1, xi], we

can approximate the i th slab Si (the part of S

that lies between the planes Pxi 1 and Pxi ) by a

cylinder with base area A(xi*) and ‘height’ ∆x.

If we choose sample points xi* in [xi - 1, xi], we

can approximate the i th slab Si (the part of S

that lies between the planes Pxi 1 and Pxi ) by a

cylinder with base area A(xi*) and ‘height’ ∆x.

16.
IRREGULAR SOLIDS

The volume of this cylinder is A(xi*).

So, an approximation to our intuitive

conception of the volume of the i th slab Si

V ( Si ) A( xi*)x

The volume of this cylinder is A(xi*).

So, an approximation to our intuitive

conception of the volume of the i th slab Si

V ( Si ) A( xi*)x

17.
IRREGULAR SOLIDS

Adding the volumes of these slabs, we get an

approximation to the total volume (that is,

what we think of intuitively as the volume):

n

V A( xi *) x

i 1

This approximation appears to become

better and better as n → ∞.

Think of the slices as becoming thinner and thinner.

Adding the volumes of these slabs, we get an

approximation to the total volume (that is,

what we think of intuitively as the volume):

n

V A( xi *) x

i 1

This approximation appears to become

better and better as n → ∞.

Think of the slices as becoming thinner and thinner.

18.
IRREGULAR SOLIDS

Therefore, we define the volume as the limit

of these sums as n → ∞).

However, we recognize the limit of Riemann

sums as a definite integral and so we have

the following definition.

Therefore, we define the volume as the limit

of these sums as n → ∞).

However, we recognize the limit of Riemann

sums as a definite integral and so we have

the following definition.

19.
DEFINITION OF VOLUME

Let S be a solid that lies between x = a

and x = b.

If the cross-sectional area of S in the plane Px,

through x and perpendicular to the x-axis,

is A(x), where A is a continuous function, then

the volume of S is:

n b

V lim A( xi *) x A( x) dx

x a

i 1

Let S be a solid that lies between x = a

and x = b.

If the cross-sectional area of S in the plane Px,

through x and perpendicular to the x-axis,

is A(x), where A is a continuous function, then

the volume of S is:

n b

V lim A( xi *) x A( x) dx

x a

i 1

20.
When we use the volume formula

b

V A( x)dx , it is important to remember

a

that A(x) is the area of a moving

cross-section obtained by slicing through

x perpendicular to the x-axis.

b

V A( x)dx , it is important to remember

a

that A(x) is the area of a moving

cross-section obtained by slicing through

x perpendicular to the x-axis.

21.
Notice that, for a cylinder, the cross-sectional

area is constant: A(x) = A for all x.

So, our definition of volume gives:

b

V A dx A b a

a

This agrees with the formula V = Ah.

area is constant: A(x) = A for all x.

So, our definition of volume gives:

b

V A dx A b a

a

This agrees with the formula V = Ah.

22.
SPHERES Example 1

Show that the volume of a sphere

of radius r is

3

V r .

4

3

Show that the volume of a sphere

of radius r is

3

V r .

4

3

23.
SPHERES Example 1

If we place the sphere so that its center is

at the origin, then the plane Px intersects

the sphere in a circle whose radius, from the

Pythagorean Theorem,

2 2

y r x

If we place the sphere so that its center is

at the origin, then the plane Px intersects

the sphere in a circle whose radius, from the

Pythagorean Theorem,

2 2

y r x

24.
SPHERES Example 1

So, the cross-sectional area is:

2 2 2

A( x) y (r x )

So, the cross-sectional area is:

2 2 2

A( x) y (r x )

25.
SPHERES Example 1

Using the definition of volume with a = -r and

b = r, we have:

r r

V A( x) dx r x dx 2 2

r r

r

2 2

2 (r x ) dx (The integrand is even.)

0

3 r 3

2 x 3 r

2 r x 2 r

3 0 3

3

r

4

3

Using the definition of volume with a = -r and

b = r, we have:

r r

V A( x) dx r x dx 2 2

r r

r

2 2

2 (r x ) dx (The integrand is even.)

0

3 r 3

2 x 3 r

2 r x 2 r

3 0 3

3

r

4

3

26.
The figure illustrates the definition of volume

when the solid is a sphere with radius r = 1.

From the example, we know that the volume of

the sphere is 43 4.18879

The slabs are circular cylinders, or disks.

when the solid is a sphere with radius r = 1.

From the example, we know that the volume of

the sphere is 43 4.18879

The slabs are circular cylinders, or disks.

27.
The three parts show the geometric

n n

of the Riemann sums

2 2

A( xi )x (1 xi )x when n = 5, 10,

i 1 i 1

and 20 if we choose the sample points xi*

to be the midpoints xi .

n n

of the Riemann sums

2 2

A( xi )x (1 xi )x when n = 5, 10,

i 1 i 1

and 20 if we choose the sample points xi*

to be the midpoints xi .

28.
Notice that as we increase the number

of approximating cylinders, the corresponding

Riemann sums become closer to the true

of approximating cylinders, the corresponding

Riemann sums become closer to the true

29.
VOLUMES Example 2

Find the volume of the solid obtained by

rotating about the x-axis the region under

the curve y x from 0 to 1.

Illustrate the definition of volume by sketching

a typical approximating cylinder.

Find the volume of the solid obtained by

rotating about the x-axis the region under

the curve y x from 0 to 1.

Illustrate the definition of volume by sketching

a typical approximating cylinder.

30.
VOLUMES Example 2

The region is shown in the first figure.

If we rotate about the x-axis, we get the solid

shown in the next figure.

When we slice through the point x, we get a disk

with radius x .

The region is shown in the first figure.

If we rotate about the x-axis, we get the solid

shown in the next figure.

When we slice through the point x, we get a disk

with radius x .

31.
VOLUMES Example 2

The area of the cross-section is:

A( x) ( x ) 2 x

The volume of the approximating cylinder

(a disk with thickness ∆x) is:

A( x)x xx

The area of the cross-section is:

A( x) ( x ) 2 x

The volume of the approximating cylinder

(a disk with thickness ∆x) is:

A( x)x xx

32.
VOLUMES Example 2

The solid lies between x = 0 and x = 1.

1

So, its volume is: V A( x)dx

0

1

xdx

0

2 1

x

2 0 2

The solid lies between x = 0 and x = 1.

1

So, its volume is: V A( x)dx

0

1

xdx

0

2 1

x

2 0 2

33.
VOLUMES Example 3

Find the volume of the solid obtained

by rotating the region bounded by y = x3,

Y = 8, and x = 0 about the y-axis.

Find the volume of the solid obtained

by rotating the region bounded by y = x3,

Y = 8, and x = 0 about the y-axis.

34.
VOLUMES Example 3

As the region is rotated about the y-axis, it

makes sense to slice the solid perpendicular

to the y-axis and thus to integrate with

respect to y.

Slicing at height y,

we get a circular

disk with radius x,

where x 3 y

As the region is rotated about the y-axis, it

makes sense to slice the solid perpendicular

to the y-axis and thus to integrate with

respect to y.

Slicing at height y,

we get a circular

disk with radius x,

where x 3 y

35.
VOLUMES Example 3

So, the area of a cross-section through y is:

2 2 2/3

A( y ) x ( y ) y

3

The volume of the approximating

cylinder is:

2/3

A( y )y y y

So, the area of a cross-section through y is:

2 2 2/3

A( y ) x ( y ) y

3

The volume of the approximating

cylinder is:

2/3

A( y )y y y

36.
VOLUMES Example 3

Since the solid lies between y = 0 and

y = 8, its volume is:

8

V A( y ) dy

0

8

23

y dy

0

8

5

3

96

5 y

3

0 5

Since the solid lies between y = 0 and

y = 8, its volume is:

8

V A( y ) dy

0

8

23

y dy

0

8

5

3

96

5 y

3

0 5

37.
VOLUMES Example 4

The region R enclosed by the curves y = x

and y = x2 is rotated about the x-axis.

Find the volume of the resulting solid.

The region R enclosed by the curves y = x

and y = x2 is rotated about the x-axis.

Find the volume of the resulting solid.

38.
VOLUMES Example 4

The curves y = x and y = x2 intersect at

the points (0, 0) and (1, 1).

The region between them, the solid of rotation, and

cross-section perpendicular to the x-axis are shown.

The curves y = x and y = x2 intersect at

the points (0, 0) and (1, 1).

The region between them, the solid of rotation, and

cross-section perpendicular to the x-axis are shown.

39.
VOLUMES Example 4

A cross-section in the plane Px has the shape

of a washer (an annular ring) with inner

radius x2 and outer radius x.

A cross-section in the plane Px has the shape

of a washer (an annular ring) with inner

radius x2 and outer radius x.

40.
VOLUMES Example 4

Thus, we find the cross-sectional area by

subtracting the area of the inner circle from

the area of the outer circle:

2 2 2

A( x) x ( x )

2 4

( x x )

Thus, we find the cross-sectional area by

subtracting the area of the inner circle from

the area of the outer circle:

2 2 2

A( x) x ( x )

2 4

( x x )

41.
VOLUMES Example 4

1

Thus, we have: V A( x) dx

0

1

2 4

( x x ) dx

0

3 5 1

x x

3 5 0

2

15

1

Thus, we have: V A( x) dx

0

1

2 4

( x x ) dx

0

3 5 1

x x

3 5 0

2

15

42.
VOLUMES Example 5

Find the volume of the solid obtained

by rotating the region in Example 4

about the line y = 2.

Find the volume of the solid obtained

by rotating the region in Example 4

about the line y = 2.

43.
VOLUMES Example 5

Again, the cross-section is a washer.

This time, though, the inner radius is 2 – x

and the outer radius is 2 – x2.

Again, the cross-section is a washer.

This time, though, the inner radius is 2 – x

and the outer radius is 2 – x2.

44.
VOLUMES Example 5

The cross-sectional area is:

2 2 2

A( x) (2 x ) (2 x)

The cross-sectional area is:

2 2 2

A( x) (2 x ) (2 x)

45.
VOLUMES Example 5

So, the volume is:

1

V A( x) dx

0

1

2 2

2 x (2 x) dx

2

0

1

x 5 x 4 x dx

4 2

0

5 3 2 1

x x x 8

5 4

5 3 2 0 5

So, the volume is:

1

V A( x) dx

0

1

2 2

2 x (2 x) dx

2

0

1

x 5 x 4 x dx

4 2

0

5 3 2 1

x x x 8

5 4

5 3 2 0 5

46.
SOLIDS OF REVOLUTION

The solids in Examples 1–5 are all

called solids of revolution because

they are obtained by revolving a region

about a line.

The solids in Examples 1–5 are all

called solids of revolution because

they are obtained by revolving a region

about a line.

47.
SOLIDS OF REVOLUTION

In general, we calculate the volume of

a solid of revolution by using the basic

defining formula

b d

V A( x) dx or V A y dy

a c

In general, we calculate the volume of

a solid of revolution by using the basic

defining formula

b d

V A( x) dx or V A y dy

a c

48.
SOLIDS OF REVOLUTION

We find the cross-sectional area

A(x) or A(y) in one of the following

two ways.

We find the cross-sectional area

A(x) or A(y) in one of the following

two ways.

49.
WAY 1

If the cross-section is a disk, we find

the radius of the disk (in terms of x or y)

and use:

A = π(radius)2

If the cross-section is a disk, we find

the radius of the disk (in terms of x or y)

and use:

A = π(radius)2

50.
WAY 2

If the cross-section is a washer, we first find

the inner radius rin and outer radius rout from

a sketch.

Then, we subtract the area of the inner disk from

the area of the outer disk to obtain:

A = π(outer radius)2 – π(outer radius)2

If the cross-section is a washer, we first find

the inner radius rin and outer radius rout from

a sketch.

Then, we subtract the area of the inner disk from

the area of the outer disk to obtain:

A = π(outer radius)2 – π(outer radius)2

51.
SOLIDS OF REVOLUTION Example 6

Find the volume of the solid obtained

by rotating the region in Example 4

about the line x = -1.

Find the volume of the solid obtained

by rotating the region in Example 4

about the line x = -1.

52.
SOLIDS OF REVOLUTION Example 6

The figure shows the horizontal cross-section.

It is a washer with inner radius 1 + y and

outer radius 1 y .

The figure shows the horizontal cross-section.

It is a washer with inner radius 1 + y and

outer radius 1 y .

53.
SOLIDS OF REVOLUTION Example 6

So, the cross-sectional area is:

2 2

A( y ) (outer radius) (inner radius)

2 2

1 y 1 y

So, the cross-sectional area is:

2 2

A( y ) (outer radius) (inner radius)

2 2

1 y 1 y

54.
SOLIDS OF REVOLUTION Example 6

The volume is:

1

V A( y )dy

0

1 2

2

1 y

0

1 y dy

1

2 y y y dy

0

2

3 1

4 y 2 y 2 y3

3 2 3

0

2

The volume is:

1

V A( y )dy

0

1 2

2

1 y

0

1 y dy

1

2 y y y dy

0

2

3 1

4 y 2 y 2 y3

3 2 3

0

2

55.
In the following examples, we find

the volumes of three solids that are

not solids of revolution.

the volumes of three solids that are

not solids of revolution.

56.
VOLUMES Example 7

The figure shows a solid with a circular base

of radius 1. Parallel cross-sections

perpendicular to the base are equilateral

Find the volume of the solid.

The figure shows a solid with a circular base

of radius 1. Parallel cross-sections

perpendicular to the base are equilateral

Find the volume of the solid.

57.
VOLUMES Example 7

Let’s take the circle to be x2 + y2 = 1.

The solid, its base, and a typical cross-section

at a distance x from the origin are shown.

Let’s take the circle to be x2 + y2 = 1.

The solid, its base, and a typical cross-section

at a distance x from the origin are shown.

58.
VOLUMES Example 7

2

As B lies on the circle, we have y 1 x

So, the base of the triangle ABC is

2

|AB| = 2 1 x

2

As B lies on the circle, we have y 1 x

So, the base of the triangle ABC is

2

|AB| = 2 1 x

59.
VOLUMES Example 7

Since the triangle is equilateral, we see

that its height is 3 y 3 1 x 2

Since the triangle is equilateral, we see

that its height is 3 y 3 1 x 2

60.
VOLUMES Example 7

Thus, the cross-sectional area is :

2 2

A( x) 2 1 x 3 1 x

1

2

2

3(1 x )

Thus, the cross-sectional area is :

2 2

A( x) 2 1 x 3 1 x

1

2

2

3(1 x )

61.
VOLUMES Example 7

The volume of the solid is:

1

V A( x) dx

1

1 1

2 2

3(1 x ) dx 2 3(1 x ) dx

1 0

3 1

x 4 3

2 3 x

3 0 3

The volume of the solid is:

1

V A( x) dx

1

1 1

2 2

3(1 x ) dx 2 3(1 x ) dx

1 0

3 1

x 4 3

2 3 x

3 0 3

62.
VOLUMES Example 8

Find the volume of a pyramid

whose base is a square with side L

and whose height is h.

Find the volume of a pyramid

whose base is a square with side L

and whose height is h.

63.
VOLUMES Example 8

We place the origin O at the vertex

of the pyramid and the x-axis along its

central axis.

Any plane Px that

passes through x and

is perpendicular to

the x-axis intersects

the pyramid in a

square with side of

length s.

We place the origin O at the vertex

of the pyramid and the x-axis along its

central axis.

Any plane Px that

passes through x and

is perpendicular to

the x-axis intersects

the pyramid in a

square with side of

length s.

64.
VOLUMES Example 8

We can express s in terms of x by observing

x s 2 s

from the similar triangles that

h L 2 L

Therefore, s = Lx/h

Another method is

to observe that the

line OP has slope

L/(2h)

So, its equation is

y = Lx/(2h)

We can express s in terms of x by observing

x s 2 s

from the similar triangles that

h L 2 L

Therefore, s = Lx/h

Another method is

to observe that the

line OP has slope

L/(2h)

So, its equation is

y = Lx/(2h)

65.
VOLUMES Example 8

Thus, the cross-sectional area is:

2

2 L 2

A( x) s 2 x

h

Thus, the cross-sectional area is:

2

2 L 2

A( x) s 2 x

h

66.
VOLUMES Example 8

The pyramid lies between x = 0 and x = h.

h

So, its volume is: V A( x) dx

0

2

L 2

h

2 x dx

0 h

2 3 h

L x L2 h

2

h 3 0 3

The pyramid lies between x = 0 and x = h.

h

So, its volume is: V A( x) dx

0

2

L 2

h

2 x dx

0 h

2 3 h

L x L2 h

2

h 3 0 3

67.
In the example, we didn’t need to place

the vertex of the pyramid at the origin.

We did so merely to make the equations

simple.

the vertex of the pyramid at the origin.

We did so merely to make the equations

simple.

68.
Instead, if we had placed the center of

the base at the origin and the vertex on

the positive y-axis, as in the figure, you can

verify that we would have

obtained the integral:

2

L

h

2

V 2 (h y ) dy

0 h

2

Lh

3

the base at the origin and the vertex on

the positive y-axis, as in the figure, you can

verify that we would have

obtained the integral:

2

L

h

2

V 2 (h y ) dy

0 h

2

Lh

3

69.
VOLUMES Example 9

A wedge is cut out of a circular cylinder of

radius 4 by two planes. One plane is

perpendicular to the axis of the cylinder.

The other intersects the first at an angle of 30°

along a diameter of the cylinder.

Find the volume of the wedge.

A wedge is cut out of a circular cylinder of

radius 4 by two planes. One plane is

perpendicular to the axis of the cylinder.

The other intersects the first at an angle of 30°

along a diameter of the cylinder.

Find the volume of the wedge.

70.
VOLUMES Example 9

If we place the x-axis along the diameter

where the planes meet, then the base of

the solid is a semicircle

with equation

2

y 16 x , -4 ≤ x ≤ 4

If we place the x-axis along the diameter

where the planes meet, then the base of

the solid is a semicircle

with equation

2

y 16 x , -4 ≤ x ≤ 4

71.
VOLUMES Example 9

A cross-section perpendicular to the x-axis at

a distance x from the origin is a triangle ABC,

2

whose base is y 16 x and whose height

2

is |BC| = y tan 30 =

° 16 x 3.

A cross-section perpendicular to the x-axis at

a distance x from the origin is a triangle ABC,

2

whose base is y 16 x and whose height

2

is |BC| = y tan 30 =

° 16 x 3.

72.
VOLUMES Example 9

Thus, the cross-sectional area is:

21 2

A( x) 16 x

1

2 16 x

3

2

16 x

2 3

Thus, the cross-sectional area is:

21 2

A( x) 16 x

1

2 16 x

3

2

16 x

2 3

73.
VOLUMES Example 9

The volume is:

4

V A( x) dx

4

2

4 16 x 1 4

4

2 3

dx

3

0

16 x 2

dx

3 4

1 x 128

16 x 3

3 0 3 3

The volume is:

4

V A( x) dx

4

2

4 16 x 1 4

4

2 3

dx

3

0

16 x 2

dx

3 4

1 x 128

16 x 3

3 0 3 3