APPLICATIONS OF INTEGRATIONAPPLICATIONS OF INTEGRATION

6

6.2

Volumes

APPLICATIONS OF INTEGRATION

In this section, we will learn about:

Using integration to find out

the volume of a solid.

In trying to find the volume of a solid,

we face the same type of problem as

in finding areas.

VOLUMES

We have an intuitive idea of what volume

means.

However, we must make this idea precise

by using calculus to give an exact definition

of volume.

VOLUMES

We start with a simple type of solid

called a cylinder or, more precisely,

a right cylinder.

VOLUMES

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.

CYLINDERS

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

as:

V = Ah

CYLINDERS

In particular, if the base is a circle with

radius r, then the cylinder is a circular

cylinder with volume V = πr2h.

CYLINDERS

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.

RECTANGULAR PARALLELEPIPEDS

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.

IRREGULAR SOLIDS

We start by intersecting S with a plane

and obtaining a plane region that is called

a cross-section of S.

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.

IRREGULAR SOLIDS

The cross-sectional area A(x) will vary

as x increases from a to b.

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.

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 and ) by a

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

1ixP ix

P

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

is: ( ) ( *) iV S A xi x

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):

This approximation appears to become better and better as n → ∞.

Think of the slices as becoming thinner and thinner.

1

( *)n

ii

V A x x

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.

IRREGULAR SOLIDS

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:

1

lim ( *) ( )n b

i axi

V A x x A x dx

DEFINITION OF VOLUME

When we use the volume formula

, it is important to remember

that A(x) is the area of a moving

cross-section obtained by slicing through

x perpendicular to the x-axis.

VOLUMES

( )b

aV A x dx

Notice that, for a cylinder, the cross-sectional

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

So, our definition of volume gives:

This agrees with the formula V = Ah.

b

aV Adx A b a

VOLUMES

Show that the volume of a sphere

of radius r is 34

3 .V r

Example 1SPHERES

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,

is:2 2y r x

Example 1SPHERES

So, the cross-sectional area is:2 2 2( ) ( )A x y r x

Example 1SPHERES

Using the definition of volume with a = -r and

b = r, we have:

(The integrand is

even.)

2 2

2 2

0

3 32 3

0

343

( )

2 ( )

2 23 3

r r

r r

r

r

V A x dx r x dx

r x dx

x rr x r

r

Example 1SPHERES

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

The slabs are circular cylinders, or disks.

43 4.18879

SPHERES

The three parts show the geometric

interpretations of the Riemann sums

when n = 5, 10,

and 20 if we choose the sample points xi*

to be the midpoints .

2 2

1 1

( ) (1 )n n

i ii i

A x x x x

ix

SPHERES

Notice that as we increase the number

of approximating cylinders, the corresponding

Riemann sums become closer to the true

volume.

SPHERES

Find the volume of the solid obtained by

rotating about the x-axis the region under

the curve from 0 to 1.

Illustrate the definition of volume by sketching

a typical approximating cylinder.

y x

Example 2VOLUMES

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 .

VOLUMES

x

Example 2

The area of the cross-section is:

The volume of the approximating cylinder

(a disk with thickness ∆x) is:

2( ) ( )A x x x

( )A x x x x

Example 2VOLUMES

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

So, its volume is:1

0

1

0

12

0

( )

2 2

V A x dx

xdx

x

VOLUMES Example 2

Find the volume of the solid obtained

by rotating the region bounded by y = x3,

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

Example 3VOLUMES

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

VOLUMES Example 3

3x y

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

The volume of the approximating

cylinder is:

2 2 2/33( ) ( )A y x y y

2/3( )A y y y y

Example 3VOLUMES

Since the solid lies between y = 0 and

y = 8, its volume is:

8

0

8 2 3

0

853 35

0

( )

96

5

V A y dy

y dy

y

Example 3VOLUMES

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.

Example 4VOLUMES

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.

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.

Example 4VOLUMES

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

2 4

( ) ( )

( )

A x x x

x x

VOLUMES Example 4

Thus, we have:1

0

1 2 4

0

13 5

0

( )

( )

3 5

2

15

V A x dx

x x dx

x x

Example 4VOLUMES

Find the volume of the solid obtained

by rotating the region in Example 4

about the line y = 2.

Example 5VOLUMES

Again, the cross-section is a washer.

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

and the outer radius is 2 – x2.

VOLUMES Example 5

The cross-sectional area is:

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

Example 5VOLUMES

So, the volume is:

1

0

1 22 2

0

1 4 2

0

15 3 2

0

( )

2 (2 )

5 4

85 4

5 3 2 5

V A x dx

x x dx

x x x dx

x x x

VOLUMES Example 5

The solids in Examples 1–5 are all

called solids of revolution because

they are obtained by revolving a region

about a line.

SOLIDS OF REVOLUTION

In general, we calculate the volume of

a solid of revolution by using the basic

defining formula

( ) orb d

a cV A x dx V A y dy

SOLIDS OF REVOLUTION

We find the cross-sectional area

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

two ways.

SOLIDS OF REVOLUTION

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

WAY 1

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

WAY 2

Find the volume of the solid obtained

by rotating the region in Example 4

about the line x = -1.

Example 6SOLIDS OF REVOLUTION

The figure shows the horizontal cross-section.

It is a washer with inner radius 1 + y and

outer radius

Example 6

1 .y

SOLIDS OF REVOLUTION

So, the cross-sectional area is:

2 2

2 2

( ) (outer radius) (inner radius)

1 1

A y

y y

Example 6SOLIDS OF REVOLUTION

The volume is:

1

0

21 2

0

1 2

0

13 2 32

0

( )

1 1

2

4

3 2 3 2

V A y dy

y y dy

y y y dy

y y y

Example 6SOLIDS OF REVOLUTION

In the following examples, we find

the volumes of three solids that are

not solids of revolution.

VOLUMES

The figure shows a solid with a circular base

of radius 1. Parallel cross-sections

perpendicular to the base are equilateral

triangles.

Find the volume of the solid.

Example 7VOLUMES

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.

Example 7VOLUMES

As B lies on the circle, we have

So, the base of the triangle ABC is |AB| =

21y x

22 1 x

Example 7VOLUMES

Since the triangle is equilateral, we see

that its height is 23 3 1y x

VOLUMES Example 7

Thus, the cross-sectional area is :

VOLUMES Example 7

2 212

2

( ) 2 1 3 1

3(1 )

A x x x

x

The volume of the solid is:

1

1

1 12 2

1 0

13

0

( )

3(1 ) 2 3(1 )

4 32 3

3 3

V A x dx

x dx x dx

xx

Example 7VOLUMES

Find the volume of a pyramid

whose base is a square with side L

and whose height is h.

Example 8VOLUMES

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.

VOLUMES Example 8

We can express s in terms of x by observing

from the similar triangles that

Therefore, s = Lx/h

Another method is to observe that the line OP has slope L/(2h)

So, its equation is y = Lx/(2h)

2

2

x s s

h L L

Example 8VOLUMES

Thus, the cross-sectional area is:VOLUMES Example 8

22 2

2( )

LA x s x

h

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

So, its volume is:0

22

20

2 3 2

20

( )

3 3

h

h

h

V A x dx

Lx dx

h

L x L h

h

Example 8VOLUMES

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.

NOTE

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

220

2

( )

3

h LV h y dy

h

L h

NOTE

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.

Example 9VOLUMES

If we place the x-axis along the diameter

where the planes meet, then the base of

the solid is a semicircle

with equation

-4 ≤ x ≤ 4 216 ,y x

VOLUMES Example 9

A cross-section perpendicular to the x-axis at

a distance x from the origin is a triangle ABC,

whose base is and whose height

is |BC| = y tan 30° =

216y x 216 3.x

Example 9VOLUMES

Thus, the cross-sectional area is:

2 212

2

1( ) 16 16

3

16

2 3

A x x x

x

VOLUMES Example 9

The volume is:

4

4

24 4 2

4 0

43

0

( )

16 116

2 3 3

1 12816

33 3 3

V A x dx

xdx x dx

xx

Example 9VOLUMES