Y

SOLUTION FL

We observe that the integrand is a product. Let us use the integration by parts

formula by setting u(x) = x and dv = cos x dx. Then

u(x) = x du = u (x) dx = 1 dx = dx

AM

v(x) = sin x dv = v (x) dx = cos x dx

Of course we calculate v by anti-differentiation.

TE

According to the integration by parts formula,

x · cos x dx = u dv

=u·v’ v du

= x · sin x ’ sin x dx

= x · sin x ’ (’ cos x) + C

= x · sin x + cos x + C.

Math Note: Observe that we can check the answer in the last example just by

differentiation:

d

[x · sin x + cos x + C] = 1 · sin x + x · cos x ’ sin x = x · cos x.

dx

The choice of u and v in the integration by parts technique is signi¬cant. We

selected u to be x because then du will be 1 dx, thereby simplifying the integral.

199

CHAPTER 7 Methods of Integration

If we had instead selected u = cos x and dv = x dx then we would have found

that v = x 2 /2 and du = ’ sin x dx and the new integral

x2

v du = (’ sin x) dx

2

is more complicated.

EXAMPLE 7.2

Calculate the integral

x 2 · ex dx.

SOLUTION

Keeping in mind that we want to choose u and v so as to simplify the integral,

we take u = x 2 and dv = ex dx. Then

u(x) = x 2 du = u (x) dx = 2x dx

v(x) = ex dv = v (x) dx = ex dx

Then the integration by parts formula tells us that

x 2 ex dx = u dv = uv ’ v du = x 2 · ex ’ ex · 2x dx. (—)

We see that we have transformed the integral into a simpler one (involving

x · ex instead of x 2 · ex ), but another integration by parts will be required. Now

we take u = 2x and dv = ex dx. Then

u(x) = 2x du = u (x) dx = 2 dx

v(x) = ex dv = v (x) dx = ex dx

So equation (—) equals

x 2 · ex ’ u dv = x 2 · ex ’ u · v ’ v du

= x 2 · ex ’ 2x · ex ’ ex · 2 dx

= x 2 · ex ’ 2x · ex + 2ex + C.

We leave it to the reader to check this last answer by differentiation.

You Try It: Calculate the integral

x 2 log x dx.

200 CHAPTER 7 Methods of Integration

EXAMPLE 7.3

Calculate

2

log x dx.

1

SOLUTION

This example differs from the previous ones because now we are evalu-

ating a de¬nite integral (i.e., an integral with numerical limits). We still use

the integration by parts formula, keeping track of the numerical limits of

integration.

We ¬rst notice that, on the one hand, the integrand is not a product. On the

other hand, we certainly do not know an antiderivative for log x. We remedy

the situation by writing log x = 1 · log x. Now the only reasonable choice is to

take u = log x and dv = 1 dx. Therefore

u(x) = log x du = u (x) dx = (1/x) dx

v(x) = x dv = v (x) dx = 1 dx

and

2 2

1 · log x dx = u dv

1 1

2 2

= uv ’ v du

1

1

2 2 1

= (log x) · x ’ x· dx

x

1

1

2

= 2 · log 2 ’ 1 · log 1 ’ 1 dx

1

2

= 2 · log 2 ’ x

1

= 2 · log 2 ’ (2 ’ 1)

= 2 · log 2 ’ 1.

You Try It: Evaluate

4

x 2 · sin x dx.

0

We conclude this section by doing another de¬nite integral, but we use a slightly

different approach from that in Example 7.3.

201

CHAPTER 7 Methods of Integration

EXAMPLE 7.4

Calculate the integral

2π

sin x cos x dx.

π/2

SOLUTION

We use integration by parts, but we apply the technique to the corresponding

inde¬nite integral. We let u = sin x and dv = cos x dx. Then

u(x) = sin x du = u (x) dx = cos x dx

v(x) = sin x dv = v (x) dx = cos x dx

So

sin x cos x dx = u dv

= uv ’ v du

= (sin x) · (sin x) ’ sin x cos x dx.

At ¬rst blush, it appears that we have accomplished nothing. For the new

integral is just the same as the old integral. But in fact we can move the new

integral (on the right) to the left-hand side to obtain

sin x cos x dx = sin2 x.

2

Throwing in the usual constant of integration, we obtain

12

sin x cos x dx =sin x + C.

2

Now we complete our work by evaluating the de¬nite integral:

2π

2π 1 1 1

sin x cos xdx = sin2 x = sin2 2π ’ sin2 (π/2) = ’ .

2 2 2

π/2 π/2

We see that there are two ways to treat a de¬nite integral using integration by

parts. One is to carry the limits of integration along with the parts calculation. The

other is to do the parts calculation ¬rst (with an inde¬nite integral) and then plug

in the limits of integration at the end. Either method will lead to the same solution.

You Try It: Calculate the integral

2

e’x cos 2x dx.

0

202 CHAPTER 7 Methods of Integration

7.2 Partial Fractions

7.2.1 INTRODUCTORY REMARKS

The method of partial fractions is used to integrate rational functions, or quotients

of polynomials. We shall treat here some of the basic aspects of the technique.

The ¬rst fundamental observation is that there are some elementary rational

functions whose integrals we already know.

I Integrals of Reciprocals of Linear Functions An integral

1

dx

ax + b

with a = 0 is always a logarithmic function. In fact we can calculate

1 1 1 1

dx = dx = log |x + b/a|.

ax + b x + b/a

a a

II Integrals of Reciprocals of Quadratic Expressions An integral

1

dx,

c + ax 2

when a and c are positive, is an inverse trigonometric function. In fact we can use

what we learned in Section 6.6.3 to write

1 1 1

dx = dx

c + ax 2 1 + (a/c)x 2

c

1 1

= √ dx

1 + ( a/cx)2

c

√

1 1

a

=√ √ √ dx

1 + ( a/cx)2

ac c

1

= √ arctan a/cx + C.

ac

III More Integrals of Reciprocals of Quadratic Expressions An integral

1

dx

ax 2 + bx + c

203

CHAPTER 7 Methods of Integration

with a > 0, and discriminant b2 ’ 4ac negative, will also be an inverse trigono-

metric function. To see this, we notice that we can write

b

ax 2 + bx + c = a x 2 + x + +c

a

b2 b2

b

=a x + x+ 2 + c’

2

4a

4a

a

2

b2

b

=a· x+ + c’ .

2a 4a

Since b2 ’ 4ac < 0, the ¬nal expression in parentheses is positive. For simplicity,

let » = b/2a and let γ = c ’ b2 /(4a). Then our integral is

1

dx.

γ + a · (x + »)2

Of course we can handle this using II above. We ¬nd that

1 1

dx = dx

ax 2 + bx + c γ + a · (x + »)2

√

1 a

=√ · arctan √ · (x + ») + C.

aγ γ

IV Even More on Integrals of Reciprocals of Quadratic Expressions

Evaluation of the integral

1

dx

ax 2 + bx + c

when the discriminant b2 ’ 4ac is ≥ 0 will be a consequence of the work we do

below with partial fractions. We will say no more about it here.

7.2.2 PRODUCTS OF LINEAR FACTORS

We illustrate the technique of partial fractions by way of examples.

EXAMPLE 7.5

Here we treat the case of distinct linear factors.

Let us calculate

1

dx.

x 2 ’ 3x + 2

204 CHAPTER 7 Methods of Integration

SOLUTION

We notice that the integrand factors as

1 1

= (—)

.

(x ’ 1)(x ’ 2)

x 2 ’ 3x + 2

[Notice that the quadratic polynomial in the denominator will factor precisely

when the discriminant is ≥ 0, which is case IV from Subsection 7.2.1.] Our

goal is to write the fraction on the right-hand side of (—) as a sum of simpler

fractions. With this thought in mind, we write

1 A B

= + ,

(x ’ 1)(x ’ 2) x’1 x’2

where A and B are constants to be determined. Let us put together the two

fractions on the right by placing them over the common denominator (x ’ 1)—

(x ’ 2). Thus

A(x ’ 2) + B(x ’ 1)

1 A B

= + = .

(x ’ 1)(x ’ 2) x’1 x’2 (x ’ 1)(x ’ 2)

The only way that the fraction on the far left can equal the fraction on the far

right is if their numerators are equal. This observation leads to the equation

1 = A(x ’ 2) + B(x ’ 1)

or

0 = (A + B)x + (’2A ’ B ’ 1).

Now this equation is to be identically true in x; in other words, it must hold for

every value of x. So the coef¬cients must be 0.

At long last, then, we have a system of two equations in two unknowns:

A+B =0

’2A ’ B ’ 1 = 0

Of course this system is easily solved and the solutions found to be A =

’1, B = 1. We conclude that

’1

1 1

= + .

(x ’ 1)(x ’ 2) x’1 x’2

What we have learned, then, is that

’1

1 1

dx = dx + dx.

x’1 x’2

x 2 ’ 3x + 2

205

CHAPTER 7 Methods of Integration

Each of the individual integrals on the right may be evaluated using the

information in I of Subsection 7.2.1. As a result,

1

dx = ’ log |x ’ 1| + log |x ’ 2| + C.

x 2 ’ 3x + 2

You Try It: Calculate the integral

4 dx

.

x 2 + 5x + 4

1

Now we consider repeated linear factors.

EXAMPLE 7.6

Let us evaluate the integral

dx

.

x 3 ’ 4x 2 ’ 3x + 18

SOLUTION

In order to apply the method of partial fractions, we ¬rst must factor the

denominator of the integrand. It is known that every polynomial with real

coef¬cients will factor into linear and quadratic factors. How do we ¬nd this

factorization? Of course we must ¬nd a root. For a polynomial of the form

x k + ak’1 x k’1 + ak’2 x k’2 + · · · + a1 x + a0 ,

any integer root will be a factor of a0 . This leads us to try ±1, ±2, ±3, ±6, ±9

and ±18. We ¬nd that ’2 and 3 are roots of x 3 ’ 4x 2 ’ 3x + 18. In point of

fact,

x 3 ’ 4x 2 ’ 3x + 18 = (x + 2) · (x ’ 3)2 .

An attempt to write

1 A B

= +

x+2 x’3

x 3 ’ 4x 2 ’ 3x + 18

will not work. We encourage the reader to try this for himself so that he will

understand why an extra idea is needed.

In fact we will use the paradigm

1 A B C

= + + .

x + 2 x ’ 3 (x ’ 3)2

x 3 ’ 4x 2 ’ 3x + 18

Putting the right-hand side over a common denominator yields

A(x ’ 3)2 + B(x + 2)(x ’ 3) + C(x + 2)

1

= .

x 3 ’ 4x 2 ’ 3x + 18 x 3 ’ 4x 2 ’ 3x + 18

206 CHAPTER 7 Methods of Integration

Of course the numerators must be equal, so

1 = A(x ’ 3)2 + B(x + 2)(x ’ 3) + C(x + 2).

We rearrange the equation as

(A + B)x 2 + (’6A ’ B + C)x + (9A ’ 6B + 2C ’ 1) = 0.

Since this must be an identity in x, we arrive at the system of equations

A+B =0

’6A ’ B + C =0

9A ’ 6B + 2C ’ 1 = 0

This system is easily solved to yield A = 1/25, B = ’1/25, C = 1/5.

As a result of these calculations, our integral can be transformed as follows:

’1/25

1 1/25 1/5

dx = dx + dx + dx.

x+2 x’3

x 3 ’ 4x 2 ’ 3x + 18 (x ’ 3)2

The ¬rst integral equals (1/25) log |x + 2|, the second integral equals ’(1/25)

log |x ’ 3|, and the third integral equals ’(1/5)/(x ’ 3).

In summary, we have found that

log |x + 2| log |x ’ 3|

1 1

dx = ’ ’ + C.

5(x ’ 3)

x 3 ’ 4x 2 ’ 3x + 18 25 25

You Try It: Evaluate the integral

4 x dx

.

x 3 + 5x 2 + 7x + 3

2

7.2.3 QUADRATIC FACTORS

EXAMPLE 7.7

Evaluate the integral

x dx

.

x 3 + 2x 2 + x + 2

SOLUTION

Since the denominator is a cubic polynomial, it must factor. The factors of

the constant term are ±1 and ±2. After some experimentation, we ¬nd that

x = ’2 is a root and in fact the polynomial factors as

x 3 + 2x 2 + x + 2 = (x + 2)(x 2 + 1).

207

CHAPTER 7 Methods of Integration

Thus we wish to write the integrand as the sum of a factor with denominator

(x + 2) and another factor with denominator (x 2 + 1). The correct way to do

this is

Bx + C

x x A

= = +2 .

x+2

x 3 + 2x 2 + x + 2 (x + 2)(x 2 + 1) x +1

We put the right-hand side over a common denominator to obtain

A(x 2 + 1) + (Bx + C)(x + 2)

x

= .

x 3 + 2x 2 + x + 2 x 3 + 2x 2 + x + 2

Identifying numerators leads to

x = (A + B)x 2 + (2B + C)x + (A + 2C).

This equation must be identically true, so we ¬nd (identifying powers of x) that

A+ B =0

2B + C = 1

+ 2C = 0

A

Solving this system, we ¬nd that A = ’2/5, B = 2/5, C = 1/5. So

’2/5 (2/5)x + (1/5)

x dx

= dx + dx

x+2

x 3 + 2x 2 + x + 2 x2 + 1

’2 1 2x 1 dx

= log |x + 2| + dx + dx

x2 + 1 x2 + 1

5 5 5

’2 1 1

= log |x + 2| + log |x 2 + 1| + arctan x + C.

5 5 5

You Try It: Calculate the integral

1 dx

.

x 3 + 6x 2 + 9x

0

You Try It: Calculate the integral

dx

.

x3 + x

7.3 Substitution

Sometimes it is convenient to transform a given integral into another one by means

of a change of variable. This method is often called “the method of change of

variable” or “u-substitution.”

208 CHAPTER 7 Methods of Integration

To see a model situation, imagine an integral

b

f (x) dx.

a

If the techniques that we know will not suf¬ce to evaluate the integral, then we might

attempt to transform this to another integral by a change of variable x = •(t). This

entails dx = • (t) dt. Also

x = a ←’ t = • ’1 (a) x = b ←’ t = • ’1 (b).

and

Thus the original integral is transformed to

• ’1 (b)

f (•(t)) · • (t) dt.

• ’1 (a)

It turns out that, with a little notation, we can make this process both convenient

and straightforward.

Y

We now illustrate this new paradigm with some examples. We begin with an

inde¬nite integral.

FL

EXAMPLE 7.8

AM

Evaluate

[sin x]5 · cos x dx.

TE

SOLUTION

On looking at the integral, we see that the expression cos x is the derivative

of sin x. This observation suggests the substitution sin x = u. Thus cos x dx =

du. We must now substitute these expressions into the integral, replacing all

x-expressions with u-expressions. When we are through with this process, no

x expressions can remain. The result is

u5 du.

This is of course an easy integral for us. So we have

u6

[sin x] · cos x dx = u du = + C.

5 5

6

Now the important ¬nal step is to resubstitute the x-expressions in place of

the u-expressions. The result is then

sin6 x

[sin x] · cos x dx = + C.

5

6

209

CHAPTER 7 Methods of Integration

Math Note: Always be sure to check your work. You can differentiate the answer

in the last example to recover the integrand, con¬rming that the integration has

been performed correctly.

EXAMPLE 7.9

Evaluate the integral

3

2x x 2 + 1 dx.

0

SOLUTION

We recognize that the expression 2x is the derivative of x 2 +1. This suggests

the substitution u = x 2 + 1. Thus du = 2x dx. Also x = 0 ←’ u = 1 and

x = 3 ←’ u = 10. The integral is thus transformed to

10 √

u du.

1

This new integral is a bit easier to understand if we write the square root as

a fractional power:

10

2 · 103/2 2

10 u3/2 103/2 13/2

du = = ’ = ’.

1/2

u

3/2 3/2 3/2 3 3

1 1

You Try It: Evaluate the integral

5 dx

.

x · log |x|

3

Math Note: Just as with integration by parts, we always have the option of ¬rst

evaluating the inde¬nite integral and then evaluating the limits at the very end. The

next example illustrates this idea.

EXAMPLE 7.10

Evaluate

π/2

cos x

dx.

sin x

π/3

SOLUTION

Since cos x is the derivative of sin x, it is natural to attempt the substitution

u = sin x. Then du = cos x dx. [Explain why it would be a bad idea to let

u = cos x.] We ¬rst treat the improper integral. We ¬nd that

cos x du

dx = = log |u| + C.

sin x u

210 CHAPTER 7 Methods of Integration

Now we resubstitute the x-expressions to obtain

cos x

dx = log | sin x| + C.

sin x

Finally we can evaluate the original de¬nite integral:

π/2

π/2 cos x

dx = log | sin x|

sin x

π/3 π/3

√

3

= log | sin π/2| ’ log | sin π/3| = log 1 ’ log

2

1

= ’ log 3 + log 2.

2

You Try It: Calculate the integral

3 t dt

.

(t 2 + 1) log(t 2 + 1)

’2

7.4 Integrals of Trigonometric Expressions

Trigonometric expressions arise frequently in our work, especially as a result of

substitutions. In this section we develop a few examples of trigonometric integrals.

The following trigonometric identities will be particularly useful for us.

I We have

1 ’ cos 2x

sin2 x = .

2

The reason is that

cos 2x = cos2 x ’ sin2 x = 1 ’ sin2 x ’ sin2 x = 1 ’ 2 sin2 x.

II We have

1 + cos 2x

cos2 x = .

2

The reason is that

cos 2x = cos2 x ’ sin2 x = cos2 x ’ 1 ’ cos2 x = 2 cos2 x ’ 1.

Now we can turn to some examples.

211

CHAPTER 7 Methods of Integration

EXAMPLE 7.11

Calculate the integral

cos2 x dx.

SOLUTION

Of course we will use formula II. We write

1 + cos 2x

cos2 x dx = dx

2

1 1

= dx + cos 2x dx

2 2

1

x

= + sin 2x + C.

24

EXAMPLE 7.12

Calculate the integral

sin3 x cos2 x dx.

SOLUTION

When sines and cosines occur together, we always focus on the odd power

(when one occurs). We write

sin3 x cos2 x = sin x sin2 x cos2 x = sin x 1 ’ cos2 x cos2 x

= cos2 x ’ cos4 x sin x.

Then

sin3 x cos2 dx = cos2 x ’ cos4 x sin x dx.

A u-substitution is suggested: We let u = cos x, du = ’ sin x dx. Then the

integral becomes

u3 u5

’ u ’ u du = ’ + + C.

2 4

3 5

Resubstituting for the u variable, we obtain the ¬nal solution of

cos3 x cos5 x

sin x cos dx = ’ + + C.

3 2

3 5

You Try It: Calculate the integral

sin2 3x cos5 3x dx.

212 CHAPTER 7 Methods of Integration

EXAMPLE 7.13

Calculate

π/2

sin4 x cos4 x dx.

0

SOLUTION

Substituting

1 ’ cos 2x 1 + cos 2x

sin2 x = cos2 x =

and

2 2

into the integrand yields

2 2

1 ’ cos 2x 1 + cos 2x

π/2

· dx

2 2

0

π/2

1

= 1 ’ 2 cos2 2x + cos4 2x dx.

16 0

Again using formula II, we ¬nd that our integral becomes

1 + cos 4x 2

π/2

1

1 ’ [1 + cos 4x] + dx

16 2

0

π/2

1 1

= 1 ’ [1 + cos 4x] + 1 + 2 cos 4x + cos2 4x dx.

16 0 4

Applying formula II one last time yields

1 + cos 8x

π/2

1 1

1 ’ [1 + cos 4x] + 1 + 2 cos 4x + dx

16 4 2

0

π/2

1 1 1 1 sin 8x

x

= ’ sin 4x + x + sin 4x + +

16 4 4 2 2 16 0

1 1π 1

π

= ’0 + +0+ +0 ’ ’0 + (0 + 0 + 0 + 0)

16 42 4 4

3π

= .

256

You Try It: Calculate the integral

π/3

sin3 s cos3 s ds.

π/4

You Try It: Calculate the integral

π/3

sin2 s cos4 s ds.

π/4

213

CHAPTER 7 Methods of Integration

Integrals involving the other trigonometric functions can also be handled with

suitable trigonometric identities. We illustrate the idea with some examples that are

handled with the identity

sin2 x + cos2 x

sin2 x 1

tan x + 1 = +1= = = sec2 x.

2

2x 2x 2x

cos cos cos

EXAMPLE 7.14

Calculate

tan3 x sec3 x dx.

SOLUTION

Using the same philosophy about odd exponents as we did with sines and

cosines, we substitute sec2 x ’ 1 for tan2 x. The result is

tan x sec2 x ’ 1 sec3 x dx.

We may regroup the terms in the integrand to obtain

sec4 x ’ sec2 x sec x tan x dx.

A u-substitution suggests itself: We let u = sec x and therefore du =

sec x tan x dx. Thus our integral becomes

u5 u3

u ’ u du = ’ + C.

4 2

5 3

Resubstituting the value of u gives

sec5 x sec3 x

tan x sec x dx = ’ + C.

3 3

5 3

EXAMPLE 7.15

Calculate

π/4

sec4 x dx.

0

214 CHAPTER 7 Methods of Integration

SOLUTION

We write

π/4 π/4

sec x dx = sec2 x · sec2 x dx

4

0 0

π/4

= (tan2 x + 1) sec2 x dx.

0

Letting u = tan x and du = sec2 x dx then gives the integral

1

1 u3

u + 1 du = +u

2

3

0 0

4

=.

3

You Try It: Calculate the integral

2π

sin6 x cos4 x dx.

π

Further techniques in the evaluation of trigonometric integrals will be explored

in the exercises.

Exercises

1. Use integration by parts to evaluate each of the following inde¬nite

integrals.

log2 x dx

(a)

x · e3x dx

(b)

x 2 cos x dx

(c)

(d) t sin 3t cos 3t dt

(e) cos y ln(sin y) dy

x 2 e4x dx

(f)

215

CHAPTER 7 Methods of Integration

2. Use partial fractions to evaluate each of the following inde¬nite integrals.

dx

(a)

(x + 2)(x ’ 5)

dx

(b)

(x + 1)(x 2 + 1)

dx

(c)

x 3 ’ 2x 2 ’ 5x + 6

x dx

(d)

x4 ’ 1

dx

(e)

x 3 ’ x 2 ’ 8x + 12

x+1

(f)

x3 ’ x2 + x ’ 1

3. Use the method of u-substitution to evaluate each of the following inde¬nite

integrals.

(1 + sin2 x)2 2 sin x cos x dx

(a)

√

sin x

√

(b) dx

x

cos(ln x) sin(ln x)

(c) dx

x

etan x sec2 x dx

(d)

sin x

(e) dx

1 + cos2 x

sec2 x

(f) dx

1 ’ tan2 x

4. Evaluate each of the following inde¬nite trigonometric integrals.

sin x cos2 x dx

(a)

sin3 x cos2 x dx

(b)

tan3 x sec2 x dx

(c)

216 CHAPTER 7 Methods of Integration

tan x sec3 x dx

(d)

sin2 x cos2 x dx

(e)

sin x cos4 x dx

(f)

5. Calculate each of the following de¬nite integrals.

1

ex sin x dx

(a)

0

e

x 2 ln x dx

(b)

1

(2x + 1) dx

4

(c)

x3 + x2

2

π

sin2 x cos2 x dx

(d)

0

π/3

(e) tan x sec x dx

π/4

π/4 tan x

(f) dx

cos x

0

CHAPTER 8

Applications of

the Integral

8.1 Volumes by Slicing

8.1.0 INTRODUCTION

When we learned the theory of the integral, we found that the basic idea was that

one can calculate the area of an irregularly shaped region by subdividing the region

into “rectangles.” We put the word “rectangle” here in quotation marks because the

region is not literally broken up into rectangles; the union of the rectangles differs

from the actual region under consideration by some small errors (see Fig. 8.1). But

the contribution made by these errors vanishes as the mesh of the rectangles become

¬ner and ¬ner.

We will now implement this same philosophy to calculate certain volumes. Some

of these will be volumes that you have heard about (e.g., the sphere or cone), but

have never known why the volume had the value that it had. Others will be entirely

new (e.g., the paraboloid of revolution). We will again use the method of slicing.

8.1.1 THE BASIC STRATEGY

Imagine a solid object situated as in Fig. 8.2. Observe the axes in the diagram, and

imagine that we slice the ¬gure with slices that are vertical (i.e., that rise out of

the x-y plane) and that are perpendicular to the x-axis (and parallel to the y-axis).

Look at Fig. 8.3. Notice, in the ¬gure, that the ¬gure extends from x = a to x = b.

217

Copyright 2003 by The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

218 CHAPTER 8 Applications of the Integral

y = f (x)

a b

Fig. 8.1

Y

FL

AM

Fig. 8.2

TE

Fig. 8.3

If we can express the area of the slice at position x as a function A(x) of x,

then (see Fig. 8.4) the volume of a slice of thickness x at position x will be about

A(x) · x. If P = {x0 , x1 , . . . , xk } is a partition of the interval [a, b] then the

volume of the original solid object will be about

V= A(xj ) · x.

j

219

CHAPTER 8 Applications of the Integral

Fig. 8.4

As the mesh of the partition becomes ¬ner and ¬ner, this (Riemann) sum will tend

to the integral

b

A(x) dx.

a

We declare the value of this integral to be the volume V of the solid object.

8.1.2 EXAMPLES

EXAMPLE 8.1

Calculate the volume of the right circular cone with base a disc of radius 3

and height 6.

SOLUTION

Examine Fig. 8.5. We have laid the cone on its side, so that it extends from x =

0 to x = 6. The upper edge of the ¬gure is the line y = 3 ’ x/2. At position x,

y

3

x

6

Fig. 8.5

220 CHAPTER 8 Applications of the Integral

the height of the upper edge is 3 ’ x/2, and that number is also the radius of

the circular slice at position x (Fig. 8.6). Thus the area of that slice is

2

x

A(x) = π 3 ’ .

2

thickness ,x

_

x/2

3

Fig. 8.6

We ¬nd then that the volume we seek is

6

2(3 ’ x/2)3

6 6 2

x

V= A(x) dx = π 3’ dx = ’π = 18π.

2 3

0 0 0

You Try It: Any book of tables (see [CRC]) will tell you that the volume of a right

circular cone of base radius r and height h is 1 π r 2 h. This formula is consistent

3

with the result that we obtained in the last example for r = 3 and h = 6. Use the

technique of Example 8.1 to verify this more general formula.

EXAMPLE 8.2

A solid has base the unit disk in the x-y plane. The vertical cross section at

position x is an equilateral triangle. Calculate the volume.

SOLUTION

Examine Fig. 8.7. The unit circle has equation x 2 +y 2 = 1. For our purposes,

this is more conveniently written as

y = ± 1 ’ x 2. ()

Thus the endpoints of the base of the equilateral triangle at position x are the

√

points (x, ± 1 ’ x 2 ). In other words, the base of this triangle is

b = 2 1 ’ x 2.

Examine Fig. 8.8. We see that an equilateral triangle of side b has height

√ √2

3b/2. Thus the area of the triangle is 3b /4. In our case then, the equilateral

221

CHAPTER 8 Applications of the Integral

Fig. 8.7

√3

b b b

2

b

Fig. 8.8

triangular slice at position x has area

√

√

3 2

A(x) = · 2 1 ’ x 2 = 3(1 ’ x 2 ).

4

Finally, we may conclude that the volume we seek is

1

V= A(x) dx

’1

1√

= 3(1 ’ x 2 ) dx

’1

√ 1

x3

= 3 x’

3 ’1

√ ’1

1

= 3 1’ ’ (’1) ’

3 3

√

43

= .

3

222 CHAPTER 8 Applications of the Integral

EXAMPLE 8.3

A solid has base in the x-y plane consisting of a unit square with center at

the origin and vertices on the axes. The vertical cross-section at position x

is itself a square. Refer to Fig. 8.9. What is the volume of this solid?

Fig. 8.9

SOLUTION

It is suf¬cient to calculate the volume of the right half of this solid, and to

√

double the answer. Of course the extent of x is then 0 ¤√ ¤ 1/ 2. At position

x

x, the height of the upper edge√ the square base is 1/ 2 ’ x. So the base of

of

the vertical square slice is 2(1/ 2 ’ x) (Fig. 8.10). The area of the slice is then

√ √

2 2

A(x) = 2 1/ 2 ’ x = 2 ’ 2x .

2 (1/√2 _ x)

2(1/√2 _ x)

Fig. 8.10

It follows that

√

1/ 2

V =2· A(x) dx

0

√

√

1/ 2

2

=2 2 ’ 2x dx

0

223

CHAPTER 8 Applications of the Integral

√

√ 1/ 2

3

2 ’ 2x

=2 ’

6

0

√

03 22

=2 ’ ’ ’

6 6

√

22

= .

3

You Try It: Calculate the volume of the solid with base in the plane an equilateral

triangle of side 1, with base on the x-axis, and with vertical cross-section parallel

to the y-axis consisting of an equilateral triangle.

EXAMPLE 8.4

Calculate the volume inside a sphere of radius 1.

SOLUTION

It is convenient for us to think of the sphere as centered at the origin in the

x-y plane. Thus (Fig. 8.11) the slice at position x, ’1 ¤ x ¤ 1, is a disk. Since

we are working with base the unit circle, we may calculate ( just as in Example

√ √

8.2) that the diameter of this disk is 2 1 ’ x 2 . Thus the radius is 1 ’ x 2 and

the area is

2

A(x) = π · 1 ’ x2 = π · (1 ’ x 2 ).

Fig. 8.11

In conclusion, the volume we seek is

1

V= π(1 ’ x 2 ) dx.

’1

224 CHAPTER 8 Applications of the Integral

We easily evaluate this integral as follows:

1

x3

V =π· x’

3 ’1

’1

1

=π · 1’ ’ ’1 ’

3 3

4

= π.

3

You Try It: Any book of tables (see [CRC]) will tell you that the volume inside a

sphere of radius r is 4π r 3 /3. This formula is consistent with the answer we obtained

in the last example for r = 1. Use the method of this section to derive this more

general formula for arbitrary r.

8.2 Volumes of Solids of Revolution

8.2.0 INTRODUCTION

A useful way”and one that we encounter frequently in everyday life”for generat-

ing solids is by revolving a planar region about an axis. For example, we can think

of a ball (the interior of a sphere) as the solid obtained by rotating a disk about

an axis (Fig. 8.12). We can think of a cylinder as the solid obtained by rotating

a rectangle about an adjacent axis (Fig. 8.13). We can think of a tubular solid as

obtained by rotating a rectangle around a non-adjacent axis (Fig. 8.14).

Fig. 8.12

Fig. 8.13

225

CHAPTER 8 Applications of the Integral

Fig. 8.14

There are two main methods for calculating volumes of solids of revolution:

the method of washers and the method of cylinders. The ¬rst of these is really an

instance of volume by slicing, just as we saw in the last section. The second uses a

different geometry; instead of slices one uses cylindrical shells. We shall develop

both technologies by way of some examples.

8.2.1 THE METHOD OF WASHERS

EXAMPLE 8.5

A solid is formed by rotating the triangle with vertices (0, 0), (2, 0), and (1, 1)

about the x-axis. See Fig. 8.15. What is the resulting volume?

(1,1)

(0,0) (2,0)

Fig. 8.15

SOLUTION

For 0 ¤ x ¤ 1, the upper edge of the triangle has equation y = x. Thus

the segment being rotated extends from (x, 0) to (x, x). Under rotation, it will

generate a disk of radius x, and hence area A(x) = π x 2 . Thus the volume

generated over the segment 0 ¤ x ¤ 1 is

1

V1 = π x 2 dx.

0

Similarly, for 1 ¤ x ¤ 2, the upper edge of the triangle has equation y =

2 ’ x. Thus the segment being rotated extends from (x, 0) to (x, 2 ’ x). Under

226 CHAPTER 8 Applications of the Integral

rotation, it will generate a disk of radius 2’x, and hence area A(x) = π(2’x)2 .

Thus the volume generated over the segment 1 ¤ x ¤ 2 is

2

V2 = π(2 ’ x)2 dx.

1

In summary, the total volume of our solid of revolution is

V = V1 + V2

1 2

’(2 ’ x)3

x3

=π +

3 3

0 1

1 1

=π ’ 0 + ’0 ’ ’

3 3

2π

= .

3

EXAMPLE 8.6

The portion of the curve y = x 2 between x = 1 and x = 4 is rotated about

the x-axis (Fig. 8.16). What volume does the resulting surface enclose?

Fig. 8.16

SOLUTION

At position x, the curve is x 2 units above the x-axis. The point (x, x 2 ), under

rotation, therefore generates a circle of radius x 2 . The disk that the circle bounds

227

CHAPTER 8 Applications of the Integral

has area A(x) = π · (x 2 )2 . Thus the described volume is

4

4 x5 1023π

V= π · x dx = π · =

4

.

5 5

1 1

Math Note: The reasoning we have used in the last two examples shows this:

If the curve y = f (x), a ¤ x ¤ b, is rotated about the x-axis then the volume

enclosed by the resulting surface is

b

V= π · [f (x)]2 dx.

a

You Try It: Calculate the volume enclosed by the surface obtained by rotating

√

the curve y = x + 1, 4 ¤ x ¤ 9, about the x-axis.

EXAMPLE 8.7

The curve y = x 3 , 0 ¤ x ¤ 3, is rotated about the y-axis. What volume

does the resulting surface enclose?

SOLUTION

It is convenient in this problem to treat y as the independent variable and x as

the dependent variable. So we write the curve as x = y 1/3 . Then, at position y,

the curve is distance y 1/3 from the axis so the disk generated under rotation will

have radius y 1/3 (Fig. 8.17). Thus the disk will have area A(y) = π · [y 1/3 ]2 .

Also, since x ranges from 0 to 3 we see that y ranges from 0 to 27. As a result,

Fig. 8.17

228 CHAPTER 8 Applications of the Integral

the volume enclosed is

27

27 y 5/3 729π

V= π ·y dy = π · =

2/3

.

5/3 5

0 0

Math Note: The reasoning we have used in the last example shows this: If the

curve x = g(y), c ¤ y ¤ d, is rotated about the y-axis then the volume enclosed

by the resulting surface is

d

V= π · [g(y)]2 dy.

c

You Try It: Calculate the volume enclosed when the curve y = x 1/3 , 32 ¤ x ¤

243, is rotated about the y-axis.

EXAMPLE 8.8

Set up, but do not evaluate, the integral that represents the volume gener-

ated when the planar region between y = x 2 + 1 and y = 2x + 4 is rotated

Y

about the x-axis.

FL

SOLUTION

When the planar region is rotated about the x-axis, it will generate a donut-

AM

shaped solid. Notice that the curves intersect at x = ’1 and x = 3; hence

the intersection lies over the interval [’1, 3]. For each x in that interval, the

segment connecting (x, x 2 + 1) to (x, 2x + 4) will be rotated about the x-axis.

It will generate a washer. See Fig. 8.18. The area of that washer is

TE

A(x) = π · [2x + 4]2 ’ π · [x 2 + 1].

[Notice that we calculate the area of a washer by subtracting the areas of two

circles”not by subtracting the radii and then squaring.]

It follows that the volume of the solid generated is

3

V= π · [2x + 4]2 ’ π · [x 2 + 1] dx.

’1

8.2.2 THE METHOD OF CYLINDRICAL SHELLS

Our philosophy will now change. When we divide our region up into vertical strips,

we will now rotate each strip about the y-axis instead of the x-axis. Thus, instead

of generating a disk with each strip, we will now generate a cylinder.

Look at Fig. 8.19. When a strip of height h and thickness x, with distance r

from the y-axis, is rotated about the y-axis, the resulting cylinder has surface area

2πr · h and volume about 2π r · h · x. This is the expression that we will treat

in order to sum up the volumes of the cylinders.

229

CHAPTER 8 Applications of the Integral

Fig. 8.18

y

h

r x

Fig. 8.19

EXAMPLE 8.9

Use the method of cylindrical shells to calculate the volume of the solid

enclosed when the curve y = x 2 , 1 ¤ x ¤ 3, is rotated about the y-axis.

SOLUTION

As usual, we think of the region under y = x 2 and above the x-axis as

composed of vertical segments or strips. The segment at position x has height

x 2 . Thus, in this instance, h = x 2 , r = x, and the volume of the cylinder is

2π x · x 2 · x. As a result, the requested volume is

3

V= 2π x · x 2 dx.

1

We easily calculate this to equal

3

3 x4 34 14

V = 2π · x dx = 2π = 2π ’ = 40π.

3

4 4 4

1 1

230 CHAPTER 8 Applications of the Integral

EXAMPLE 8.10

Use the method of cylindrical shells to calculate the volume enclosed when

the curve y = x 2 , 0 ¤ x ¤ 3, is rotated about the x-axis (Fig. 8.20).

y

x

Fig. 8.20

SOLUTION

We reverse, in our analysis, the roles of the x- and y-axes. Of course y

ranges from 0 to 9. For each position y in that range, there is a segment stretch-

√ √