CHAPTER 10
FURTHER TECHNIQUES AND APPLTCATIONS OF INTEGRATION
-
10.1 Trigonometric Integrals
PREREOUISITES
1.
Recall how to differentiate and integrate trigonometric functions
(Sections 5.2 and 7.1).
2.
Recall how to integrate by substitution (Section 7.2).
3.
Recall how to define trigonometric functions in terms of the sides
of a right triangle (Section 5.1).
4.
Recall how to complete a square (Section R.1).
PREREQUISITE QUIZ
1.
2
Differentiate y = sin 4x - cos x
2.
(a)
Evaluate
jcos 3x dx
(b)
Evaluate
jsin(2x
3.
tan x
.
.
3)dx
.
Consider the figure at the left
x
4.
-
+
(a)
What is
sin 6 ?
(b)
What is
cos 6 ?
Complete the square in the following:
+
2x
+
4
(a)
x2
(b)
x2 - 3x - 1
Copyright 1985 Springer-Verlag. All rights reserved.
430
Section 10.1
GOALS
1.
Be able to use trigonometric identitiesfor integrating expressions involving products of sines and cosines.
2.
Be able to use trigonometric substitution for integrating expressions
,
,
, or a2 + x2 .
involving
LT?
c'
LT?
STUDY HINTS
1.
Half-angle formulas. You should memorize or learn to derive
(1 - cos 2x)/2
2
cos x = (1
and
+
.
cos 2x)/2
If you forget which
sign goes with which formula, substitute x = 0
formulas can be derived by using
2
cos x
+
2
sin x
.
sin2x =
as a check.
2
cos 2x = cos x
-
2
sin x
These
and
1
=
Adding and subtracting yields the desired results.
The half-angle formulas are commonly used for integration.
2.
m
Integrating sin x cosnx.
2
identity cos x
+
2
sin x
=
Basically, if one exponent is odd, use the
1
whichever is the odd power
and substitute u = sin x
.
or
u = cos x ,
If both are even, use the half-angle
formulas. The trigonometric integal box on p. 458 is a good one to know.
3.
Substituting u = sec x.
When
tan x
times useful to substitute u = sec x
and
.
See Example 3(c).
Addition and product formulas. Knowing that
and that
cos 2x =
2
2
cos x - sin x
and (lb) on p. 460.
mulas, along with
appear, it is some-
Often, it is necessary to re-
write the integrand into a useful form.
4.
sec x
sin 2x
=
2 sin x cos x
may help you recall formulas (la)
Simply substitute x
sin(-x) = -sin x
for deriving the product formulas.
and
for
y
cos(-x)
.
=
The addition forcos x ,
are useful
If you choose to memorize the pro-
duct formulas, which are useful for integration, note that the angle
x-y
always appears first as it is written on p. 460.
Also, it may be
Copyright 1985 Springer-Verlag. All rights reserved.
4.
(continued)
x = y ?"
helpful to ask yourself, "What if
only
5.
sin x cos y
In addition, note that
has sine terms on the right-hand side.
Trigonometric substitutions. This technique is often used when
2 2
(+a ix
appears in the integrand, where
n = integer and
a
are
constants. Notice that this technique is based upon the Pythagorean
2
identities cos x
+
2
sin x
substitution to use for
=
x
2
sec x
Know what
in each of the three cases.
Using the
=
1
+
2
.
and
1
tan x
substitution equation, draw an appropriate triangle and label it like
those in the box on p. 4 6 1 .
After the integration is completed, use
the triangle to express your answer in the original variable.
6.
Integrating
sec x
and csc x.
Just note theinterestinp trick used in
the integration on p. 462(lines 3 - 5).
7.
Integrating
3
sec-x.
Often, when
appears in the integrand,
3
sec x
trigonometric substitution will call for the integration of
The technique is shown in the solution to Example 8(a),
that
8.
3
sec x
p. 4 9 6 .
.
Note
is integrated by parts.
Integrals involving axL
+
bx
+
c.
In many instances, the first step
is to complete the square. Then, use a trigonometric substitution.
9.
Example 5 comment.
10. Practice.
If
a = ?b ,
remember that
cos(0) = 1
.
A lot of material has been covered in this section.
placed in memory are easy to forget.
Items
Practice helps to reinforce what
has been memorized.
Copyright 1985 Springer-Verlag. All rights reserved.
432
S e c t i o n 10.1
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
j(l -
,
6
+
u 16 - u 1 4
- x/2
-
+
.
1)/2
By u s i n g t h e h i n t ,
3
!tan
2
=
+c
A
- c o s S) / c o s 61 dH
figure t o get
.
=
=
-
c o s 2x
= j ( c o s 2x - l ) d x / 2 =
-,/
[(I
+
e
0)
+
s
C
- l)d3
=
,
s i n 6x/12
+
=
so
r
1 du
=
+
j (u
du
-4
-
u
-6
C = - s e c 3x / 3
dx = 2 s e c
2 , / t a n r 8 d9
=
2(tan r
2
2
+
- s i n x dx ;
=
)du =
+ s e c 5x /
tan
d
2~(sinL3/cosi3)d0=
=
- 9 + C. We u s e d t h e
cns 9
=
1
.
Now, u s e t h e
C
.
2 sec t d~
.
2 j t a n r s e c * db = 2 s e c c
+
(
c
2 tan F
2 sec2* du
so
sin Y
and
= 2
) /u
6
115 c o s 5 x
=
=
= 2j(sec20
-
- u
2
2 s e c P t a n S da
x
-
s i n 2x14
=
+
y)]
- cos(x
(1/2)[cos(x - y)
u = cos x ,
Kow, l e t
s i n G/cos
2(tan 0
=
.
Let
2
- cbs x)dx
2 see 9
Let
2
i d e n t i t i e s tan 3
- u )du
=
3
6
b e c o m e s ,((sin x/cos x)dx =
= -113 c o s 3s
2
2: [ ( I
du
3
gives us
2
- cos 6x)/2
/ ( c o s 2x
3
x s e c x dx
6
-5/(-5)
2
cos 2x)/2
sin x sin y
- c o s x ) / c o s x] dx
-
j(u
so
.
C
/ ( c o s 2x
t h e r e f o r e , t h e i n t e g r a l becomes
-3 / ( - 3 )
=
5
=
.
C
/ s i n 4x s i n 2x dx
/[sin x(l
+
+
(1
=
Thus,
Using t h e p r o d u c t formula
we g e t
13.
cos x
3
- / ( I - u ) u du
- cos x/4
,
u = cos x
2
4
2
formula
( c o s 2x
=
s i n 2x/4
9.
6
C = cos x/6
The h a l f - a n g l e
cos x
Now, substitute
and t h e i n t e g r a l becomes
4
2
.
c o s x ) s i n x c o s x dx
3
3
, [ s i n x c o s x dx
2
- c o s x , we g e t
1
=
3
2
-sin x
5.
2
sin x
Using t h e i d e n t i t y ,
From t h e f i g u r e , we g e t t h e f i n a l a n s w e r ,
/
,
2
so
+
- cos-l(2/x))
ds
=
+
2
Then
C
C
Copyright 1985 Springer-Verlag. All rights reserved.
.
25.
1) = 4(x
+
118)
2
+
and letting
Note that
-cosnx
1;
.
+
+ ja
+
+ 1 = 4(x +
+ 118 , so
x
u = x
x)
=
, we have
-cos x ,
jtn cosnx
Hence,
cosnx dx
Let
2
cosn(x
+
1;
=
n
is odd,
cosnx dx
For
n = 0 ,
(b)
For
n
2
=
+
(1/4n)(x
(c)
For
For
4
(1/2n)jin cos x dx
+
2
cos 2x) dx
(ll8)sin 4x)
n
2
+
,fO'(1/(16n))(1
3
cos 2x)dx
1;
+
=
=
(1/16n) [(5r)
i
(1/~)1:~.100
=
sin(754T)
/in+
(112) (1
3/8
0 dx
.
0
=
.
cos 2x)dx =
+
+
2
2 cos 2x
cos 4x)ldx
2
dx =
+
2
cos 2x)dx
+
(1/8n)(x
=
=
sin 2x
+
.
+
3 cos 2x
+
lo
+
+
(3/2)(1
(312)sin 2x
+
=
=
3 cos 2x
+
2
3 cos 2x
cos 4x)
+
(1 -
+
2
2x - cos 2x sin 2x)dx
=
5/16
=
.
and a half-angle formula to get
(50~/T)l;(l
- cos(754t))dt
(50R/T) (T - (1/754)sin(754T))
. Now, substitute R
=
2.5
and
T
=
+
( 3 1 2 ) ~+
3
2
((112)sin 2x - (1/6)sin 2x)
10 sin(377t)
T
+
(1/16~)ji'(cos
2
sin (377t)dt
(1/754)sin(754t))
27
jO
(COS X)
3
2
cos 2x) dx = (l/(16n))jon(1
2
(1/16n)jOn(1
(318)iin 4x)
Substitute
dx =
for four cases.
+
2n
(1:211)j0
=
2n
(l/8n)jo (1
=
2
sin 2x)cos 2x)dx = (1/16n)(x
33.
=
2
6
2n
2 3
(1/2n)joncos x dx = (1/2r)j0 (cos x) dx =
6 ,
=
+ x)
lo
(118n)~i~jl+ 2 cos 2x
(dl
Then,
1-dx = 2nI2n = 1
2
2n
(1/2ji)jin cos x dx = (1/4n)j0 (1
2n
(1/2)sin 2x)
= 112 .
n = 4 ,
+
=
n
,
(l/8;r)jin(l
x/2
.
+
j(2 d ~ /
+ jZn c o p x
0
(1/2n)jin cos x dx = (1/2n)ji6
(-1116
11-
cosn(n
- cosnx)dx
n)dx = j:(cosnx
+
du = dx
j(dx/=l)
Therefore, we need to consider only even values of
(a)
2
Factoring out
so if
dx =
118)
.
j(du/=)
=
13 = 8 u / m
cos(n
.
15/16
j ( d x / / 4 x )
29.
4x
Completing the square, we have
=
=
P
=
(50R/T) (t -
50R - (50R1754T) *
2 ~ 1 3 7 7to get
P
=
125 -
(125-377/(2~.754)) sin 4n = 125 .
Copyright
1985 Springer-Verlag. All rights reserved.
4.34
S e c t i o n 10.1
37.
(a)
31:(x
2
+
3(t sin t
sin x
3 ~ 0 ( xs i n x
3jk(x s i n x
Let
s i n t cos t )
+
2
+
+
t
J O x s i n x dx
+
,
2
2
.
s i n x cos x)dx
s i n x cos x)dx
.
dv = s i n x dx
t
jO cos x
In
.
Thus,
~ ( 0 =) 0~ =
,
let
+
s i n x)i
u = x
.
x
j k ( l - c o s 4x)dx = ( 1 / 8 ) ( x
-
I ( t ) = -t c o s t
S1(t) = 0
cos t
whenever
sin t = 0
-1
or
< cos
2
1
t
, -1
of
S(t)
.
When
+
=
XI:
-x c o s
.
sin t
+
Since
dx = (118)
x
(114)sin 4 x ) / k = (1/8)/(t - (114)sin 4 t )
.
+
sin t
+
t / 8 - ( 1 / 3 2 ) s i n 4 t ) l 113
2
+
2
s i n t cos t = 0 ;
2
t cos t
.
=
, and l e t
t / 8 - (1132) s i n 4 t
sin t = 0
n
A(t)
I(t)
+
< sin
t
Then
du = dx
= (1/4)j;sin22x
s i n t cos t = 0
c r i t i c a l v a l u e s of
.
=
sin t
s i n t cos t >1 - 1 = 0
a r e when
Then
2
2
t >1
A(t)
( ~ ( t ) =) ~
+
t sin t
+
Let
so
-t c o s t
=
0
, B(t)
=
.
Therefore,
2
B ( t ) = jOsm x c o s x dx
= -cos
3m [ 3 ( - t
.
c = 0
2
2
s i n x cos x)dx
, so v
dx = (-x cos x
so
t . 2
A(t)
=
i n t o t h e given
s i n x c o s x = (112) s i n 2x
Then
(c)
2
(s(t)13
t = 0
( S ( 0 ) ) ~ ' ( 0 )= 0 = S ( 0 )
2
=
Substitute
2
and l e t
B(t)
.
s i n x c o s x)dx
+
3(s(t))'s1(t)
=
Now, i n t e g r a t e t o g e t
2
I ( t ) = jk(x s i n x
A(t)
.
2
equation t o get
(b)
( d / d t ) [ T s ( t ) ) 31
Differentiation yields
.
Since
<1 .
Thus,
.
i . e . , whenever
< s i n t < 1 and
if t > 1 , t +
-1
Thus,
s'(t)
T h e r e f o r e , t h e o n l y z e r o s of
, i.e.,
t = nn
S(t) =
.
for
These a r e
, c o r r e s p o n d i n g t o r e l a t i v e maxima and minima
i s odd,
sln t
,
and c o r r e s p o n d i n g l y ,
~ ' ( t )
c h a n g e s from p o s i t i v e t o n e g a t i v e , i n d i c a t i n g a r e l a t i v e maximum
excursion.
T h e r e i s no a b s o l u t e maximum, s i n c e
S(t
+
27)
> S(t) .
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.1
435
SECTION QUIZ
1.
Use the technique of trigonometric substitution to evaluate
x
2.
3.
/
2
.
Did you get the expected result?
Evaluate the following integrals:
(c)
~ s e c ~
tan
~ x dx
(d)
j[sin(3x/2)cos
(e)
jsin 3x sin 4x dx
(f)
j[3/(1 +
2x
t2)3'2~
+
l/Jx2]dx
dt
x-axis, and the lines x = 0
,
2
helpful to integrate x cos ax
4.
x = n/2
.
the
[Hint: You may find it
for a constant
u
.I
Evaluate the following integrals:
a
j[(pX
+xd]-/)i
(b)
/[(4x
+
+
2)/(x2
j[(x - l)/(x
(c)
5.
y = x cos2x ,
Find the center of mass of the region bounded by
2
6x
+
10)~'~]dx
- 2x - 5)ldx
Underwater divers have recently discovered a sea monster at the ocean
floor.
(4
-
Its shape has been described as follows:
x2)ll4
and
sin x
If each unit represents
of
1500 kg/m3
,
on
The region between
[O,n/41 , revolved around the x-axis.
10 meters
and the sea monster has a density
how much does it weigh?
ANSWERS TO PREREQUISITE QUIZ
1.
4 cos 4 x + 2 s i n x c o s x + s e c2x
2.
(a)
sin 3x13 4- C
(b)
-cos(2x - 3)/2
+
C
Copyright 1985 Springer-Verlag. All rights reserved.
436
Section 10.1
3.
(a)
1
/
6
(a)
(X
+
112
(b)
(x
-
3 1 2 ) ~ - 1314
(b)
4.
7
llm
+
3
ANSWERS TO SECTION QUIZ
+
.
, a s expected
1.
sin-lx
2.
(a)
3
5
7
s i n 0 - s i n 9 + 3 s i n 615 - s i n 8 / 7 + C
(b)
5t/16
(e)
s i n x / 2 - s i n 7x114
C
+
sin2t/4
+
3 s i n 4t/64
+
-
3
s i n 2t/48
+
C
C
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.2 437
10.2 Partial Fractions
PREREQUISITES
1.
Recall how to factor a polynomial (Section R.l).
2.
Recall how to integrate by the method of trigonometric substitution
(Section 1 0 . 1 ) .
3.
Recall how to integrate by the method of substitution (Section 7 . 2 ) .
PREREQUISITE QUIZ
1.
Evaluate
J [ x 2 / L 7 + x + 2bx
2.
Evaluate
j1'jl/ 1
x +
3.
Factor the following polynomials:
(a)
x3 - 27
(b)
x3 - x2
7+ 2 i)dx
2x
.
.
GOALS
1.
Be able to integrate rational expressions by the technique of partial
fractions.
STUDY HINTS
1.
Beginning partial fractions. Look before you leap!
easier method.
See Example 7.
There may be an
If partial fractions is the method of
choice, be sure the degree of the denominator is larger than the degree
of the numerator.
2.
If not, begin with long division.
Denominator factorization. All factors should be of degree one or two.
If not, the denominator can be factored further. Check to be sure quadratic factors do not factor further (by using the quadratic formula,
Copyright 1985 Springer-Verlag. All rights reserved.
438
Section 10.2
2.
(continued)
if necessary).
x = x - 0 ,
Don't forget that
the linear factors
(x
-
O).(x
-
0)
.
so
x2
See Example 4 .
is composed of
Also, remember
that a factor raised to the nth power must be represented
3.
times.
Determination of coefficients. This method is called comparing coefficients:
Multiply so that both sides of the equation are over a common
denominator.
Rather thanexpanding, it is best to leave the expression
as a sum of factored terms.
Then, substitute values of
many terms as possible become zero.
linear equations.
4.
n
If no such x's
x
so that as
The result should be a few, simple
are left, choose any other constants.
Differentiating to determine coefficients. To solve Example 2, this
author prefers Method 1.
I find that there's a greater chance for error
with the differentiation process; you may disagree.
5.
Comparing integrands.
After you have found the coefficients, use a cal-
culator and an arbitrary number to compare the original integrand with
your new one.
6.
Rationalizing substitutions. If
[f(x)] P"
appears in the integrand,
you might be able to make a simplification by substituting u
i.e.,
7.
uq = f(x)
.
=
[f (x)] l / q
,
See Example 8 .
_Integrals of rational trigonometric expressions. The technique used in
Examples 10 and 11 is useful for rational functions in
Rather than memorizing equations ( 8 ) ,
sin x
and
cos 2x
=
cos x
( 9 ) , and (lo), it is suggested that
you reproduce Fig. 10.2.2, and use the identities sin 2x
2
2
cos x - sin x
and
=
2 sin x cos x
to complete the substitution. Often, partial
fractions may be necessary to finish the integration.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.2
439
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
+
~/(x - 2)2(x2
Let
+ F)/(x 2 + I ) ~. Then we need
2 2
~(x' + 1)2 + (CX+D)(X - 2) (x +
(Ex
1
B(2
=
2
+
112
,
i.e.,
0 ;
=
gives:
and
+
-2A
.
- 212 + (8x +
F
and
5
!
(
x
151125
=
+
+
+
ll21 dx
=
1/25
+
15:[dx/(x2
.
=
81125 , D
+
(x2
+
1)
812
+
sin 2814
+
2(1
5.
2
x )
1
:(sec2q
=
+
4
dA/sec 8)
C
=
+ C = 012 + sin 8
. Hence, j[dx/(x
Since the discriminant
(b
2
+
1)
2
1
+
I)]
(x
+
-
C)/(x
2) (x2
+
2x
+
+
2x
+
2)
equations gives A
=
2)
.
=
2A
+
E
=
201125 ,
- 2) +
~ ) ~ ] d =x (1/125)
11 tan-'x
+
10/ [2x/
. The last integral can be evaluated
tan 0
2
jcos 8 dB =
cos R/2
- 2)2(x2
- 4ac)
=
1 [(I
[(A+ B)X
+
+
i.e.,
/[dx!
cos 28)/2]d1
=
C
+
1) 21 = (1/125){-8 1 n / x - 21 -
for x2
2
x ,
+
+
(1/2)tan-'x
=
2x + 2
we cannot factor it further. Thus, x /(x - 2)(x
(Bx
and
(1/125)j[-8/(x
=
+
2
2
x = 2 ,
D = -1125 ;
111125 ,
=
15)/(x2
by using trigonometric substitution with
2
+
Solving this set of equations
(20x
2
1) 1 1
+
0 ; -2A - 4C
=
+ 4j [2x dx/(x2 +
i-8 In / x - 2 / - 5/(x - 2)
(x2
Substituting in B
C
+
1)
+
If
(Ex+F)(x
2 2
jdx/ [(x - 2) (x
Il)/(x
~)/(x~
+ 1) +
.
4F = 24/25
2
+
(CX
- 2)
C
Thus,
+
+
1)
+
A = -81125, B = 5/125 ,
2
1 = A(x - 2) (x2
.
+
4D
B/(X - 2)
to solve
B = 1/25
comparing coefficients gives A
4F
+
A/(X - 2)
I)'=
(2A - 2B
+
2
is less than zero,
+ 2x
C)x
+ !x
+
+
2) = A/(x - 2)
(2A - 2~)]/
Comparing coefficients and solving simultaneous
215 ,
B
=
315 ,
and
C = 215
.
Thus,
1 [x2/
Copyright 1985 Springer-Verlag. All rights reserved.
+
440
5.
Section 10.2
(continued)
The first term gives /[2/(x - 2)l dx = In [(x - 2)
+
term, note that (3x
2
1/(x
+
2x
+
1
want to have
+
d(x
2
2)/(x
.
1)
+
2
+
2x + 2)
.
Factoring
x4
2x
+
2)
+
+
in the numerator and whatever is left con-
2)ldx
2
+ 2x + 2)] dx = (312) x
2)/(x
2
2
1) + l)]d~ = (3/2)ln(x
+ 2x + 2)
j[(3x +
Thus,
- j[l/((x+
+
(3/2)ln(x2
2x2 - 3
2x
+
2
(x
+
1)
+ D/(x 2
(x - l)]/(x
+
3
results: x :
0
-
gives
gives A = -114
,
-
3)]dx
[-ln(x2
13.
+
+
B
=
+
x/(x4
B)(x~ - 1)
+
1) (x
+
2x2
+
C(x
(1
+
.
+
2u - 2 tan-'u
2
x :
+
+
3)
+
l/(x
+
l n / x - 111
1 - 1)/(u2
C = 2&
- 2 tan-'&
j( [2du/(l +
2J(u
+
u )
2
u ) ] /[1
and
2)ldx =
.
1))
+
C
-
1)
.
Using the tech-
+
3)
(Ax
=
-
3)(x
+
B)/(x~
+
+ D ( X ~+
1)
3)
3)
x
x:-A+3C+3D=1
+
C
+
118
1)
+
l/(x - l)]dx = (118) x
=
+ 2u/(l + u
+ ~ ) - ~ d=u -2(u + 1)-1 + c
=
1
u
=
C
=
2
- 1)/(x2 + 3)/
& , so u2
+
-
21[1/(u2
2
u )
=
x ,
2
2jiu /
+
u = tan(x/2)
=
+
C
and
1)ldu =
+
, so
. Thus, j[dx/(l + sin x)]
j[2du/(l + u2 + 2u)l
-2/(1
/[x/(x4
.
Substitute
dx = 2du/(l
2
(1/8)lnI(x
Thus,
+
=
l)]du = 2jdu
+
.
D
2
J[u/(l + u )]2u du
+x)]dx=
We use the technique of Example 11.
sin x = 2u/(l
+
+
B - C + D = O ;
118 , and
=
+
2
2x
Comparing coefficients gives the following
C
Thus, j[&/(1
2
u )]du = 2j[(u2
2
+
-
Solving these four equations simultaneously
0 ,
inlx i 11
3) (x
We apply the method of Example 8 by letting
2u du = dx
17.
.
3D = 0
(1/8)/[-2x/(x2
=
3)
.
- 1)
+
A + C + D = O ;
+
2
2x
1) = [(Ax
2
3)(x
B - 3C
x (1):
2
2
j[x /(x - 2) (x
2) - tan-'(x
nique of partial fractions, we have
C/(x
For the second
2
2)/(x
+ 2x + 2)] -
+
(312) [(2x
=
All this information gives
(1/5)4:ln(x - 2)2
9.
+
2x
+
1 .
The reason we break it up chis way is because we
stitutes another term.
1[(2x + 2)/(x
-1
tan (x + 1)
2
2
tan(x/2))
=
+c .
Copyright 1985 Springer-Verlag. All rights reserved.
=
Section 10.2
21.
By the shell method, V
6
2nj5[~/(1
fractions yield
B/(1 - 2x)
x
for all x
,
x = 1
Let
,
-
B/(1
.
2x))
x = 132
,
so -A = 1
+
x)
- x)(l
.
- 2x))ldx
Now, partial
, with ~ / ( l - x) +
2x)Idx
+
Therefore, A(l - 2x)
so
B = 1 . Let
6
Therefore, V = 2nj5 [-l/(l - x) +
B/2
.
i.e., A = -1
B(l - x) =
112 ,
=
l56 =
In 4
(a)
Integrating the right-hand side and using the method of partial
2x)ldx
=
-
i.e.,
+
1/(1
25.
.
-
-
- x)(l
x/((l
=
6
2nj5[x/((l
=
441
2n[ln(x - 1) - (1/2)ln(2x
I)]
.
In 3) = n ln(2251176)
B(80
-
1/20
.
C
Let
-
l/(x
when
(b)
x) = 1
for all
x
80 ,
=
-
(1/20)j[-1/(80
=
kt
for constants
+
x)
+
,
,
B
1 ,
=
,
C
we have
(1/20)ln(4/3)
=
-
.
- 60) /
,
(1/20)ln(4/3)
(1/20)ln/ (x
=
20B = 1 ,
-1120
-
1 (x -
80)/(x
60) /
-
80)/(x
Since we assume
exp(20kt) = x
-
-80
+
If
x = 20 when
.
80exp(20kt)
(a) to get
ln(4/3)1
=
.
80
=
(1/20)ln(9/8)
part (b) to get
20kt
-
-
, we
=
=
+
80) =
0
x
60
60)
or
+
ln(4/3) =
(4/3)exp(20kt)
,
=
the formula
(4/3)(x
Thus, x = 80[ 1 - exp(20kt)] / [ 1
(1/20)1n(3/2)
=
Thus, kt
- 60)
Rearrange again to get [ (4/3)exp(20kt)
t = 10
Now, substitute k
(x - 80)/(x
B =
.
and exponentiation yields
.
- 60) 1
or
and the formula becomes
Rearrangement of the equation in part (a) gives
ln/(x
.
Since x
- 60) 1
80)/(x
+
~ ( 6 0- x)
1/(60 - x)]dx = (1/20)/[1/(x
60)I dx = (1/20)ln/ (x - 80)/(x
t = 0
=
I [A/(80 - x) +
where
so
A
or
=
,
C
and
x = 60 ,
Let
so -20A
becomes (4/3)exp(20kt)
(c)
.
x
A
+C
kt
fractions for the left-hand side, we get
B/(60 - x)]dx
277(ln 5 - l n m -
-
X
- llx
=
(4/3)exP(20kt)l
substitute into the formula In part
IOk
10h.
ln(9/8)/200
+
(1/20)1n(4/3)
or
(1/20)[ln(3/2)
.
Therefore, k
=
and
into the formula In
t = 15
x = 80[ 1 - exp(3ln(9/8)/2)1/[
80(1- (9/81~/~)/(1- (413)(9/8)~/~)
= 26.2 kg
ln(9/8)/200
-
1 - (4/3)exp(3111(9/8)/2]
.
Copyright 1985 Springer-Verlag. All rights reserved.
=
442
S e c t i o n 10.2
SECTION Q U I Z
j[dx/(2x3 +
1.
Evaluate
2.
2
Evaluate ~ : [ d ~ / ( ~
2
Evaluate
4.
Find t h e average of
t < 4 .
(Hint:
.
.
2
2) x]
-
- 2) ( x
3.
[dx/(x
+
2
4x ) ]
I)]
.
f(t) = t 8 G / ( 1
let
u
=
+
G
.
-
t6)
on t h e i n t e r v a l
2
<
)
.
5.
Evaluate
6.
When t h e m i n i s t e r asked f o r any o b j e c t i o n s , t h e b r i d e ' s g r a n d f a t h e r
i;I4[tan
6/(1
s e c €I)] d ~
s a i d , "Sonny-boy has t o prove t o me h e ' s smart enough t o be h e r husband.
y = 10 - x / ( x 2
Look a t t h a t archway.
-
2x - 3)
on
[0,21
describes
He can marry h e r i f he knows t h e a r e a under t h a t archway."
it.
What
answer would b r i n g happiness t o t h e young couple?
LYSWERS TO PREREQUISITE QUIZ
1.
2 sin-l(x12)
2.
X
3.
(a)
+
l
- x n 1 2
+
E
(x - 3 ) ( x L
+
+
3x
+ x2/2 +
2x
+c
C
+
9)
2
(b)
x ( x - 1)
(c)
(x
+ 3) ( x + 2)
ANSWERS TO SECTION Q U I Z
+
2 ) / x \ - 1/4x
+
1.
(118) l n / ( x
2.
( 1 / 8 ) 1 n ( 3 / 2 ) - 1/48
3.
[(2
4.
- ( 1 / 9 ) ( 6 6 m - 1 0 m ) +(l/l2) ln[(J66
+
f i ) / 4 ] l n / x -I-
(1146) l n [ ( m
+
+
fit +
6 ) ( m
C
[(2 - fi)/4] l n / x -
-
-
6)/(J66 -
fi/ -
1 ) ( m
+
l n l x - 11
1 ) / ( m
6)(a
+ 6)]
+
+
C
1 ) ( m
-
-56.14
2
tan (n/8)1
5.
ln[l
6.
20-1116
Copyright 1985 Springer-Verlag. All rights reserved.
l ) ] -t
Section 10.3
443
10.3 Arc Length and Surface Area
PREREQUISITES
1.
Recall how to integrate by using trigonometric substitution (Section 10.1).
2.
Recall how to compute the distance between two points in the plane (Section R.4).
PREREQUISITE QUIZ
1.
What is the distance between the points
2.
Evaluate
j(dx/&
(3,6)
and (4,2) 7
.
)
GOALS
1.
Be able to express the length of a curve as an integral and perform the
integration, if possible.
2.
Be able to express the area of a surface of revolution as an integral
and perform the integration, if possible.
STUDY HINTS
1.
Definition. The notation
ds
is introduced in this section.
infinitesimal length defined by
2.
+
Ad,)'
(dy)2
Arc length. You should become familiar with
Either memorize this or learn to derive it.
ddx)
2
+ (dy)'
, substitute dy
=
L
.
=
It is an
See Fig. 10.3.1.
.IbA
+
[f '(x)j
Starting with
dx
.
ds =
, factor out dx , and
f '(x)dx
integrate. Since the integrand is positive, you made a mistake if
your answer is negative or zero
3.
Constant with no effect.
f
The length of
f
+
C
is the same as that of
. Shifting a graph along the y-axis does not change its length.
pute the length of
(x
-
L ) ~ ' on
~
[1,2]
Com-
and compare with Example 2
Copyright 1985 Springer-Verlag. All rights reserved.
444
4.
Section 10.3
Tricks to find arc length.
In general, many textbook and exam questions
are chosen for their simplicity. Thus, the expression under the radical
If
will often simplify; for example, it may be a perfect square.
it is not a perfect square, a trigonometric substitution will often
be helpful.
5.
Square roots in arc length problems.
The appearance of square roots in
arc length questions may sometimes present a special problem. Wnen in
doubt, take the absolute value of the expression inside the square root.
For exampie, consider finding the arc length of x2I3 on [-8,-21
+
j::/(9x2/3
4)19~'/~ dx
u = 9 ~ "+ ~6
yields
=
-13x~'~jdx
jpz&
dull8 = (1/27)u
Step function derivation (pp. 480-482).
We get
and substituting
I
3
1
which is negative. The correct method is to use
6.
.
2
l
m
3/x1I3
=
.
Except in honors classes, most
instructors will not emphasize these theoretical aspects on their exams.
7.
Area of revolution. Learn ts derive fornula (2) in the box on p. 483.
It may be easier to think of the infinitesimal frustums as cylinders, so
the area is circumference times width.
2nf(x)
8.
and the width is simply the arc length. Thus,
2 ~ r r f ( ~6)
f I ( ~ ) ] dx .
a
Revolution around y-axis. Here, the circumference is
than
2ny
x h
is that
9.
The circumference is
+
.
A
2nx
A
The width is still the arc length. Thus,
[f '(x)I2
dx
.
=
2nf(x)Ldx
=
rather than
=
The only difference with the preceding formula
f(x) appears in one and
Mathematical illusion?
=
2nr
b
x
appears in the other.
The bands in Fig. 10.3.8 are equal in area.
The smaller radius is compensated for by a larger
ds
.
Copyright 1985 Springer-Verlag. All rights reserved.
Why?
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
4
f ( x ) = x 18
Given t h a t
+
2
1/4x
,
Thus t h e l e n g t h of t h e g r a p h i s
I
5.
+
3 3
1 / 4 x dx = ,f1(x 1 2
+
112
f(x)=xn,
If
+
I:/l
Thus, t h e l e n g t h i s
(0
[ ( I - 0)
=
2
- I)~]
[0,1],
-1
+
=
on
I f , m d x
Given t h a t
+
,6+ fi + +'%.
( 2 - 0)211/2
, and
[1,2]
+
3
+
112x ) d x
and
L
=
+
(x
4
(x 18
=
]
is
a
i
p
+
ll
d
+
x
on
[2,5]
.
f '(x)
=
9219
.
.
f '(x)
+
-
[(5
2
=
2)
2
+
on
Thus, t h e l e n g t h i s
1
6xI0
+
2
6xll
(1/2)(x
+
1)
=
1)li2/l
+
(1 - Z ) ~ I
On t h e o t h e r h a n d ,
= 2nli(x
2 3
II4x )
-
=
Thus, t h e a r e a of t h e s u r f a c e o b t a i - n e d by r e v o l v i n g
about t h e x-axis
.
Ey d i r e c t c o m p u t a t i o n , we h a v e
, we have
1)ll2
3
3 2
j t X d o i d x l+
d- : ]
f(x)
1/2x
A + [ f ' ( x ) ~ ~ =
=
[ ( 2 - 112
-113
-
( x 1 2 - 1 / 2 x ) dx =
3
T h i s e x e r c i s e i s a n a l o g o u s t o Example 5 .
L
13.
=
n- 1
fl(x)=nx
then
h.
9.
L
6
3
f l ( x ) = x I2
we h a v e
f(x)
[ ( 1 / 2 ) (x
+
+
-112
on
.
[0,2]
1 ) - l i 2 1 2dx
=
J V
+
Z r r ~ i [ ~ / ? ] =
d x( i / b ) ( 4 x
17.
p f
= s i n x
u = sin x
, so rhe a r e a
so
a s i n Example 3 t o g e t
21.
.
Then
A
=
A = 271 [ ( u / 2 ) K
T h i s e x e r c i s e i s a n a l o g o u s t o Example 9 .
[0,1]
and
f(x)
=
+
2)dxI
=
2&1[(x
2
12)
(lI2)lnu
f(x)
idi;-Iil
=
x
on
Thus, t h e a r e a i s
A =
)
m
d
x
] = 2fin[!AX
+
2
jl(-x
2
(-x / 2
+
21:)
, 2I
on
2
=
2~%[(1/2)
dx
+
(-312
.
Let
+
.
+
(f, +
.
[1,2]
-x+2
2 ~ [ / k x m d x
i
+
We h a v e
18.7
m dx
A = 2 i j ~ / ~ 2 c o sx
is
du = c o s x dx
5 ) 3 i 2 i = ( 1 / 6 ) ( 1 3 ~-~ 5312)
~
+
+
2)] =
Copyright 1985 Springer-Verlag. All rights reserved.
=
446
25.
Section 10.3
2
[ f '(x)~
[i/L+
(8127a2) (1
+
+
c
2
9a (x
Let
.
Note that
=
2
sec x
+
4
sec x
jbI2&4-
2n \;l2(tan
x
+
2
+
,
.
+
+
s
+
.
.
+
so
Changing
c
-.":I
+
2
9a (1
so
du =
+
-
b)/4)312
s = (1/27a2) [(4 +
has no effect, since
[f '(x)]
And the area is
2
4 sec x dx
,
b)l4
s =
=
jx='
4u1I2du/9a2 =
x= 0
=
,
8
=
so the length is
4
sec x
2
9a (x
2
(8127a ) [(I
f '(x)
.
, so the length
Then
4312
2 312
9a b)
1
2
4sec x dx
2x)L
1
=
=
dropped out when we found
f '(x)
+ b)/4
u
b)14)~'~ 1;
+
- (4 +
b))312
.
2
dx = 4 du/9a
9a2b/4)312]
+
2
9a (x
=
2
9a ( x i b)/4 dx
i.e.,
2
9.3 (1
33.
(3a-12)~
2
9a dxl4 ;
(1
29.
=
2
dx
=
2
dx
2 1 ~ j ~ ~ f ( x [f
)~
'(x+
)]
.
2rrlbf(x)h + [f '(x)] 2dx =
Ja
For a small arc length where f(x) is almost constant,
Thk area of n surface of revolution is
znj:f
(x)ds
.
this is approximately
by the small length.
2n
multiplied by the function value multiplied
To find the area of the surface obtained by re-
volving the given curve, divide the curve into
1 mm
segments.
On
each segment, measure the distance from the x-axis to each endpoint, and
about
37.
Let
fi(1
s =
1
j0.l
- 0.1)
Thus,
41.
(a)
=
LZZZ
sin x
3
0 . 9 ~ 4Z= 1.28
>sl
.
dx
(0.1,1.0001) to (1,2)
from
Using this method, our answer is
16 cmL
s1 be the length of
Then
.
f(x)
then take the average value as
and
since
.
Now
,
so
s2
be the length of
cos x
<1 ,
s1 6
1
+
4
x
1;. 1~'2dx
.
=
s2 must be longer than the line
s
2
> 40.9) 2 +
(0.9999)~Z= 1.34
.
.
Cutting and unrolling the frustum as in Fig. 10.3.12, we found the
area to be
the x-axis.
ns(rl
+
r2)
if a linear function is revolved around
Since the formula is independent of
x
and
y ,
the
formula for the area obtained by revolving a linear segment around
Copyright 1985 Springer-Verlag. All rights reserved.
=
41.
(a)
(continued)
the y-axis is also
n(a
(b)
+b
-
) m ( b
ns(rl
,
a)
+
r2) , which works out to be
since
From part (a), the area is
2
a 12)
=
~ T ~ ( x ' / Z );1
A
we have
=
s = m
. m ( b 2
(
- a2)
=
.
j ; 2 m m d x
=
.
- a)
b
2.rm(b2/2 -
Since m
=
fl(x)
,
/-.--T-T
2njax 1 + [f (x)] dx . Let the entire surface be
b
composed of a finite sequence of frustums of cones and use the
additivity property of the integral to obtain formula (3)
SECTION QUIZ
1.
Compute the arc length of the following functions:
2.
5
x 12
+
(a)
y
(b)
y =
(c)
y = (2~+5)~"+ 3
=
on
[l,t] ,
1/30x3 on
[ 1,2]
t
>1
(Hint: Substitute
on
Suppose the arc length of
about the arc length of
.)
=
[0,31
f
g =
u
on
f
+
[a,bl
k
on
is L
.
LJat can you say
[a,b] , where
k
is a
constant?
3.
Find the area of the surfaces obtained by revolving the curves in
Question 1 around the y-axis.
(Hint: Lise formulas 65 and 66 of the
integral table for part (b).)
4.
Find the area of the surfaces obtained b>- revolving tlle curves in
Question 1, parts (a) and (b) , around the x-axis.
5.
Upon close inspection, you notice that your pet tarantula's legs can
be described by
(a)
y = 1
- t3I2
on
[O,l] .
How long are the tarantula's legs?
One day, while you were walking your tarantula, a cyclone suddenl?.
appeared and twirled the spider around so fast that it seemed lihe
3
Copyright 1985 Springer-Verlag. All rights reserved.
5.
(continued)
s o l i d s u r f a c e was b e i n g formed by r e v o l v i n g o n e o f t h e l e g s a r o u n d t h e
y-axis.
(b)
What was t h e a p p a r e n t s u r f a c e a r e a ?
ANSWERS TO PREREQUISITE QUIZ
1.
fi
ANSWERS TO SECTION Q U I Z
1.
5
+
1 / 3 0 t 3 - 7/15
(a)
t 12
(b)
(1/4)[ 6 f i - 2 6
(c)
(1000 - 46vZCiI27
+
ln((&
2.
They a r e e q u a l .
3.
(a)
6
T ( 5 t /6 - 1 / 1 0 t 2 - 1 1 / 1 5 )
(b)
( 5 1 f i - 9&)/16
(c)
2n(35000 - 2116&?)/1215
(a)
2
? ( t 1 ° / 4 - t 130
(b)
( " / 6 ) ( 2 7 - 5&)
(a)
( 1 3 m - 8)/27
(b)
1 6 ( 8 8 4 m - 4)/1215
4.
5.
+
3)/(&
- (1/64)1n(17
+
+
+
2))1
12Ji7)/(9
+
4&))
1 / 9 0 0 t 6 - 491225)
Copyright 1985 Springer-Verlag. All rights reserved.
10.4 Parametric Curves
PREREQUISITES
1.
Recall how to sketch curves described by parametric equations (Section
2.4).
2.
Recall how to compute arc lengths (Section 10.3).
PREREQUISITE QUIZ
1.
Sketch the curve described by
x = 2t
2.
Sketch the curve described by
3.
Compute the length of the graph of
4.
+
x = t2
1
and
and
y = 2x
4
y = t
+
3 on
Write a formula for the length of the graph of
.
[1,2]
y = t + 2
.
.
[0,1]
y = x
3
- x
.
2
on
(Do not evaluate.)
GOALS
1.
Be able to find the tangent line of a parametric curve.
2.
Be able to relate speed to arc length for parametric curves.
3.
Be able to express the length of a parametric curve as an integral
and perform the integration, if possible.
STUDY HINTS
1.
Direction of motion.
Parametric curves move in a specified direction.
This is illustrated in Example 1.
2.
Converting parametric equations. One of the simplest ways to convert
a parametric equation into the form
for
3.
t
y
=
f(x)
is to solve one equation
and then substitute into the other equation.
Tangent lines.
You should definitely know how to compute a tangent line
for parametric equations.
Remember that for
y = f(x) ,
the tangent
Copyright 1985 Springer-Verlag. All rights reserved.
450
Section 10.4
3.
(continued)
l i n e i s g i v e n by
metric
and
4.
xo
by
f(tO)
ddxldt)
.
For t h e analogous paraby
+
, we m u l t i p l y and d i v i d e by
(dyldt)
Adxldt)'
+
,
g(tO)
.
2
dt
.
dt
Start-
t o get
ds =
I n t e g r a t i o n y i e l d s the a r c length formula.
Speed. I t i s p r o b a b l y b e s t t o memorize t h a t s p e e d
mula
, y0
g ' ( t0) / f l ( t0 )
i s r e p l a c e d by
ds =
2
fi(xo)(x - x )
0
The i n f i n i t e s i m a l argument i s v e r y e a s y t o f o l l o w .
Arc l e n g t h .
ing with
5.
f ' (xo)
form,
+
y = yo
, which
(dyldt)'
i s g i v e n by t h e f o r -
is the a r c length integrand.
SOLUTIONS TO EVERY OTHER ODD EXERCISE
From
and s o
(x
+
9)
slope
5.
x
1
=
=
becomes
and
t
.
cos t
.
114
=
2 = (x
+
1)/4
t
+
g(1)
x = t ,
+
x
1
1
=
m
1
.
then
,
1)/4
2 = (114)
x
The g r a p h i s a s t r a i g h t l i n e w i t h
and
y-intercept
x
9/4
.
t
.
=
2x
2
+
.
A n o t h e r one i s
x = sin
tlfi
and
y
=
,
the equation
y = t 3 + 1 .
Wecanalsolet
2
x = + t
,
so
Several other solutions a r e possible.
f ( t ) = (112)t
and
=
+
+
Many o t h e r s o l u t i o n s a r e p o s s i b l e .
f t
Given
t = (x
we g e t
y = i
m
, so a parametric representation is
y = i
Ifwelet
y
13.
+
y = t
,
1
One o f t h e s i m p l e s t p a r a m e t r i c r e p r e s e n t a t i o n s i s t o l e t
y2
9.
-
x = 4t
2
+
t
g f ( t ) = (2/3)t1I3
f '(I) = 2
y
=
, and
and
.
At
y = g(t) =
to = 1
g f ( l ) = 213
[ ( 2 / 3 ) / 2 ] ( x - 312)
+
1 ,
.
t2I3 ,
, we h a v e
we h a v e
fl(t) =
f ( 1 ) = 312
.
So t h e t a n g e n t l i n e h a s
or
y = (113) ( x
+
312)
.
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.4
17.
451
The parametric form of the tangent line equation is: x = f '(to)
(t - to) +
2(2
-
1 ,
f (to)
3t)(-3)
and
and
ff(l) = 6 ,
6(t- 1)+ 1
.
g'(t) = -3
g(1) = -1
and
+
y = gr(tO) (t - to)
and
g'(l) = -3 ;
.
Since
At
sin t
t
=
y+
Here,
f '(t) =
to = 1 , we have
Thus, at
1) - 1
y=-3(t-
.
g(tO)
therefore,
=
3 ,
+J(I -
f(1) =
=
the bead is at
2t)/2
COS
, Y
=
The tangent line should be horizontal when
X
at these points.
izontal.
21.
Since
=
j i 2 t m d t
ji/ldu
.
2t
.
Let
u
=
n
u
and
cos €I)/(-2 sin 8)
.
and
=
0
=
nn
,
i.e.,
y
also
2t2
t = nn/2
If
so
t = mn
=
n/2 - 1
dy/d0
.
is
for all integers m
the length is
.
du = 4t dt
1;
+I
When
n
.
yf/x' shows that the tangent is not
s =
Then
s
j k m
=
(1/2)
Integrate as in Example 3, Section 10.3, to get
dx/d8 = -2 sin 8
0 ,
2t
so there is a vertical tangent.
dy/dt = 4t3 ,
and
P--+ 1+ 2
(a)
,
and the limit of
[(u/2) u
29.
is an integer. However, x'
So vertical tangents occur when
dx/dt
t = (2n
l)n/2 , where
n
The graph should be vertical at
odd, y' = 0 ,
vertical.
i.e.,
+
=
From the graph, we see that the tangent is never hor-
is even, y' # 0
If n
0 ,
y'
=
=
(112) [/5 + (112) in(2
1 - cos 8
8 = n/2 ,
.
x
s = (112)
+ &)I
Then dy/dx = (1 -
dy/dx = I/(-2) = -112 ,
The tangent line is
dt =
x
[y - ( ~ / 2- l ) ] /x
=
=
Copyright 1985 Springer-Verlag. All rights reserved.
x
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.4
453
SECTION QUIZ
1.
2.
x
,
Sketch t h e curve
(b)
What i s t h e t a n g e n t l i n e when
A c u r v e i s d e s c r i b e d by
t
=
x
t
2
(a)
=
t5
y
+
t
=
t
3t2
3
.
=
0 ?
- t
+
7
3
y = 3 t 4 - 2 t 13
and
- 4 . What i s t h e e q u a t i o n of t h e t a n g e n t l i n e when
t
=
+
1 ?
<t <1
3.
What i s t h e l e n g t h of t h e c u r v e i n Q u e s t i o n 1 f o r
4.
What i s t h e s p e e d o f a p a r t i c l e moving a c c o r d i n g t o t h e p a r a m e t r i c e q u a t i o n s g i v e n i n Q u e s t i o n 2 when
5.
-1
t = 0 ?
The human c a n n o n b a l l ' s p a t h c a n b e p a r a m e t r i c a l l y d e s c r i b e d by
and
y
=
5 6 t - it2
.
?
x = 5r
She it; s h o t t h r o u g h a r i n g of f i r e a t t h e h i g h -
e s t p o i n t of h e r f l i g h t and e n d s h e r f e a t by d i v i n g i n t o a n a r r o w t u b e
of w a t e r a t
y = 0
.
The cannon i s l o c a t e d a t
(0,O)
.
(a)
How f a s t i s s h e g o i n g t h r o u g h t h e r i n g of f i r e ?
(b)
A t what s p e e d and a t what a n g l e d o e s s h e e n t e r t h e w a t e r ?
ANSWERS TO PREREQUISITE Q U I Z
Copyright 1985 Springer-Verlag. All rights reserved.
454
S e c t i o n 10.4
ANSWERS TO SECTION Q U I Z
+
350
2.
30y
3.
~(13~/~-8)/27
4.
"5
5.
(a)
5
(b)
10 ; -713 r a d i a n s from g r o u n d
33x
-
=
0
Copyright 1985 Springer-Verlag. All rights reserved.
10.5 Length and Area in Polar Coordinates
PREREQUISITES
1.
Recall how to compute arc lengths of parametric curves (Section 10.4).
2.
Recall how to convert from cartesian to polar coordinates (Section 5.1).
3.
Recall how to sketch graphs described in polar coordinates (Section 5.6).
4.
Recall the trigonometric half-angle formulas for sine and cosine (Section
5.1 and inside.front cover).
PREREQUISITE QUIZ
1.
Which of the following is equivalent to
(a)
2 sin 8 cos 8
(b)
(1
+
cos 28)/2
(c)
(1
-
cos 28)/2
2
sin 6 ?
2.
Sketch the graph of
3.
What is the length of the curve described by
for
4.
0
8
<n
?
r = sin 0
in the xy-plane.
x
=
2 cos 4 , y
=
sin 0
(Write a formula, but don't evaluate it.)
Convert the cartesian coordinates
(3,3)
to polar coordinates.
GOALS
1.
Be able to compute the arc lengths of curves described in polar
coordinates.
2.
Be able to compute the area of a region described by polar coordinates.
STUDY HINTS
1.
Arc length.
Simply use
The derivation of formula (1) on p. 500 is not difficult.
x
=
r cos 8 = f(8)cos
0 and
y
=
r sin 6
=
differentiate by the product rule and substitute into
f(8)sin
L
" ,
=
Copyright 1985 Springer-Verlag. All rights reserved.
456
1.
Section 10.5
(continued)
b
ja/(dx/dt)2
+
2
(d~ldt) dt
. Don't forget to change
You may find it easier just to memorize
Arfa.
la,Li
.
.
e
A = (l/2)l~r2d8
You should remember that
to
L = /
~
d
+
1
2.
[a,b]
.
A sketch is usually
helpful and noting symmetries will save much work.
3.
Choosing limits.
In many instances, you will have to find the limits
of integration. For arc length problems, be careful the limits are
chosen so that the curve is traversed only once.
of the circle
not 0 to
-
r = 1
47 .
should be found by integrating from
0
to
2r
,
For most problems about area, as in Example 3, you will
need to find where
4.
For example, the length
.
f(b) = 0
Area between curves.
This is done by computing the larger area and then
subtracting the smaller area.
[f
(e)]
= v%
*
Thus, the integrand is
2
-
SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
We have
j~'h
dr/d8 = 3 cos 8 ,
2
cos 0
+
s(1
+
sin 8)
/ 2 T ~cos(F:
- ~ - 112) d ?
'0
=
I,
27i
J18
'I2
'0
+
by the identity
the half-angle formula, this becomes
12 sin[(^ - i./2)/2]
=
L
so the length of
2
12JF
is
18;ms
de
sin 8
=
cos(P - r.12)
v%~i''fi
cos [(1/2)
(0
.
By
- d2)] d: =
.
Using the formula A
(1/2)j:(3
(9/2)($/2
sin 3)*d3
-
=
sin 29/4) ;1
, we get
(1/2)j:r2di
=
(9:?)1;[(1
=
-
cos 2~)/2]d!
9 ~ / 4.
Copyright 1985 Springer-Verlag. All rights reserved.
=
Section 10.5
A
The area is
2
sin 8)d8
.
+
+
+
8 sin 6
Using the half-angle
formula yields
8 sin R
+
(1/2)(in(4
=
2
2
sin 8) dB = (1/2)j 0'(16
457
(1
(l/2)jin [I6
-
+
cos 28)/2] de = (112)
[336/2 - 8 cos 8 - sin 28/41
27i
x
-
lo
.
33n/2
The length is L
n/2
2
j-T12dan (912)
~'-7
=
dB =
2
sec (9/2)/4 d6
+
.
The area
2
is A = (l/2)1Br2d8 = (1/2)j:/;~tan
(i/Z)di =
n/2
2
(1/2)1-~/~(sec (812) - l)d6 = (tan(812) 912)
17.
The length is
The area is
A
L
=
=
+
(1/4)i cos 28
21.
2
21 cos 8
(1/2)j0/~(8~+
2 .
(28 sln 9
-
I - 7il4 + 1
=
J~/~/(I + con i - i sin 0)'
2
~ ~ ~ ~ ( R2 +9 F ~cos
~ 0
24/2)d81
s
3
(112) [n 116
1/j2
+ 48
.
+
2
2
3
8 cos d)di = (112) [(i 13)
cos O - 4 sin
(114) sin lo)]
=
e +
3
(112) [n 116
cos E
=
2
cos 8
6 ~ / 6
=. (2
(112) [j;l63
+
+
+
8 = n/6 ,
when
+ cos2R db
716
2
3 sin 8 do = n/2 - 1116+
.
&)n/6
sin2B di
(112) [(3/2)joi6(l
+
A
=
jT16cos2d di]
=
The area is
n/2
- cos 2E)d0
=
.
2
8
=
A
+
2
n 12 - 4 - 1/81
6 sin e
the length is L
j:'6h
2 .
(1/4)9 sln 20
dB
10'~+
Integrate each term by parts to get
Since
n/2
jnI6(l
+
- ill4 = 2 - -12 .
+ 2 cos R + cos28)
02 (1
+
(112)
+
x
+ (c + (l/Z) 1
cos 2a)del = (114) [(so - (312)sin 28)
n/2
sir. 28)
= (114) [n/2 - 3 6 / 4 + n/2 - n/6 - 6/41 = (1/4)(5i;/6 +
6) .
Copyright 1985 Springer-Verlag. All rights reserved.
.
458
Section 10.5
25.
Substituting into the formula L = j B j f ( 0 )
+
2 di
if '(13)l
,
we get
SECTION QUIZ
1.
(a)
Find the area bounded by the curve described by
and the rays
(b)
8 = 271 and
8 = 377
r = 8 cos 8
.
Write the length of the curve as an integral, but do not evaluate
it.
2.
Find the area and the perimeter of the region enclosed by each of the
following:
3.
+
(a)
r = sin 0
(b)
r = 12 cos 8 /
cos 8
The region inside both
2
A = (112) [jbacos (8
+
r = sin R
n/3)d8
+ J:
and
r = cos(8
sin2? d91
.
+
n/3)
has area
Find the limits of inte-
gration and compute the area.
4.
An astropirate on the surface of Mars was about to accelerate into
hyperspace when he noticed that the space patrol seemed confused and
was flying circles around him.
relesing a fence described by
(a)
In reality, the space patrol was
r
=
1
and
r = 1
+
cos 0
.
1 , how much fence material was used?
If the fence has height
(Assume that the space patrol may penetrate their own fences.)
(b)
The astropirate was caught in one of theregionsinside
outside
r = 1
+
cos A
.
r = 1
and
How much material was needed to make the
top and bottom of the space prison?
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.5
459
ANSWERS T O PREREQUISITE QUIZ
ANSWERS TO SECTION QUIZ
1.
2.
(a)
1 9 ~ ~ 1 1+2n/8
(b)
1::h2
(a)
Area =
(b)
Area = n/2 ; perimeter
+
;
T
3.
a = n/12 ; b
4.
(a)
8+2n
(b)
n/2
+
2
cos I -
=
e
s i n 28 dc
perimeter = 2fi11
7716 ;
=
c = 0 ;
2n
d = n/12 ;
area = n/l2 - 114
4
Copyright 1985 Springer-Verlag. All rights reserved.
460
S e c t i o n 10.R
10.R Review E x e r c i s e s f o r C h a p t e r 10SOLUTIONS TO EVERY OTHER ODD EXERCISE
1.
Substitute
u
,
sin x
=
so
du
,
c o s x dx
=
2
j 3 s i n x c o s x dx
and
=
2
3
3
J'3udu=u
+ ~ = s i n x + C .
/-'I
.
j(1
Let
2
.
- c o s 6 ) s i n 8 d3
we g e t
-f(l
2
- u )du
x
Now, l e t
=
+
-u
u /3
9.
13.
Let
2
(X
17.
u = x 2 + l ,
+
2
1) Idx
=
j[(u
- l)du/2u 1
+ c
= (1/2) [ l n l x
Let
u
&,
=
-2JG c o s
Using
vG +
so
.
2 sin&
2
+
u2
2
+
=
x
+
11
and
+
tan x
+
C
C
.
+
2
s e c 6 do
x
+
211
and
Using t h e
+
2u du = dx
Thus
2
- l / u )du
c
.
and
J' ldu/!u2 +
=
d u = 2 x d x .
111
,
x 2 + x + 2 = ( x +
=
= (1/2)j(l/u
1/(x2
and
du
I n t e g r a t e by p a r t s t o g e t
sin x = 1
lnlsec x
~ h i ~ s ,/, [ d x / ( x
x 2 = u - 1 ,
so
so
2
and
6-7
+ C .
tan 0
,
dB
-du = s i n 3 do
- cos
C = c o s 913
d%%
2
l/uI
21u s i n u du
21.
so
so
3
- x2)3'2 -
(1
dx = c o s
,
cos @
=
u = x + 1 / 2 ,
.
sec 8
+
so
Let
J7/4
=
,
sin B
u
3
f i g u r e , t h e a n s w e r becomes
/?+ 714
=
=
714)l
=
l[x3/
(112) [ l n l u l
+
.
Then
-2u c o s u
jsin&dx
+
2 sin u
=
+
C
=
+c .
2
- c o s x , we g e t
- sin x
+
C
.
2
j ( s i n x/cos x)dx = I ( s e c x
The t e c h n i q u e f o r i n t e g r a t i n g
- c o s x)dx
sec x
shown i n Example 6 ( b ) , S e c t i o n 1 0 . 1 .
Copyright 1985 Springer-Verlag. All rights reserved.
is
=
S e c t i o n 10.R
25.
Let
3
I=j[x/(x
-9)ldx
andlet
2
( x 2 + a x + a ) ) l d x = j [ ~ / ( x- a )
a(aB
-
1 ;
and
.
D)
.
+
.
aB - D = 0
+
Thus,
aB = 1
Substitute
, so B
C
.
= 1/3a
+
+
(afi)]] l + C = (1/3a) [ l n i x - a / - l n L 2
29.
+
/(l
ex)-'dx
tuting
33.
j [ l - ex(ex
=
+
u = ex
+
1
1)- ] d x
+
+
C
& ,
Let
u =
41.
Let
u = x2 + 3
45.
Let
u
Let
u
2
aB
=
Therefore,
(-x/a
+
1)/(x2
-1/3a
=
+
ax
a2
+6
+
x - ln(ex
1)
+ ax +
C
+
and
D
=
2
a )]dx
+
1
tan- ((2xIa
+
=
a ( B - C)
into
C
D
=
a/2)/
+
I)/
, by s u b s t i -
+
s i n ( x - y)]
s i n x)dx
=
t o get
- ( l / l O ) c o s 5x - ( 1 / 2 ) x
+
so
,
du = ( l / 2 & ) d x
.
Then
(eh/&)dx
so
.
Then
j[x/(x2
du = 2x dx
C = (1/2)ln(x2
( l n 3x)
+5
, so
+
3)
+
+
C = ln&
du = ( 3 1 3 x ) d x = d x / x
.
+
=
2 j e u d u = 2eu
3)ldx
=
(1/2)j(du/u) =
C = (1/2)in(x
Then
+
2
+
[ ( l n 3x
3)
+
+
=
, so
cos 9
cos e ) ] de =
1
-jl
[l/(l
-du = s i n 8 dB
+
u
+
2
u )I du
.
.
Then
jg2n [ s i n
0/(1
+
cos
C
5 )a /
.
186.1
49.
=
D
+
a ( B - C)
C = 0 ;
and
, where
.
37.
(1/2)ln/uj
+
=
+
y)
j s i n 3x c o s 2x dx = ( 1 / 2 ) / ( s i n 5 x
cos x
+
2
a x + a )]dx
1
s i n x c o s y = (112) [ s i n ( x
Use
+
I=/[x/((x-a)x
2
a ) - (3a)(~/a&)tan-' [2(x
+
ax
B
-B
=
I = (1/3)j[(l/a)/(x - a)
( 1 / 3 a ) i l n / x - a / - (112) [ l n ( x 2
Then
(Cx + D ) / ( x '
E q u a t i n g c o e f f i c i e n t s , we g e t
D = 1 = a(2B)
113
3fi.
a =
461
:+
S i n c e t h e l i m i t s of i n t e g r a t i o n a r e
b o t h 1 , t h e1985
i n t e g r aSpringer-Verlag.
l is 0 .
Copyright
All rights reserved.
.
462
S e c t i o n 10.R
b
/-4
+
L =
2
[ f l ( x ) l dx
,
53.
Use
57.
S i n c e t h e graph i s revolved around t h e y-axi*,
+
2rjb x h
2nj;:O
xJi
I2
[ i t (x)
+
.
dx
( i / x i n 10)
where
Here,
2
dx
f x
A
10) ) ~ n +
x
(112(ln
+
51100
L2 + i / ( l n
61.
Solving f o r
into
y
.
,
x = 1
i.e.,
,
+
we g e t
t o get
y
213
=
2x13
and
y
2
3t /4t3
In t h i s case, it is
=
The l e n g t h
.
y
=
2
3x14
+
514
and
L
+
.
r2
+
.
R&
Let
The a r e a
A =
+
A
TI2 4
(1/2)j0 i dt
The l e n g t h
L
2
I. = r
0'
2
2
+4
,
6 (3
i s g i v e n by
=
5
(8 110)
3 cos ( ? / 4 ) s i n ( t / 4 )
2
,
9 c o s (814) [ c o s ( 0 1 4 )
so
+
+
L
r2
2
4)
+
+
i / ( i n 10)
2
2
l / ( i n 10) ) /
.
1
.
Substitute
This i s a l i n e
1
.
dyldx = (dy/dt)/
t = 1
,
,
d y l d x = 314
- 2 ) / ( x - 1)
(y
+
di
=
314
,
and
L = (112)
;:&iF-+[ r '(.)I
'd:
=
(1/2),(:
5
i
(r')'
1320
x
In t h i s case,
.
8
=
In this care,
becomes
.
=
.
c'
and
2
[ r ( t ? ) ] dH
A
=
,
2
)
du = 2P d3 ,
so
lbI2
i s g i v e n by
3
6
=
=
2
.
46'
u
+
y-intercept
When
=
~ntegrate
l / ( l n 10)
t = x/3
Thus, the t a n g e n t l i n e i s
i s g i v e n by
4
( r ~ =) 8~
.
+
+
A0000
A
=
A =
= 2n [ 5 O h 0 0 0 0
J '"
73.
2
For p a r a m e t r i c e q u a t i o n s , t h e s l o p e of t h e g r a p h i s
(dxldt)
69.
:0
t
with slope
65.
1
so
l / ( l n 10) 2 dx
2
( l l 2 ( l n 10) ) l n [ ( 1 0 0
+
l / ( l n 10)'
10)
2
,
10
2n[(x/2)/x
=
L
Thus,
use t h e equation
1
+
271 I o 0 J x 2
10
=
a s i n Example 3 , S e c t i o n 1 0 . 3 t o g e t
2
x
=
.
- l/4x2
f l ( x ) = x2
9 cos ( r / 4 )
6
s i n ( 8 / 4 ) 1 = 9 c o s (314)
=
+
.
Here,
r
6
9 cos ("/4)sin
2
9 cos (0/4)(1 -
Copyright 1985 Springer-Verlag. All rights reserved.
=
2
(-14) =
S e c t i o n 10.R
73.
(continued)
2
sin (914))~
.
Then
u = sin(@/&),
The a r e a
:j
+
+
128)Jn(1
0
+
128)(3
+
914
3 sin 8
1:
+
= 9n(1/8
This exercise uses the formulas
s i n ( x - y)]
and
- ( 1 / 2 ~ )[ ( 1 / ( 3
+
unless
.
m = 3
=
1
.
31x1
lo
=
m)) s i n ( 3
s i n (812))
+
914
20)dB
+
(1
9718
+
11256)
(112) [ c o s ( x
+
+ m)x +
y)
+
+
4 C O S ( ~ / Z+)
+
+
+
cosa) /4]de
1; +
914
=
+
=
(91
+
914 - 914
914
(91
9 ~ 1 1 2 8+
+
y)
+
.
a
=
c o s ( 3 - m)xldx
( 1 / ( 3 - m ) ) s i n ( 3 - m)xl .l o2 n
hm = ( 1 1 n ) ~ i ~ c o3x
s s i n mx dx
+
3))cos(m
+
3)x
+
=
.
3 1 5 1 ~ 1 2 5 6+ 914
cos(x - y)]
+ m)x +
1: +
2
8 s i n (0/2)]
27~132
=
A = (112) x
8 sin(0/2)1
s i n x cos y = (1/2)[sin(x
=
-(1/27) [(l/(rn
=
Thus,
3
-
8 sin(5/2)
11128
=
=
=
0
a 3 = ( 1 1 2 n ) ~ ~ ~ ( c6ox s + 1 ) d x = ( l / Z n ) ( ( l / 6 ) s i n 6 x
Then
Also,
sin(m - 3)xldx
2n
cos x cos y
2
2n
cos 3x c o s mx dx = ( 1 1 2 r ) J ' ~ [ ( C O S ( 3
(l/i:)j;IT
x)lin
+
+
+
10
c o s (8/2)1dA = ( 9 1 3 2 )
+ c o s 8)dA = 9 ~ 1 3 2+
0 ) 1; + ( 9 / 2 5 6 ) j ; ( l + c o s
s i n 2812)
.
(9/32)j: [ i
=
4
-
2
- u )du
.
4
C O S ( ~ / ~do116
))
2
2 cos 8
+
5fi
Let
12'fi12(1
10
=
2
4 cos(8/2)(1
+
( 9 1 3 2 ) [38
2 sin
(9/256)(8
+
cos 5)/2
=
L
A = ( 1 1 2 ) ~ : [ r ( B ) ] d8
+
+
4 c o s (812)
+
(9/32)j:[6(1
97/32
3
+
6 cos (e/2)
- hiiz)
i s g i v e n by
A
, and
du = ( 1 / 4 ) c o s ( 8 / 4 ) d B
= 12(fi12
.
2
- c o s ( 0 / 4 ) s i n ( 8 / 4 ) ] d8
3j;[cos(8/4)
=
8
9 c o s ( e / i ) d e = (9/2)j:(l
2
81.
L
so
- u313) :I2
12(,
77.
463
(l/2n)iET[sin(m
+
+
3)x
(l/(rn - 3))cos(m -
.
0
We u s e t h e p r o d u c t f o r m u l a s l i s t e d on p . 460 and t h e h a l f - a n g l e f o r m u l a
2
sin x
2
a = ( l / n ) j i n s i n x c o s mx d x = ( 1 1 2 ~ )x
m
2n
~ ~ ' [ c omx
s - c o s 2x c o s m l d x = ( 1 / 2 n ) j 0 [ c o s mx - ( 1 / 2 ) ( c o s ( 2 + m)x
cos(2
=
-
(1
-
m)x)] dx = ( 1 1 4 ~ [) ( 2 / m ) s i n mx
m ) ) s i n ( 2 - m)x]
2T
jO
.
cos 2x)/2
[COS 4x
+
( l / 2 7 r ) j i n l dx
l o2 ,r
lldx
+
+
,
0
=
unless
-
m = 0
(1/(2
or
0 = - ( 1 / 4 ~ ) ( ( 1 / 4 ) s i n 4x
0 = 1
.
Also,
bm
=
+
+ m ) ) s i n ( 2 + m)x - ( 1 / ( 2 m = 2 . Then a 2 = - ( 1 / 4 ~ ) .
+ x ) I 02*I
=
-112 ;
2
2
( ~ / T ) J ' ~s i' n x s i n ( m x ) d x
and
=
a.
(l/27)
=
x
Copyright 1985 Springer-Verlag. All rights reserved.
+
464 Section 10.R
81.
(continued)
85.
2cos 2x sinim)] dx = (112n)j0 [sin(mx) - (112) (iin(2
-
[sin(m)
2
2
( 4 1 ~'0
)
[Om(-2 sin 0m sin 8
2
- 2 sin
=
(Llr)!>(l
+
2
2 sin m )-'I2dc
2 ~in~@~)-''~dl
=
J,(Llsin $m)(l
2
- sin fi)-li2di = ( 2v2/r)j0
r
m
' [l/(sin 0 cos i)] d:
m
corn
tiate
c
sin
@ =
sin
@ =
@
m
sin B
to get
0 ,
cos
,
=
.
sin @ cos B d? -1
m
89.
I
Substituting the~givenformulas yields
.
B = 0
m
2
= sin
here k
When
and when
@
=
to get
=
( ~ / T ~ ~ ) j ~ ' 2 dr>
[ .l / ~ ~
(a)
n+ 1
a
)/(n
+
b - a
(b)
+
1) , unless
A
n
=
.
-1
The iength of this graph is
I.
=
/:
dx
=
J?(h
then the length is
b
~:m
dx = jlfi
b
ja(l
+
=
.
ln(b/a)
.I
=
If
n
(2k
J:A
2)
and
+ 1 i2k
dx
-
=
-1 ,
A
=
h
-
- a)
.
=
a
. If n
If n
L
=
jb/l +
.
k
is a nonnegative integer, we have
( 9 1 4 ) ~, we get
Since
2)12 x"(~+')
dx
=
cos f
$
(bn+'
+
lb(l
'a
. If
L = (S/Zi)(l
+
-
xW1)dx
n
0 ,
=
L
=
=
L
=
. Vow, jet u
=
=
(2k
+
2 Ln-2
n x
ll(l-n)du/~(217 - 2 ) x
1
.
Then
=
g x j ~ ) ~ =' ~ ~
For the more genpiai case where n
+
Differen-
( 9 1 4 ) ~dx
2
(3-2n)/(2n-2)
+
I,
x
.
x
3)/(2k
sin
2 , then
=
+
? = 712 ;
l , then
=
+
=
.
.
so
b - a
=
r
lbL
+ n2x2n-2 dx
312 , we have
=
x = n
SO
(U
+
1
I
Substituting u
1 / 2 4 + 9
n
xn)dx
+
Ja
n = 2 case is Example 3 of Section 10.3
For the case
=
Substitute all this and cancel
The area beneath the graph is
- L
, we get dJ
, sin 6
0
=
(n
m)x
(26/i)
cos G dE
sin Q
@ dq =
=
$
+
1+n2b2n-L
Id =/l+n2a2n-2[
T
l/(l-n)d,, ,
Copyright 1985 Springer-Verlag. All rights reserved.
1)/
465
Section 10.R
89.
(b) (continued)
k
(2n - 2)(u
1 .
- 1)
(u - l)k
Recall that the binomial expansion for
k
li=O[r]ui
is
)
1 /(i +
312)
.
Vx
4b(1
+
Around the x-axis,
(c)
n+ 1
a
)/(n
+
n
.
-112
=
2 n
+
1)
x-')dx
If n
n [b - a
=
+ xn+l)dx
2n;:(x
Around the x-axis, Ax
:12
x
n
=
2 2n-2 i+3/2 [(l+nb
1
k
(Note: recall that ( . ) = k!/
2xn
-1
=
-
I
+
4&
=
7,
Vx
-
+
4&
[h2 - a2
.
+
+
=
x
Note that
2n
)dx = nib -
2
l
If
ax
=
2nj:
=
0
,
n
xnk=
then
a
n+ I
=
2x-I
+ xW2)dx
=
n[b - a
+
Vx = r!~:(l
-1
-
or
+
+
2x-'"
Around the y-axis,
w dx
i
the arc length which was analyzed in part (b).
iook at
2(b
n
n+2
- a
)/(n
)
+
unless
.
n
i
,
-112 ,
ln(b/a)l
2(bn+'
x
1)l
+
= r!~;(l
. If n
2ilb(l
=
m dx
,
+
+
(bznf1 - aZn+l)/(2n
a - a
b
(nl/(l-n) /(2n-2))~
=
k-i i+1/2
- 1
u
du = (nl/(l-n)/ (2n-2))x
2 2n-2 it312
(1 + n a
.I
and this gives us
[]
k
+
=
V
Y
unless n
2)l
2
i
i
~ dx ~is
b
=
=
m dx
+
217
X
times
Thus, we only need to
dx.
=
a)
.
If n
b 2
J
2
2n13(x
1 + 4x )dx
.
The integral
2.;/:
dx
=
2r(b
-
=
1
,
then a
=
a
If n
=
2 , then a_
~b[ix~/J;+ 4x2)dx.
u
=
x3
, dv
=
=
The latter integral is integrated by parts with
4 x / m - d x , du = 3x2 dx , and
v
=
. Therefore,
Copyright 1985 Springer-Verlag. All rights reserved.
466
Section 10.R
89.
(d) (continued)
~
:
(
~
~
m
b
= )ja(x
d
aeariangement yields
2
2
/~J
1 + 4x )dx + [x31;;1;11:
41: ( x 2 m ) d x = b
3
.
in Example h(b),
(112) tan
=
a
so
dx
2 lnjsec 0
+
tan
-1
011
.
/E:z-l;:
Now if
and similarly, if tan 13
tan 6
m+ 'a/)
/
n = (2k
+
. Finally, we get a
+
3)/(2k
2) where
let x = ((u - l)/n ) 1/(2n-2)
(u - 1)
for
(4k
(3-2n)/ (2n-2)
n
+
du
l)n (n+l) / (n- 1)
-
[n/(n
in
-
2
.
=
u
-
1
k
A
k
, i=O i (-l)k-iui+3/2 /(i
get
a
=
k
=
+
-
+
312)
+C .
a
+
m
2)n l/(n-1))
ax = 2 n j ; x n X d x
=
Substitute
+
4/(2k + 1)
-
2)
x=b
k & du
- ~)]j~=~(u-l)
(u - lIk = Zi=,,
JE:=~[:]
.
1))/(2
[nn(n+l)'(l-n)/(n
=
2
is any nonnegative integer,
dx = 1/[ ((2n
I
2bl -
( h z+
+
Then
=
(r/2)[ (1/16) in]
2) = ((3 - 1 - 2)/(2k
.
-
. This
=
-
=
tan 81
2b, then sec 0
=
u du
a
+
lnjsec 0
x=a
Hence
u
+
@
1
. Substitute this into
jx-b
(3-n)/(2n-2)
(U -
(3 - n)/(2n
2)/4 = k
so
From the binomial expansion formula,
Ler
+
=
2a, then sec 0
=
substitution gives us (1116) ( 2 b h x - ln
If
m
Section 10.1 and Example 3, Section 10.3, respectively.
Therefore, , / ~ ( r 2 / ~ ) d =x (L/I6)[scc 0 tan
n
3
2
(112) sec Ode
-1
2
3
tan e sec e do = (1/8)jE:z-1::(sec
@ 3
The integration of sec 0 and sec e are shown in detail
x
metric substitution. Let
-1
This yields (118)j::z-1;;
1
x - a
A
. The latter integral can be done with a trigono-
1b(x'/J1~)dx
sec 8:de
.
- 3]i(x2Az)dx]
r]
i (-l)k-iui
(-i)k-iui+1/2
Substitute for
u
.
du =
and
k
to
nn (n+l) 1 (1-n)
/
2 211-2 i+3/2 2 211-2 i+3/2
((l+nb
)
(l+na
)
)/(i+3/2).
Copyright 1985 Springer-Verlag. All rights reserved.
.
=
S e c t i o n 10.R
89.
(d) (continued)
Around t h e y - a x i s ,
A" = 2nj:
A = 2n~:xdx=r(b
then
1 + 4x2
yields
( n / 6 ) [ (1
+
.
a x dx
A
Y
,
Then
(1
3
b
A
Y
2
K
n = (k
If
+
(2 - n ) / ( n
then
k .
Let
1) = k
If
,
If
n = 2
--
n = 1
,
A = 2njb x
y
2 "
= (n/3)~::2
3)/(k
1)
+
2)
= (2
m dx
-
1 - l/(k
(u
-
+
la
Let
+
+
l))/(l/(k
.
i s any i n t e g e r , t h e n f o r
,
and
sin(n - a) = -sin(n
about t h e x-axis.
sin u
Since
the point
Also, since
+
a)
=
.
n
1 - I)/] =
and
A
=
-
(u
A
/ (n-1)-1
1)
k
=
Y
/ ( i +3/2)1
x
.
but notice that i f
.
2
Y
Since (2 - n)/
.
du
P l o t t h e curves point-by-point,
(-1,O)
+
1 ) ) = (k
Then
-
2 2 n - 2 i + 3 / 2 - ( 1 + n a2 2n-2 ) i + 3 / 2 1
[ ( l + n b
)
is
a
i s any n o n n e g a t i v e i n t e g e r ,
(2-n)
, P
=
3b2
- 3
-)I
/ (2n - 2 ) ) du
t = 0
du
1 3a2
this into the integral t o get
1
t = n
=
+
l ) k w i t h t h e binomial formula:
[l]
( - ~ ) ~ - ~ u lS .u b s t i t u t e
=
u = 3x2 s o
(l(2)lnlu
n[ n
(a)
u
2 312 b
4x )
x = ( u - 1 ) 1/(2n-2)nl/(l-n)
(3-2n)/ (211-2)
expand
k
.
+
+
M+1)/(3a2
, s o
=
Y
I n t e g r a t e a s i n Example 3 ,
du
where
-
A
then s u b s t i t u t i o n of
A = (n/3)[ (u/2)&
Y
+ I n [ (3b2 +
-
, then
,
n = 0
If
I.
4a2)3'2
2 2n-2
l+n b
- I)]! 1+~2,2n-2 ( U
[ nn 2'(1-n)/(n
'i=o
+
2 2n-2
u = l + n x
dx = ( n l / (1-n) ( u
-
.
.
.
m dx
then
Section 10.3 t o get
(n:6) (
x
2
- a )
2 n f b x h -t 4 x 2 dx = ( n / 6 ) ( l
=
-
4b ) 3 / 2
n = 3
If
(n
2
Y
b
2
2
2 n f a f i x dx = n f i ( b - a )
93.
467
cos(n
for a l l a
P
-
is
m
=
1
and
;
(1,O)
a ) = cos(n
+
n
for
a)
, t h e c u r v e s a r e symmetric
c o s ( n - a) = - c o s u
and
sin(^
-
a) =
, t h e c u r v e s a r e a l s o s y m m e t r i c a b o u t t h e y - a x i s and t h e r e -
Copyright 1985 Springer-Verlag. All rights reserved.
468 Section 10.R
93.
(a)
(continued)
fore, the origin.
the x-axis at
n
times, but as
t
Finally, since
+
1 points.
goes from n
yinln)
=
(Actually, each curve crosses
to
2n
2n
+
2
the curve will recross the
x-axis at t:-.e same points it crossed when
(b)
0 , each curve crosses
Consequently, each curve will consist of
t
n
went from
loops, for
0 to
n
r: .)
odd or
even
Copyright 1985 Springer-Verlag. All rights reserved.
Section 10.R
93.
469
(d)
TEST FOR CHAPTER 10
1.
True or false.
(a)
Every polynomial is a product of linear and/or quadratic factors
(b)
The area between
on the interval
(c)
,
is
r
(112);:
=
2
+
(Cx
>0
on
+ II)/x2
, where
g(8)
[f (9) - g(d)]
f(3)
2
l/(x - llLxL
for some constants
> g(5)
.
=
A , B , C
A/(x - 1) +
,
and
D
(d)
[a,b) , the surface area obtained by revolution
lb
2
around the x-axis is 21Ja&]2+
[fl(x)f (x)] dx .
(e)
In polar coordinates, the arc length of
If
The line
lution.
3.
[a,:]
and
Using the method of partial fractions,
B/(x - 1)
2.
r = f(0)
f(x)
y = -x
+
1
on
[0,31
r = f(8)
.
over the interval
is revolved to form a surface of rev,)-
Find the surface area if the line is revolved around:
(a)
the y-axis
(b)
the x-axis
Find the arc length of the curves described by the following equations
on the given intervals:
2
sin 9
(b)
r
(c)
y = t2+3
=
in polar coordinates for
and
x=2t+l
for
0
<C
2;;
O<t<l
Copyright 1985 Springer-Verlag. All rights reserved.
470
4.
Section 10.R
Evaluate the following integrals:
1;[(x3
(b)
j[dt/t(t2
5.
Show that
6.
Suppose
b
1)l
and
=
2
~ i ~ / i~ dO
~ o
= nn/4
s
.
y2 = f(x) + 3
If
for all integers n 2 0
f (x)
0 on
(b)
The surface areas of revolution obtained by revolving
[a,h]
yl
and
y2
on
[a,b]
The surface areas of revolution obtained by revolving
[a,b]
f(t)
f (t)
-
1) t
f(t)
=
l/(t
=
2
3
sin (t/2)cos (t/2)
and
Y2
Y1
and
Y2
for the given interval:
on the interval
2 <t < 3
on the interval
Find the area of the region inside
cos 2a
Y1
around the y-axis.
2
r
=
.
0 G t G 712
sin B cos
e
.
and outside
r
=
.
Find the area between the following curves and the x-axis in the given
intervals:
10.
(a)
y
=
4
sin x
(b)
y
=
x sin x cos x
on
2
10, n/81
on
1-14> 7/21
Dragster Debbie, the 80-year-old grandmother, needed some excitement in
her life.
She decided to take her Ferrari for a little spin down at the
speedway.
As she floors the gas petal, her position can be described
parametrically by
x = 2 cos t
(a)
What is her speed at time
(h)
At
t
=
8n
,
and
y
=
4 sin t cos t
.
to ?
she realized she had to hurry to her karate lessons,
so she sped off along the tangent line.
Describe the tangent line
with a set of parametric equations
Copyright 1985 Springer-Verlag. All rights reserved.
.
where
around the x-axis.
Compute the average value of
(b)
,
.
The arc lengths of
(a)
[a,b]
, then which of the following are equal, if any?
(a)
on
9.
+
igni2sin28 di
>a >0
(c)
8.
- t
yl = f(x)
on
7.
2
2
7x - 8)/(x
- 4)ldx
+
(a)
S e c t i o n 10.R
471
ANSWERS TO CHAPTER TEST
1.
(a)
True
is)
False;
it i s
(1/2)jEi[f(e)]
+ D)/x~
should be
2
.
- [ g ( e ) ] Id:
+
(d)
True
(e)
F a l s e ; t h e given formula i s f o r Cartesian coordinates
(a)
9nfi
(b)
57ifi
(a)
1512
-
(b)
l
l
3 In 3 - 4 In 2
t
/
G
/
+
(l/fi)tan-I [(2t - 1)/3]
) % / 2 s i n 2 ~ = j ( l / 2 - c o s 2812)d.
:
~
D/x
.
False;
~
C/x
2
(c)
n
(Cx
2
2
0/ = ~/ ( 1 /c2
+~ c o~s
+
C
2f3/4)l o11712 - ni/4 ;
= (t?/2 - s i n
2 8 / 2 ) d 8 = (012
+
s i n 2614)
l:n12
= n ~ / 4.
( a ) and ( c )
(a)
ln(314)
(b)
7fi130n
7i
+
112
- 415
(b)
+
( 3 n 6+ 2 0 a -
(a)
2jsin2t0
(b)
x = 2
(a)
3n164 - a 1 8
+
and
1/32
24n)/144
4
4 s i n to
+
4
4 cos t o - 2
COS
2
2
t s i n to]
112
0
y=4(t-an)
Copyright 1985 Springer-Verlag. All rights reserved.