TEACH YOURSELF VISUALLY - CALCULUS

Algebraic 11Techniques of Integration chapter
Substitution

Sometimes you encounter an integral that doesn’t seem to fit any of the more common integration
forms. In this case, the u-substitution technique may be useful.

EXAMPLE 1 let u = x - 1, or in another form u = ^x - 1h1/2 to be used later
then u2= x - 1
͐x x - 1 dx so that u2+ 1 = x
and 2u du = dx
1 Let u be the most
complicated part of
the integrand; in this
case, let u = x - 1.
Now find dx and solve
for x in terms of u.

2 Now you can make a large series of substitutions, plugging data ͐x x - 1 dx
from Step 2 into the original integral.
= ͐ _u2+ 1i $ u $ 2u du

3 Expand and simplify the integrand in Step 2. $= ͐ _u2+ 1i 2u2 du
$= 2 ͐ _u4+ u2i du

4 Use the Simple Power Rule on both terms of the integrand. $= 2 < u5 + u3 F + c
5 3

= 2 u5+ 2 u3+ c
5 3

237

Algebraic Substitution = 2 a^ x - 1h1/2 5 + 2 a^ x - 1h1/2 3 + c
5 3
(continued) k k

5 You need to replace the u terms with their = 25^x - 1h5/2+ 23^x - 1h3/2+ c
appropriate x terms, this time using the
substitution u = ^x - 1h1/2 found in Step 1.

or in another form

= 6 ^ x - 1h5/2 + 10 ^ x - 1h3/2 + c
15 15

= 2 ^ x - 1h3/2 83 ^ x - 1h + 5B
15

= 2 ^ x - 1h3/2^3x + 2h
15

EXAMPLE 2

͐1 x dx
-2 x+ 3

1 Let’s work Example 2, but this letting u = x + 3 for x = 1:u = 1 + 3 = 2
time change the x limits to u then u2= x + 3 for x = -2:u = -2 + 3 = 1
limits. Letting u = x + 3, so that u2- 3 = x
proceed as before in finding dx and then 2u du = dx
and x in terms of u.

238

11Techniques of Integration chapter

2 Make your long list of substitutions ͐1 x dx % this has x lim. of int.
into the original integral, changing -2 x+ 3
all x terms and limits to u terms and
limits. ͐2 _u2- 3i
1 u
$= 2u du % u terms and u lim. of int.

= ͐2 3i du

2 1 _u2-

3 Integrate and then plug in the u limits. $= <2 u3 - 2
3
6u F

1

= ; 2 u3 - 6u 2
3
E

1

= ; 2 $ 2 3 - 6 $ 2 E - ; 2 $ 13- 6 $ 13 E
3 3

= ;136 - 12E - ; 2 - 6E
3

= 16 - 12 - 2 + 6
3 3

= -43

TIP

If we had not changed from the x limits to the u limits
of integration the computation at the end of the problem
would have been much different — yet the final answer
would be the same. Try it and see what happens.

239

Solving Variables Separable
Differential Equations

When you find the derivative of some function, the resulting equation is also known as a

differential equation. For example, dy = xy + 7x , or yl= x2 , or (3 + x3)y' – 2xy = 0. Our goal in this
dx y

section is to determine from which original equation, or function, the differential equation was

derived.

GENERAL SOLUTION 1

Using this technique, you will separate the variables (hence the name for the technique) so that all the y
and dy terms are on one side of the equation and all the x and dx terms are on the other. Then, by
integrating both sides, you will arrive at an equation involving just x and y terms.

Solve the differential equation dy = x2 and write the solution in the form “. . . = some constant.”
dx y

1 Cross-multiply to get the dy and y terms on the left side and the dx and x dy = x2
terms on the right. dx y

y dy = x2 dx

2 Now that you have separated the variables, integrate both sides of the ͐y dy = ͐x2 dx
equation.
y2 = x3 + c
Notice that is only one “+ c.” If you were to put “+ d ” on the left and 2 3
“+ m” on the right, they would eventually combine to make some
third constant “+ c.”

3 Put all the y and x terms on the left (the answer form y2 - x3 = c or just y2 - x3 = c
that was requested). 2 3 2 3

240

11Techniques of Integration chapter

4 If you chose to, or if it was requested of you, you could y2 - x3 = c
eliminate the fractions by multiplying all terms of the equation 2 3
by 6.
6 $ y2 - 6 $ x3 = 6 $ c
m = 6c is just another constant. 2 3

3y2- 2x3= 6c

3y2- 2x3= m

GENERAL SOLUTION 2 dy
dx
For the differential equation = xy +3x, write its solution in the form “y = . . . .”

1 This problem takes a little more creativity to get the variables dy = xy + 3x
separated. dx

dy = x _ y + 3i
dx

2 Multiply both sides by dx. dy = x(y + 3)dx

3 Divide both sides by y + 3, and the variables will finally be y dy 3 = x dx
separated. +

4 You’re ready to find the integral of each side. ͐ dy = ͐x dx
+
y 3

ln y+3 = x2 +c
2

241

Solving Variables Separable y + 3 = ex2'2+c
Differential Equations (continued)
$y + 3 = ex2 ' 2 ec
5 Rewrite the last natural log equation as an exponential one instead,
and then simplify the right-hand side. 0

TIP $y + 3 = ex2 '2 m
$y + 3 = m ex2 '2
If ln y = x, then y = ex.
y + 3 = ±m • ex2 ÷ 2
6 Get rid of the absolute value symbol on the left and then simplify y = kex2 ÷ 2 – 3
the constant on the right.

GENERAL SOLUTION 3
Solve the differential equation (3 + x2)y' – 2xy = 0 and write the solution in the form “y = . . . .”

1 Add 2xy to both sides and then replace the y' with dy _3 + x2i yl- 2xy = 0
dx

_3 + x2i yl= 2xy

_3 + x 2i dy = 2xy
dx

_3 + x 2i dy = 2xy
dx

2 Multiply both sides by dx. (3 + x2)dy = 2xydx

3 Divide both sides by y. _3 + x 2i dy = 2xdx
y

242

11Techniques of Integration chapter

4 Divide both sides of the equation by 3 + x2. dy = 2x dx
y + x2
3

5 With the variables ͐ dy = ͐ 3 2x 2 dx
separated, you can y +x
integrate both sides.
ln y = ln 3 + x2 + c

ln y = ln _3 + x2i +c Note that 3 + x2 ias always positive.

6 There are two ways to deal with this double natural log ln y = ln _3 + x2i + ln ec
situation. We’ll use one method here and then demonstrate
the other method later in the problem. $ln y = ln 9_3 + x2i ecC
Replace the c with lnec.
$y = _3 + x2i ec
7 Use your log properties to rewrite the right-hand side of the
equation as a single natural log. 0

8 Since the natural log of left quantity equals the natural log of the y = _3 + x2i $ m
right quantity, you can get rid of the ln on both sides.
y = m_3 + x2i
TIP

ec is just another constant, say m.

9 Take out the absolute value symbols on the left and insert a ± sign y = ! m_3 + x2i
on the right. 0

TIP y = k_3 + x2i

±m is just another constant, say k. 243

Solving Variables Separable
Differential Equations (continued)

0 Here’s the promised ͐ dy = ͐ 3 2x 2 dx
alternate solution, y +x
beginning with Step 5.
ln y = ln 3 + x2 + c

ln y = ln _3 + x2i + c Note that 3 + x2 ais always positive.

! If m = r, then em = er must also be true. So in our case, e = eln | y | ln (3+ x2 )+ c
eleft side of equation = eright side of equation.
$e = e eln | y |
ln (3+ x2 ) c

0

$e = e mln | y |
ln (3+ x2 )

$e = m eln | y | ln (3+ x2 )

@ Using the natural log property eln x = x, simplify y = m(3 + x2) or just y = m(3 + x2)
both sides of the equation.

Particular Solution

The general solution of a differential equation will always have some constant in the solution. If you are
given additional information, often called an initial condition, you will be able to find a particular value
of the constant, and thus a particular solution to the differential equation.

Given the initial condition of y(1) = 3, find the particular solution of the differential equation yy' – 3x = 0.

1 Replace the y' with dy. y dy - 3x = 0
dx dx

244

11Techniques of Integration chapter

2 Add 3x to both sides. y dy = 3x
dx

3 Multiply both sides by dx to separate your variables. ydy = 3xdx
4 Integrate both sides.
͐y dy = ͐3x dx

͐y dy = 3 $ ͐x dx

$y2=3 x2 + c
2
2

5 Multiply both sides by 2 and simplify the new constant on the y2= 3x2+ 2c
right side. 0

y2= 3x2+ m

You just found the
general solution.

6 Using the initial condition that y(1) = 3 (that is, y2 = 3x2 + m
when x = 1, then y = 3), you can find a particular (3)2 = 3(1)2 + m
value of the constant m and thus a particular –6 = m
solution to the differential equation.
Therefore, the particular solution is
y2 = 3x2 – 6.

245

12chapter

Applications of
Integration

The chapter opens with integration of functions Acceleration, Velocity, and
related to the motion of an object and then Position . . . . . . . . . . . . . . . . . . . . . . . . 247
moves on to finding the area of a region bounded
by the graphs of two or more functions using an Area between Curves:
appropriate integral. Using Integration . . . . . . . . . . . . . . . . 250

Revolving a bounded region about a given Volume of Solid of Revolution:
vertical or horizontal line produces a solid the Disk Method . . . . . . . . . . . . . . . . . . . . 260
volume of which you will be able to compute using
an appropriate integral. Three methods for doing Volume of Solid of Revolution:
this are introduced: disk, washer, and shell. Washer Method . . . . . . . . . . . . . . . . . 268

Volume of Solid of Revolution:
Shell Method. . . . . . . . . . . . . . . . . . . . 275

Acceleration, Velocity,
12and Position
Applications of Integration chapter

In Chapters 3–6 on derivatives, you found a If acceleration function is a(t), then
way to move from the position to the velocity
and then on to the acceleration function by velocity function is v ^th = ͐a^ th dt, and
differentiating. Now you will reverse the position function is s ^th = ͐v _t i dt.
process, going from acceleration back to the
position function by integrating.

MOTION PROBLEM: ACCELERATION TO VELOCITY

The acceleration function, a(t), for an object is given by a(t) = 36t – 168t + 120. If v(1) = 28, find the
velocity function, v(t).

1 Beginning with the acceleration function, a ^th = 36t 2- 168t + 120
find its integral to get the velocity function.
v ^th = ͐a ^th dt = ͐ _36t2- 168t + 120i dt

36t 3 168t 2
3 2
v ^th = - + 120t + c

v ^th = 12t3- 84t 2+ 120t + c

2 Using the given data that v(1) = 28, substitute 1 v ^1h = 12 ^1h3- 84 ^1h2+ 120 ^1h + c
for t and 28 for v(1) to solve for the constant c. 0
28 = 12 - 84 + 120 + c
3 Replace the c with –20 in the velocity function.
- 20 = c

v ^th = 12t3- 84t 2+ 120t + c
0

v ^th = 12t3- 84t 2+ 120t - 20

247

Acceleration, Velocity,
and Position (continued)

MOTION PROBLEM: VELOCITY TO POSITION
If s(1) = 38 and v(t) = 12t – 84t + 120t + 20, find the position function, s(t).

1 Integrate v(t) to get s(t). s ^th = ͐v ^th dt = ͐ _12t3- 84t2+ 120t + 20i dt

12t 4 84t 3 120t 2
4 3 2
s ^th = - + + 20t + k

s ^th = 3t 4- 28t3+ 60t 2+ k

Note: Since c is used for the constant in the
problem above, k is used here so that there is
no confusion about which constant goes with
which problem.

2 Use the data s(1) = 38 to make appropriate s ^1h = 3^1h4- 28^1h3+ 60 ^1h2+ k
substitutions and then solve for k. 0
38 = 3 - 28 + 60 + k
3 Replace the k with 3 in the velocity function. 3=k

s ^th = 3t 4- 28t3+ 60t 2+ k
0

s ^th = 3t 4- 28t3+ 60t 2+ 3

248

12Applications of Integration chapter

Motion Problem: Acceleration to Position

If the acceleration function of an object is given by a(t) = sin t + cos t and v(π) = 2, while s(π) = 1, find
the position function s(t).

1 Find the velocity function v(t) by integrating v ^th = ͐a ^th dt = ͐ ^sin t + cos th dt
the acceleration function a(t), and then find the
position function s(t) by integrating the velocity v ^th = -cos t + sin t + c
function v(t).

2 Use the data v(π) = 2 to find the value of c. v ^πh = -cos α+ sin π + c
0
2 = -^-1h + 0 + c
1=c

therefore v ^th = -cos t + sin t + 1

3 Find the position function s(t). s ^th = ͐v ^th dt = ͐ ^-cos t + sin t + 1h dt

s ^th = -sin t cos t t + k

4 You are given additional data, s(π) = 1, s ^πh = -sin π - cos π + π + k
which will allow you to find the value 0
1 = -^0h -^-1h + π + k
of k.
-π=k
thus s ^th = - sin t - cot t + t - π

249

Area between Curves:
Using Integration

There are two scenarios to consider when finding the area of the region bounded by the graphs of
two or more equations: Either the graphs of the equations do not intersect or the graphs of the
equations intersect at one or more points. You will use integration to compute the area of the
bounded region.

Scenario 1: The Graphs of the Equations Do Not Intersect

1 The region, the area of which you are y
trying to compute, is bounded by the
graphs of two functions that do not y1 = f(x)
intersect y1= f ^xh and y2= g ^xh and
the graphs of two vertical or horizontal h = y1−y2
lines (x = a and x = b). x

a b

dx y2 = g(x)

2 Notice the thin green “representative” area of “rep.” rect. = height • base = (y1 – y2)dx

rectangle. Its base is a small change in
x, typically labeled as ∆x, or just dx. Its
height is the difference in y coordinates

for the two functions—in this case y1 – y2.

3 If you were to add the areas of an infinite number of very thin Area = ͐b _ y1- y2i dx
such rectangles, you could find their sum by using just an a
integral.

250

12Applications of Integration chapter

4 If you substitute y1 = f(x) and y2 = g(x), you end up with Area = ͐b _ f ^ xh - g ^ xhi dx
an integral representing the area of the region bounded a
by the graphs of the two given functions and the two
given vertical lines. Notice that f(x) is the top function
and g(x) is the bottom function in
the diagram shown in Step 1. (See
p. 250.)

Scenario 2: The Graphs of the Two Functions Intersect One or More Times

1 The region, the area of which you y h = y2−y1
are trying to compute, is bounded h = y1−y2 y1 = f(x)
by the graphs of two functions
y1= f ^xh and y2= g ^xh, which y2 = g(x)
intersect at points where x = a, x = b,
and x = c. dx
x
dx
a bc

2 Each green shaded region has a thin green area of left rep. rect. = (y1 – y2)dx
rectangle with base dx, but they have different area of right rep. rect. = (y2 – y1)dx
heights. The one on the left has a height of y1 – y2,
and the one on the right has a height of y2 – y1; 251
notice that in either case, the height is just the top
function minus the bottom function.

Area between Curves:
Using Integration (continued)

3 If you were to add the areas of total area = area of left region + area of right region
an infinite number of very thin
rectangles in each region, their ͐ ͐b c
sum could be found by the sum
of two integrals. = a _ y1- y2i dx + b _ y2- y1i dx

4 Make the substitutions total area = ͐b _ f ^xh- g ^ xhi dx + ͐c _g^xh- f ^ x hi dx
y1 = f(x) and y2 = g(x) to a b
get the area in terms of the
given functions and the x
coordinates of the points of
intersection of their graphs.

EXAMPLE 1 y

Find the area of the region bounded by the
graphs of y = x, y = 0, x = 1, and x = 4.

y1 = x2

1 Find the area of the green representative rectangle. h = y1−y2
252 x

1 4 y2 = 0
dx

area rep. rect. = hb = (y1 – y2)dx

12Applications of Integration chapter

2 If you find the sum of lots of these rectangles, you get an integral A = ͐4 _ y1- y xi dx
that represents the area of the green shaded region. 1

3 Since the term at the end of the integral is dx, the integrand can A = ͐f _x2- 0i dx
contain x variables and/or numbers. Substitute y1 = x2 and y2 = 0 1
and then simplify.
A = ͐4 x2 dx
1

4 Evaluate the integral to get the area of the green shaded A = < x3 4 64 - 1 = 63
region. 3 3 3 3
F=

1

A = 21

EXAMPLE 2 3x – x2 = 0
x(3 – x) = 0
Find the area of the region bounded by the graphs of y = 3x – x2 and y = 0. x = 0, x = 3

1 You have not been given the limits of integration, so you will have to find
them by setting the equations equal to one another and then solving for x.

253

Area between Curves:
Using Integration (continued)

2 Sketch a graph showing the bounded y
region and a representative rectangle
indicating its base and its height.

0 h = y1−y2

3 Write the formula to find the area of the representative rectangle. y2 = 0 x
3
4 Set up the integral that represents the sum of lots of these dx y1 = 3x−x2
rectangles, and thus the area of the green shaded region.
area rep. rect. = hb
5 Substitute the appropriate x or number equivalents for the = (y1 – y2)dx
terms y1 and y2 and simplify (since there is a dx term at
the end, you need all x or numerical terms in the integrand) A = ͐3 _ y1- y2i dx
y1 = 3x – x2 and y2 = 0. 0

6 Evaluate the integral, finding the area of the desired A = ͐3 a_3x - x2i - 0k dx
bounded region. 0

A = ͐3 _3x - x 2i dx
0

A = < 3x 2 - x3 3 27 - 9
2 3 2
F=

0

A = 9
2

254

12Applications of Integration chapter

EXAMPLE 3 x3 – 9x = 0
Find the area of the region bounded by the graphs of y = x3 – 9x and y = 0. x(x2 – 9x) = 0
x(x + 3)(x – 3) = 0
1 Find the x coordinates of any points of intersection of the x = 0, x = –3, x = 3
graphs of the two functions. This will also aid in sketching
the graph of the first function.

Note: These will serve as
your limits of integration.

2 Sketch the graphs of the two y
functions, indicating the bounded
regions whose area you are going to h = y1−y2
compute. Also, show a representative
rectangle for each region, along with dx x1 = x3−9x
its base and appropriate height. 0 x

-3 3
dx y2 = 0

h = y2−y1

3 Find the area of the representative rectangle for area of left rep. rect. = (y1 – y2)dx
each region and then find their sum. area of right rep. rect. = (y2 – y1)dx

4 Set the sum of the two integrals: one to total area = ͐0 _ y1- y2i dx + ͐3 _y2- y1i dx
find the area of the green shaded region -3 0
from x = –3 to x = 0, and the other from
x = 0 to x to 3. Notice that for each integrand, it’s the top
function minus the bottom function within
each interval of integration.

255

Area between Curves:
Using Integration (continued)

5 Substitute y1 = x3 – 9x and y2 = 0 and then simplify each integrand.

Now the integrands will total area = ͐0 9xi - 0k dx + ͐3 a0 -_x3- 9xik dx
match variables with the dx 0
term at the end of each -3a_ x3-
integral. All expressions
will be in terms of the = ͐0 x3- 9x i dx + ͐3 x 3+ 9xi dx
variable x.
-3_ 0_-

6 Evaluate each integral and total area = ͐0 _x3- 9xi dx ͐3 _- x 3+ 9xi dx
find their sum. This will be -3
the sum of the areas of the +0
two green shaded regions.
x4 9x 2 0 x4 9x 2 3
4 2 4 2
= < - F + <- + F

-3 0

= < 0 - b 81 - 821lF + <b- 81 + 821l - 0F
4 4

total area = 81
2

REPRESENTATIVE RECTANGLE IS HORIZONTAL

Occasionally you encounter a situation in which you have to draw the representative rectangle
horizontally rather than vertically.

1 The region, the area of which you are y x2 = g(y)
trying to determine, is bounded by the b x1 = f(y)
graphs of x1 = f(y) and x2 = g(y), which
intersect at the points with y dy
coordinates y = a and y = b.

Note: The height of the representative
rectangle is just the right function minus
the left function for the shaded region.

h = x2−x1

x

a

256

12Applications of Integration chapter

2 The green representative rectangle has a base area “rep.” rect. = (x2 – x1)dy
of dy and a height of x2 – x1. Its area is found
at right. (Notice that the representative
rectangle has a horizontal orientation rather
than the usual vertical orientation. If you try to
make the rectangle vertical, at the left end of
the bounded region, the height would be x1 – x1 = 0.)

3 If you add up an infinite number of these very thin Area = ͐b _ x 2- x1i dy
rectangles, their sum can be found by the integral at right. a

4 Substitute x1 = f(y) and x2 = g(y) so that the variables Area = ͐b ag _ yi- f _yik dy
within the integrand match the dy term at the end of the a
integral.

EXAMPLE 4 y2 = y + 2
Find the area of the region bounded by the graphs of x = y2 and x = y + 2. y2 – y – 2 = 0
(y + 1)(y – 2) = 0
1 Find the x coordinates of any points of intersection; these will y = –1, y = 2
also serve as your limits of integration.

257

Area between Curves:
Using Integration (continued)

2 Sketch the graphs of the two functions, y x2 = y + 2
indicating the bounded region the area 2 x1 = y2
of which you are going to compute.
Also show a representative rectangle dy
for the region, along with its base and h = x2−x1
appropriate height.
x
Note: Notice that the height of the
green horizontal rectangle is the right
function minus the left function.

-1

3 Write the area of the green representative rectangle. area “rep.” rect. = (x2 – x1)dy

4 Set up the integral to find the area of the green shaded region. ͐2

A = -1 _ x2- x1i dy

258

12Applications of Integration chapter

5 Since you have a dy term at the end of the integral, you A = ͐2 a_ y + 2i - y2k dy
need to make the substitutions x1 = y2 and x2 = y + 2, so -1
that the integrands contain just y terms and numerical
A = ͐2 _ y + 2 - y2i dy
values. -1

6 Evaluate the integral and simplify the resulting A = = y2 + 2y - y3 2
computation to find the area of the green shaded 2 3
region. G

-1

A = b2 + 4 - 83 l - b 1 - 2 + 1 l
2 3

A = 9
2

259

Volume of Solid of Revolution:
Disk Method

The region bounded by the graphs of y = f(x), y = 0, x = a, and x = b is revolved about the x-axis.
Find a formula for computing the volume of the resulting solid.

1 At right is a figure showing the bounded y
region with three thin red rectangles.

y = f(x)

2 After the bounded region is revolved (or a x
rotated) about the x-axis, it creates a solid y b y=0
as shown at right. Notice that each thin
red rectangle traces out a thin red disk a r = y = f(x)
(or cylinder). y = f(x)

bx
y=0

h = dx
260

12Applications of Integration chapter

3 The “representative” disk toward the center of the solid has a Vcyl.= πr 2 h
height of dx and a radius of r = y = f(x). Using the formula for the Vdisk = πy2 dx
volume of a cylinder with radius r and height h, you end up with a or with all x terms
formula for the volume of the representative disk.
2

Vdisk= π _ f ^xhi dx

4 If you were to find the sum of the volumes of an infinite number $V = ͐b π y2 dx
of very thin disks, an integral could be used to do that a
computation.
or

$V = ͐b π 2
a
_ f ^xhi dx

5 When doing a specific problem, it is not y
necessary to try to sketch the three-
dimensional figure. A suggested sketch is r = y = f(x)
shown at right. y = f(x)

TIP a bx
y=0
The thickness of the representative disk is
really the height of this thin cylinder, in this
case h = dx. The radius of the disk is
always measured from the axis of
revolution to the graph being rotated.

h = dx

261

Volume of Solid of Revolution: Vdisk = πr 2 h
Disk Method (continued) = πy2 dx

6 The appropriate work would be shown as demonstrated to the Vdisk = πy2 dx,
right. then volume of solid is

TIP $V = ͐b π y2 dx,
a
Make sure that all variables match (x and dx or y and dy)
before attempting to integrate. or to get all x terms,

V= ͐b 2
a
π _ f ^xhi dx

EXAMPLE 1

The region bounded by the graphs of y = x2, y = 0, x = 1, and x = 2 is revolved about the x-axis. Find the
volume of the resulting solid.

1 Sketch a side view of the revolved y y = x2
region. Label the height (looks like the
thickness from the side) and the radius
of the representative disk.

r=y

a x
b y=0

h = dx

262

2 Find the volume of a representative disk. 12Applications of Integration chapter

Vdisk = πr2h
= πy2dx

3 Now you are ready to set up the volume integral. V = ͐2 πy2 dx
1
Notice that the y terms were substituted with appropriate x terms and
the constant π was moved out in front of the integral. ͐V = 2 π _ x 2 2 dx
1
i

V = ͐2 x 4 dx

π1

4 Evaluate the integral and plug in the limits of integration to V = π < x5 2 π b 32 - 1 l
compute the volume of the solid that results. 5 5 5
F=

1

V = 31π
5

EXAMPLE 2

The region bounded by the graphs of y = x2, y = 2, and x = 0 is rotated about the y-axis. Find the volume
of the solid that results.

1 Sketch a side view of the figure showing y
the height and radius of the y = x2
representative disk.
2 y=2

r=x
h = dy

x
0

263

Volume of Solid of Revolution: Vdisk = πr2h
Disk Method (continued) = πx2dy

2 Find the volume of the representative red disk. V = ͐2 πx2 dy
0
3 Set up the appropriate volume integral, noting that the dy at the end
means that the integrand must eventually be in y terms also.

V = ͐2 πydy
0

V = ͐2 ydy

π0

4 Integrate and substitute the limits to find the volume of the V= π R y2 V2 = π b 4 - 0 l
solid. S 2 W 2 2
SS WW0
TX
V = 2π

EXAMPLE 3 1
x
The region bounded by the graphs of y = , y = 0, and x = 2 is revolved about the x-axis. Find the

volume of the resulting solid.

1 Sketch a side view of the solid. Label y
the radius and height of the
representative disk. Note that there is
no right hand limit of integration, so in
this problem it will be +ϱ.

h=y

y = 1
x
to + ∞x
y=0
2
base = dx

264

2 Find the volume of the red disk. 12Applications of Integration chapter

Vdisk = πr2h
= πy2dx

3 Set up your volume integral and then substitute to get all x terms. V = ͐+3 πy2 dx
2

V = ͐+ 3 π b 1 2 dx
2 x
l

V = ͐+ 3 b 1 2 dx
x
π l
2

͐+ 3

V = π 2 x-2 dx

4 Integrate and then plug in the limits to get the volume 1 + 3 1 1
of the solid. x 3 2
V = π ;- E = π d- - b- ln
It’s interesting that the area of the region is finite,
even though its right hand limit of integration is 2
infinite!
V = π
2

265

Volume of Solid of Revolution:
Disk Method (continued)

EXAMPLE 4
(axis of revolution not x or y axis)

The region bounded by the graphs of y = x2, y = 0, and x = 1 is rotated about the line x = 1. Find the
volume of the solid that results.

1 Sketch a side view of the solid, noting x=1
the height and radius of the horizontal y
representative red disk. Notice that the
left hand limit of integration is just x =
0.

y = x2

TIP x h = dy
01 2
Be sure to measure the radius from the y=0
axis of revolution back to the original r = 1–x
curve.

1

2 Find the volume of the representative red disk. Vdixk = πr2h
3 Set up the integral used to find the volume of the solid. = π(1 – x)2dy

V = ͐1 π ^1 - xh2 dy
0

V = ͐1 ^1 - xh2 dy

π0

V = ͐1 _1 - 2x + x2i dy

π0

266

12Applications of Integration chapter

4 Since there is a dy at the end of the integral, all terms before V = π͐1 _1 - 2y1/2 + yi dy
that must in terms of y also. With 0
y = x2 , you also have y1/2= x.

5 Last, integrate and then plug in the limits. 4 y2 1
3 2
V = π = y - y 3/2 + G

0

= π <b1 - 4 + 1 l - 0F
3 2

V = π
6

267

Volume of Solid of Revolution:
Washer Method

The region bounded by the graphs of y1 = f(x), y2 = g(x), the x-axis(y = 0), x = a, and x = b is
revolved about the x-axis. Find the volume of the resulting solid.

1 Sketch the bounded region. y
y = f(x)

y = g(x)

x
ab

2 Try to envision the three-dimensional y x
solid with a “hole” through it, which a b
results from rotating the bounded
region about the x-axis. It sort of looks
like a candle holder on its side. Notice
the thin red “washer” with the hole in
it.

268

12Applications of Integration chapter

3 Let’s look at just the thin red washer for a moment. It has an
inner radius of r and outer radius of R, with a height (or
thickness) of h.

Derive a formula for the volume of the thin red washer.

TIP rR

R is the radius of the disk and r is the h
radius of the hole.
Vwasher = Vdisk – Vhole
4 In the process of doing a problem with a “hole,” = πR2h – πr2h
it is not necessary to try to sketch the three-
dimensional version. Just sketch a side view Vwasher = π(R2 – r2)h
and label the big R, the little r, and the height h.
y h = dx
TIP y1 = f(x)

Both the large radius R and the small radius r must R = y1 y2 = g(x)
be measured from the axis of revolution; R from the a
axis to the outer curve and r from the axis to the r = y2
inner curve. b

269

Volume of Solid of Revolution:
Washer Method (continued)

5 Substitute the appropriate pieces into the volume Vwasher = π _R2- r 2i h
of the washer formula.
22

V =washer π b_ y1i - _ y2i l dx

6 If you were to add the volumes of an infinite number ͐b 2 2
of thin red washers, you could compute that sum by
using an integral. V = a π b_y1i - _y2i l dx

7 With the dx term at the end of the integral, you need V = ͐b π c_ f 22
to make sure that the integrand has only numerical or a
x terms. ^xhi - _g^xhi m dx

EXAMPLE 1

The region bounded by the graphs of y = x and y = x3 in the first quadrant is rotated about the x-axis.
Find the volume of the solid that results.

1 Sketch a side view of the solid; draw in a y y1 = x3
representative red washer and label its big and
small radius as well as its height.

h = dx y2 = x

R = y2 r = y1 x
0 1

270

12Applications of Integration chapter

2 Write the formula for the thin red washer’s volume. V =washer π _R2- r 2i h

22

V =washer π b_ y2i - _ y1i l dx

3 Write the appropriate integral to compute the volume of the ͐1 2 2
solid.
V = 0 π b_y2i - _y1i l dx

͐1 2 2
V = π 0 b_y2i - _y1i l dx

4 Replace each y term with its corresponding x term V = ͐1 2 _x 3i l dx
equivalent.
π0 b _xi -

V = ͐1 _ x2- x6i dx

π0

5 Integrate, evaluate, and simplify to get the volume of the V = π < x3 - x7 1
desired solid. 3 7
F

0

= π <b 1 - 1 l - 0F
3 7

V = 4π
21

271

Volume of Solid of Revolution:
Washer Method (continued)

EXAMPLE 2

The region bounded by the graphs of y = x2, the x-axis, and x = 2 is revolved about the y-axis. Find the
volume of the resulting solid.

1 Sketch a side view of the solid; draw y
in a representative red washer, and y = x2
label its big and small radius as well as
its height.

TIP R=2
4
The large raduis R is fixed at 2. Only
the small radius r is changing. r=x
h = dy
x

-2 0 2

2 Create a formula for the volume of this particular thin red Vwasher = π(R2 – r2)h
washer. Vwasher = π(22 – x2)dy
Vwasher = π(4 – x2)dy
3 Set up the volume integral that enables you to compute the
volume of the resulting solid. V = ͐4 π_4 - x2i dy
0
4 As usual, make sure that the variable within the integrand matches
the dy at the end of the integral. V = π͐ 4 _4 - x2i dy
0

V = π͐ 4 _4 - yi dy
0

V = π͐ 4 _4 - yi dy
0

272

12Applications of Integration chapter

5 Integrate and evaluate the result using the limits of integration. V = π[4y – y2]0
= π[(16 – 8) – 0]

V = 8π

EXAMPLE 3

The region bounded by the graphs of y = x2, y = 0, and x = 1 is rotated about the line x = 3. Determine
the area of the resulting solid.

1 Sketch a side view of the figure; label y
both large and small radii, as well as
the height of the thin red washer.

TIP y = x2 hx= dy
56
Measure large radius R from the axis x r=2
of revolution (x = 3) to the outer curve 13
(y = x2). The small radius r is measured
from the axis of revolution (x = 3) to
the inner curve (x = 1).

R = 3–x

2 Write the formula for the thin red washer’s volume, Vwasher = π _R2- r 2i h
substituting for appropriate parts labeled on your figure. Vwasher = π a^3 - xh2- 22k dy

273

Volume of Solid of Revolution: V = ͐1 π a^3 - xh2- 22k dy
Washer Method (continued) 0

3 Write the integral for calculating the volume of the V = π͐1 a^3 - xh2- 22k dy
resulting solid. 0

4 Using y = x2 so that y1/2= x , substitute so that all V = ͐1 _9 - 6x + x2- 4i dy
variables within the integrand match the dy term at the
end of the integral. π0

5 Now you’re ready to integrate and evaluate using the V = π͐1 _5 - 6x + x2i dy
limits to determine the volume of the solid. 0

V = π͐10_5 - 6y1/2+ yi dy

V = π =5y - 4y 3/2 + y2 1
2
G

0

= π <b5 - 4 + 1 l - 0 F
2

V = 3π
2

274

Volume of Solid of Revolution: 12chapter
Shell Method

The region bounded by the graphs of y = f(x), the x-axis, x = a, and x = b is revolved about the
y-axis. Find an expression that represents the volume of the resulting solid.

1 To the right is a diagram of the y
bounded region before being revolved y = f(x)
about the y-axis.

x
ab

2 After being revolved about the y-axis, y
the figure shown at right results. y = f(x)

a x
b

275

Volume of Solid of Revolution: y
Shell Method (continued) y = f(x)

3 In the new technique, named the “shell a x
method,” you find the volume of the b
resulting solid by finding the sum of
an infinite number of very thin “shells” R
(or pieces of pipe). r1

4 Let’s take a look at just one of those many “shells” (or pieces h
of pipe). The shell has an outer radius, r2; an inner radius, r1;
and a height of just h. One other dimension, labeled R, is the r2
distance from the axis of revolution to the center of the thin red t
shell.
Vshell = π(r2)2h – π(r1)2h
5 Find a formula for the volume of that thin red shell.
This is just the volume
of the piece of pipe
without the hole, minus
the volume of the hole.

276

12Applications of Integration chapter

6 Play with the formula 22
a bit to rewrite it in
another form. Vshell= π _r2i h- π _r1i h

Note: Big R is really 22
just the average
radius. It is measured Vshell= π b_r2i - _r1i l h
from the axis of
revolution to the Vshell= π _r2+ r1i_r2- r1i h Factored the middle term above.
center of the shell. --
Vshell = 2π d r 2 + r1 n_r2- r1i h Mult. and then div. by 2.
2

Vshell = 2π d r 2 + r1 n h _r2- r1i Moved the h to the left one term.
2
-

Vshell = 2π . ..
R ht
-

ave. radius height shell’s thickness
Vshell= 2πRht

7 In the process of doing an actual y
problem, the figure you sketch will y = f(x)
look more like the one at right. R is the
radius from the axis of revolution to the
center of the shell, and t is the
thickness, dx or dy.

ab h=y
t = dx x

R=x

277

Volume of Solid of Revolution: Vshell = 2πRht
Shell Method (continued) Vshell = 2πxydx

8 Using the formula Vshell = 2πRht, substitute the appropriate pieces
labeled on your diagram to get a formula for the volume of the thin
red shell.

9 As in other volume techniques, if you were to add the volumes of an V = ͐b 2πRhtdx
infinite number of very thin red shells, the volume of the resulting a
solid could be determined by using an integral.

EXAMPLE 1

The region bounded by the graphs of y = x3, y = 0, and x = 1 is revolved about the y-axis. Find the
volume of the resulting solid.

1 Sketch a side view of the solid, with y
the two thin red rectangles actually y = x3
representing the side view of a shell.
Label the average radius R, the height
h, and the thickness t.

t = dx
h=y
x

-1 1 y = 0
R=x

2 Plug the pieces labeled on your diagram into the formula for the Vshell = 2πRht
volume of a generic shell. Vshell = 2πxydx

278

12Applications of Integration chapter

3 Set up the integral to find the volume of the solid, V = ͐1 2πxydx
substituting to get the integrand in terms of the same 0
variable as the dx at the end.
V = ͐1 xydx

2π 0

$V = 2π͐1 x x3 dx
0

V = ͐1 x 4 dx

2π 0

4 Integrate and evaluate the result. V = 2π < x5 1 = 2π b 1 - 0l
5 5
This problem could also have been done using the washer F
method.
0

V = 2π
5

TIP

When using either the disk or the washer method, the
“representative” disk or washer is drawn perpendicular to the
axis of revolution. In the shell method, the “representative”
shell is always drawn parallel to the axis of revolution. If you
have a choice between using the washer or the shell method,
it is usually easier to set up the shell method — you need to
find only one radius.

279

Volume of Solid of Revolution:
Shell Method (continued)

EXAMPLE 2

The region bounded by the graphs of y = x2, y = 0, and x = 2 is revolved about the x-axis. Find the
volume of the resulting solid.

1 Sketch a side view of the solid, y y = x2
showing a representative shell.

4
h = 2–x

x R=y t = dy
0 x

2 y=0

2 Plug the pieces labeled on your diagram into the formula for the Vshell = 2πRht
volume of a generic shell. Vshell = 2πy(2 – x)dy

3 Set up the integral to find the volume V= ͐4 2πy ^2- xh dy Notice the y lim . of int .
of the solid, substituting to get the 0
integrand in terms of the same
variable as the dy at the end. V= ͐4 y ^2- xh dy

2π 0

V= 2π͐ 4 y _2- y1/2i dy
0

V= 2π͐ 4 _2y- y3/2i dy
0

280

12Applications of Integration chapter

4 Integrate and then find the value of the resulting expression. V = 2π ; y2- 2 y 5/2 4
5
E

0

= 2π <b16 - 2 $ 32l - 0F
5

V = 32π
5

This problem could also have been done using the disk method.

EXAMPLE 3

The region bounded by the graphs of y = x2 and y = –x2 + 2x is revolved about the line x = 3. Find the
volume of the resulting solid.

1 Find the x-coordinates of the points of intersection of the graphs of the x2 = –x2 + 2x
two equations. 0 = –2x2 + 2x
0 = –2x(x – 1)
x=0x=1

2 Sketch a side view of the solid and y
label the appropriate pieces, both the y2 = -x2 + 2x
radii and the height.

h = y2–y1

TIP R = 3–x

The radius R must be measured from 0 t = dx 1 3 5 x
the axis of revolution (x = 3) to the y1 = x2 6
middle of the shell.

281

Volume of Solid of Revolution:
Shell Method (continued)

3 Find the volume of the thin red shell. Vshell = 2πRht
Vshell = 2π(3 – x)(y2 – yi)dx

4 Set up the integral to determine the volume of the V = ͐1 2π ^3 - x h_ y 2 - y1i dx
solid and then evaluate using the limits of 0
integration.
V = ͐1 ^3 - x h_ y 2 - y1i dx

2π 0

V = ͐1 ^3 - x h_- x 2 + 2x - x2i dx

2π 0

V = 2π͐1 ^3 - x h_- 2x 2 + 2xi dx
0

V = ͐1 _2x 3 - 8x 2 + 6xi dx

2π 0

5 Integrate and then evaluate using the limits of x4 8x 3 1
integration. 2 3
V = 2π < - + 3x 2F

0

= 2π <b 1 - 8 + 3l - 0F
2 3

V = 5π
3

282

Appendix

Common Differentiation Rules

In the following, c is a constant and a is a constant. In cases where u appears, u is some function of
another variable and the du is just the derivative of u.

GENERAL FORMS $ $6
d = f ^xh G = f l^xh g ^xh - gl^xh f ^xh
dx g ^xh
1 d ^ch = 0 2
dx
8 g ^ xhB

2 (Quotient Rule)

d ^ xh = 1 7 d
dx dx
9 f _g^xhiC = f l_g^xhi $ gl^xh

3 d 8c $ f ^xhB = c $ f l^xh (Chain Rule)
dx

4 d 8 f ^xh ! g^xhB = f l^xh ! gl^xh $8 d _ x n i = n x n- 1
dx dx

(Sum/Difference Rule) (Simple Power Rule)

5 d $ $9d ni
dx 8 f ^xh $ g^xhB = f l^xh$ g^xh+ gl^xh $ f ^xh dx _u = n un- 1 du

(Product Rule) (General Power Rule)

EXPONENTIAL FORMS LOGARITHMIC FORMS

0 d eu= eu du bin particular, d e x = exl @ d ^ln uh = du bin particular, d ^ln xh = 1 l
dx dx dx u dx x

! d a u$ ln a $du $# d = 1 du
dx au= dx _log ui ln a u
a

TRIGONOMETRIC FORMS

$ d ^sin uh = cos u du & d ^csc uh = - csc u cot u du
dx dx

% d ^cos uh = - sin u du * d ^sec uh = sec u tan u du
dx dx

^ d ^tan uh = sec2 u du ( d ^cot uh = - csc2 u du
dx dx

283

INVERSE TRIGONOMETRIC FORMS

) d ^arcsin uh = du e d ^arc csc uh = -du
dx 1 - u2 dx u2- 1
u

q d ^arccos uh = - du r d ^arc sec uh = du
dx 1 - u2 dx u2- 1
u

w d ^arctan uh = du t d ^arc cot uh = -du
dx 1 + u2 dx 1+u
2

Common Integration Formulas

In the following, k is a constant, c is a constant. In cases where u appears, u is some function of another
variable and the du is just the derivative of u.

GENERAL FORMS 4 ͐xn dx = xn+ 1 + c ^n ! -1h
n+
1 ͐dx = x + c 1
2 ͐kdx = kx + c
3 ͐ 8 f ^xh ! g^xhB dx = ͐f ^xh dx ! ͐g^xh dx 5 ͐un du = un+ 1 + c ^n ! -1h
n+
1

LOGARITHMIC FORMS EXPONENTIAL FORMS

6 ͐ du = ln u + c 7 ͐eu du = eu+ c
u
In particular ͐ex dx = ex+ c
In particular ͐ 1 dx = ln x +c
x

284

TRIGONOMETRIC FORMS # ͐ cot u du = ln sin u + c or - ln csc u + c
$ ͐ sec2 u du = tan u + c
8 ͐ sin u du = - cos u + c % ͐ csc u cot u du = - csc u + c
9 ͐ cos u du = sin u + c ^ ͐ sec u tan udu = sec u + c
0 ͐ tan u du = -ln cos u + c or ln sec u + c & ͐ csc2 u du = - cot u + c
! ͐ csc u du = ln csc u - cot u + c
@ ͐ sec u du = ln sec u + tan u + c

INVERSE TRIGONOMETRIC FORMS

∫* du = arcsin u + c ∫) du = 1 u +c
arc sec
a2 − u2 a u u2 + a2 a a

∫( du = 1 arctan u + c
a2 + u2 a a

Unit Circle and Some Common Trigonometric Identities

y

(– 1 , √3 ) (0, 1) ( 1 , √3 )
2 2 90° 2 2
π
(–√22 , √2 ) 2π 2 π (√22 √2
2 3 120° 60° 3 π 2
, )
3π 4
√3 1 45°
(– 2 , 2 ) 4 π ( √3 , 1 )
5π 135° 2 2

65
150° 30°

(–1, 0) 180° 0° 0
π x

360° 2π (1, 0)

210° 330°
11π
7π
315° 6
√3 6 225° (√23 1
(– 2 , – 1 ) 5π 7π ,– 2 )
2 4 4
240° 300°
√2 √2 4π 5π (√22 , – √2 )
2 2 2
(– , – ) 3 3π 3
270° 2
1 √3
(– 1 , – √3 ) ( 2 , – 2 )
2 2
(0, –1)

285

PYTHAGOREAN IDENTITIES tan2x + 1 = sec2x 1 + cot2x = csc2x
cos2x + sin2x = 1

SUM AND DIFFERENCE IDENTITIES ± sin(x ± y) = sin x cos y ± cos x sin y
cos(x ± y) = cos x cos y sin x sin y

DOUBLE-ANGLE IDENTITIES

sin2x = 2sin x cos x cos2x = cos2x – sin2x cos2x = 2cos2 x – 1 cos2x = 1 – 2sin2x

HALF-ANGLE IDENTITIES

cos b x l = ! 1 ^1 + cos xh sin b x l ! 1 ^1 - cos xh
2 2 2 2

286