In this chapter we explore some of the applications of the definite integral by using it for
1. Computing the area between curves
2. Computing the volumes of solids
3. Computing the work done by a varying force
4. Computing average value of a function
The common theme is the following general method, which is similar to the one we used to find areas under curves:
We break up a Q quantity into a large number of small parts. We next approximate each small part by a quantity of the form and thus approximate Q by a Riemann sum. Then we take the limit and express Q as an integral
Finally we evaluate the integral using the Fundamental
Theorem of Calculus or the Midpoint Rule.

• http://www.math.tamu.edu/AppliedCalc/Cl asses/Riemann/Diff.html

x
2
 x
2 
2 x

x x
3

y
2
  x y
1 x
2
How can we find the area between these two curves?
We could split the area into several sections, use subtraction and figure it out, but there is an easier way.


Consider a very thin vertical strip.
y
2
  x y
1 x
2
The length of the strip is: y
1
 y
2 or

2
 x 2
  
Since the width of the strip is a very small change in x
, we could call it dx
.


y
1 y
1 x
2 y
1
 y
2 y
2 dx y
2
  x
Since the strip is a long thin rectangle, the area of the strip is:
 

2
 x
2 

If we add all the strips, we get:


1
2
2
 x
2  x dx


y
2
  x y
1 x
2


1
2
2
 x
2  x dx
2 x

1 x
3 
1
3 2 x
2
2

1
4
8 1 1
2 2
3 3 2
6
3
2
1 1
3 2
6

27
6

9
2


The formula for the area between curves is:
Area
  a b
1
   f
2
  dx
We will use this so much, that you won’t need to “memorize” the formula!


Some regions are best treated by regarding x as a function of y

dx y
 x dx y x 2
If we try vertical strips, we have to integrate in two parts:

0
2 x dx
 
2
4 x

2
 dx y
 x y
 x y
2  x dy y y x 2 x y x 2
We can find the same area using a horizontal strip.
Since the width of the strip is find the length of the strip by solving for x in terms of y
.
dy
, we


y
 x
We can find the same area using a horizontal strip.
y
 x y
2  x dy y y x 2 x y x 2

0
2  y

2 y dy length of strip width of strip
Since the width of the strip is dy
, we find the length of the strip by solving for x in terms of y
.
1
2
2 
2

1 y y y
3
3
2
0
2 4
8
3

10
3


1
2
General Strategy for Area Between Curves:
Sketch the curves.
Decide on vertical or horizontal strips. (Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.)
3 Write an expression for the area of the strip.
(If the width is dx
, the length must be in terms of x
.
If the width is dy
, the length must be in terms of y
.
4
Find the limits of integration. (If using dx
, the limits are x values; if using dy
, the limits are y values.)
5 Integrate to find area.
p

3
3
0 h
3 s
Find the volume of the pyramid:
Consider a horizontal slice through the pyramid.
3 dh
The volume of the slice is s 2 dh
.
If we put zero at the top of the pyramid and make down the positive direction, then s=h
.
V slice
 2 h dh
  3
2
V h dh
0
3

1 h
3
3
0
This correlates with the formula:

9

1
V Bh
3 3
9 3

9


1
Method of Slicing:
Sketch the solid and a typical cross section.
2 Find a formula for
V ( x )
.
(Note that I used
V ( x ) instead of
A ( x )
.)
3 Find the limits of integration.
4
Integrate
V ( x ) to find volume.


A 45 o wedge is cut from a cylinder of radius 3 as shown.
Find the volume of the wedge.
y x
You could slice this wedge shape several ways, but the simplest cross section is a rectangle.
If we let h equal the height of the slice then the volume of the slice is:
  
2 y h dx h
Since the wedge is cut at a 45 o angle:
45 o x
Since x
2  y
2 
9 y

9
 x
2 h
 x


y x
Even though we started with a cylinder, p does not enter the calculation!
V
V
 
0
3
  
9
0
2 x
1

2 9
 x
2  
9
 
 2

2 x dx y h dx du
  x
2
2 u u
 
 


9
0 h
 x

2
3 u 2
3
9
0 y

2
9

3
27 x
2 
18


Cavalieri’s Theorem:
Two solids with equal altitudes and identical parallel cross sections have the same volume.
Identical Cross Sections p

2
1
0 y
 x
1 2 3 4
Suppose I start with this curve.
My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape.
So I put a piece of wood in a lathe and turn it to a shape to match the curve.


2
1
0 y
 x
1 2 3 4
How could we find the volume of the cone?
One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes.
The volume of each flat cylinder (disk) is: p p r
2  x
the thickness
 
2 dx
In this case: r= the y value of the function thickness
= a small change in x = dx 

2
1
0 y
 x
1 2 3 4
The volume of each flat cylinder (disk) is: p r
2 
the thickness p
 
2 dx
If we add the volumes, we get:

0
4 p
 
2 x dx
 
0
4 p x dx
 p
2 x
2
0
4

8 p


This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk.
If the shape is rotated about the x-axis, then the formula is:
V
 p  a b
2 y dx
Since we will be using the disk method to rotate shapes about other lines besides the x-axis, we will not have this formula on the formula quizzes.
A shape rotated about the y-axis would be: V
 p  a b
2 x dy


• Consider a function f(x) on the interval [a, b]
• Now consider revolving that segment of curve about the x axis a
• What kind of functions generated these solids of revolution?
b f(x)
31

• We seek ways of using integrals to determine the volume of these solids
• Consider a disk which is a slice of the solid
– What is the radius
– What is the thickness
– What then, is its volume?
Volume of slice = p 
( )

2 dx dx f(x)
32

• To find the volume of the whole solid we sum the volumes of the disks
• Shown as a definite integral
V
 p  b a
 
2
( ) dx a b f(x)
33

The region between the curve , and the
1 y y
-axis is revolved about the y
-axis. Find the volume.
4 y x
1 1
1
2
2

.707
3
1
3

.577
4
1
2
4
3
1
0
2 dy
1
We use a horizontal disk.
The thickness is dy
.
.
The radius is the x value of the function

1 y
V
 
1
4 p


1 y


2 dy
 
1
4 p
1 y dy volume of disk
 p ln y
1
4  p  
0
 p ln 2
2



500 ft y The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis: x

.000574
y
2

.439
y

185 x
The volume can be calculated using the disk method with a horizontal disk.
p 
0
500

.000574
y
2 
.439
y

185

2 dy

24, 700, 000 ft
3


y
 x
2 y

2 x y
 x y
 x
2
The volume of the washer is:
0
V
 
0
4 p


 
2

2

 dy
1
1
2
4
3 y

2 x y
 x
2
The region bounded by y
 x the y-axis.
y

2 x
Find the volume.
2
If we use a horizontal slice:
The “disk” now has a hole in it, making it a “washer”.
 p
R
2  p r
2

 thickness p

R
2  r
2
 dy
V
 
0
4 p 
1
2 y y dy
4
V
 p 
0
4 
1
2 y y dy
4 outer radius
 p


1 y
2 
1
2 12 y
3


4
0 inner radius
 p

 8

16
3



8 p
3 

This application of the method of slicing is called the washer method.
The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.
The washer method formula is: V
 p  a b
2  2
R r dx
Like the disk method, this formula will not be on the formula quizzes. I want you to understand the formula.


2
1
4
3 y

2 x y
 x
2 y
 x
0 y

2 x y
2
 x
1
R r
2
V
 p 
0
4
R
2  r dy
 p 
0
4
2
 y
2
2 
2 y
2
 dy
 p 
0
4


 y
 y
4
2



4 4 y

 y
 x
2
If the same region is rotated about the line x
=
2
:
The outer radius is: y
R 2
2
The inner radius is: r y
 p 
0
4  y
 y
4
2
4 4 y
 y dy
 p 
0
4
3 y
1
4 y
2 
4
1
2 p p




3
2 y
2 
1
12 y
3 
8
3
24
16

64
3 3

 y 2

3


0
4
8 p
3 p

5
4
3 y
 x
2 
1 axis.
p y


1

4
1
5
 y
 y


1 x

2 dy


1
5
 y dy

4 p
4 p x

2 y
-
2
1
0 1 2 p

5

1 y y
2
2


5
1

4 p p
We can use the washer method if we split it into two parts: y x
2 x
 y

1 p




25

25
2

 
5

1
2




4 p

1
5
2
2 
 y

1
2
 dy p inner outer radius radius thickness
Japanese Spider Crab of slice
Georgia Aquarium, Atlanta cylinder p



25

9
2 2




4 p p 
16

2
4 p
8 p 
4 p 
12 p


The solids in Examples 1 –5 are all called solids of revolution because they are obtained by revolving a region about a line. In general, we calculate the volume of a solid of revolution by using the basic defining formula
We find the cross-sectional area in one of the following ways:

5
4
3
2
1
0 1 2 y
 x
2 
1
If we take a vertical slice we get a cylinder.
Here is another way we could approach this problem: cross section and revolve it about the y-axis
If we add all of the cylinders together, we can reconstruct the original object.


5
4
3
2
1 y
 x
2 
1
0 1 2 cross section
The volume of a thin, hollow cylinder is given by:

=2 p  
=2 p thickness

2 
 dx r circumference h thickness r is the x value of the function.
h is the y value of the function.
thickness is dx
.


5
4
3
2
1 y
 x
2 
1
This is called the shell method because we use cylindrical shells.
0 1 2 cross section
=2 p  
=2 p thickness

2 
 dx r circumference h thickness
If we add all the cylinders from the smallest to the largest:

0
2
2 p

2 
1
 dx
2 p 
0
2 x
3
 x dx
2 p


1
4 x
4

1
2 x
2


2
0
2 p   
12 p


Find the volume generated when this shape is revolved about the y axis.
4
3
2
1
0 1 2 3 4 5 6 y
 
4
9
 x
2 
10 x

16

7 8
We can’t solve for x, so we can’t use a horizontal slice directly.


If we take a vertical slice and revolve it about the y-axis we get a cylinder.
Shell method:
Lateral surface area of cylinder
4
3
2
1
0 1 2 3 4 5 6 y
 
4
9
 x
2 
10 x

16

7 8
=2 p 
Volume of thin cylinder

2 p r h dx


4
3
2
1
2 3 4 5 6 y
 
4
9
 x
2 
10 x

16

7

160 p

502.655 cm
3
8 0
Volume of thin cylinder

2 p r h dx
1

2
8
2 p r x



9
4  x
2 
10 x

 
16
 dx circumference h thickness
Note: When entering this into the calculator, be sure to enter the multiplication symbol before the parenthesis.


• Based on finding volume of cylindrical shells
– Add these volumes to get the total volume
• Dimensions of the shell
– Radius of the shell
– Thickness of the shell
– Height
55

• Consider the shell as one of many of a solid of revolution dx f(x) f(x) – g(x) x g(x)
• The volume of the solid made of the sum of the shells
V

2 p  b a

( )

( )

56

• Consider the region bounded by x = 0, y = 0, and y

8
 x
2
V

2 p
2 2

0 x

8
 2 x dx
57

• Sketch the graph over the limits of integration
• Draw a typical shell parallel to the axis of revolution
• Determine radius, height, thickness of shell
• Volume of typical shell
2 p  
• Use integration formula
p  b a
58

• Rotate the region bounded by y = 4x and y = x 2 about the x-axis thickness = dy radius = y
• What are the dimensions needed?
– radius
– height
– thickness
V

2 p 
16
0 y


 y
 y
4

 dy height = y
 y
4
59

• Possible to rotate a region around any line f(x) g(x) x = a
• Rely on the basic concept behind the shell method
s p
60

• What is the radius?
a – x
• What is the height?
f(x) – g(x) f(x) r x = c x = a
• What are the limits?
c < x < a
• The integral:
V
  a
( a
 x )

( )

( ) c
 g(x)
61

• Rotate the region bounded by 4 – x 2 , x = 0 and, y = 0 about the line x = 2 r = 2 - x
4 – x 2
• Determine radius, height, limits

2
0
62

• Integral for the volume is
V

2
2 p
(2 x ) (4 x
2
) dx
0

63

Review Problems
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line. Sketch the region, the solid, and a typical disk or washer
1.

2.

3.

4.

Set up, but do not evaluate, an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
1.
2.
3.

4.

Use the method of cylindrical shells to find the volume generated by rotating the region bounded by the given curves about the Specific line. Sketch the region and a typical shell.
1.
2.

3.

4.

5.

The average value of function f on the interval [a, b] is defined as f ave

1 b
 a
 a b f ( x ) dx
Note: For a positive function, we can think of this definition as saying area/width = average height
Example: Find the average value of f(x)=x 3 on
[0,2].
f ave

1
2

0

0
2 x
3 dx

1
2

 x
4
4


2
0

1
2

2
4
4

2

Theorem: If f is continuous on [a, b], then there exists a number c in [a, b] such that f ( c )
 f ave
 b
1
 a
 a b f ( x ) dx that is,
 a b f ( x ) dx
 f ( c )( b
 a )
Example: Find c such that f ave
=f(c) for f(x)=x 3 on
[0,2].
From previous slide,
2 f(c)=f ave
=2.
Thus, c 3 =2, so

4. Computing average value of a function

Find the average value of the function on the given interval .
1.
2.
3.

1.
2.