Volume
Our first example is the volume of a solid body. Before proceeding,
let’s recall that the volume of a right cylinder is Ah. Here we use the
“right cylinder” in the general sense; the base does not have to be
circular, but the sides are perpendicular to the base.
Suppose that the solid body extends from height y = a to y = b along
the y-axis. Let A(y) be the area of the horizontal cross section at
height y.
Volume as the Integral of Cross-Sectional Area Let A(y) be the
area of the horizontal cross section at height y of a solid body
extending from y = a to y = b. Then
b
Volume of the Solid Body   A  y  dy
a
Volume of a Pyramid Calculate the volume V of
a pyramid of height 12 m whose base is a square
of side 4 m.
We need a formula for the horizontal cross
section A(y).
1
s
2
12  y
y
2
Base
1 0.5s
 
 Write s in terms of y.
Height 6 12  y
12  y
 s  y 
 Write A in terms of y.
3
b
1
2
V  A  y  dy
 A  y   12  y 
9
a

12
1
2
V   12  y  dy u  12  y  du  dy
90
0
12
1 2
1 2
   u du   u du
9 12
90
12
1 u 
1
3
3
   
12

64
m


9  3  0 27
3
Compute the volume V of the solid, whose base is the region between
the parabola y = 4 − x2 and the x-axis, and whose vertical cross sections
perpendicular to the y-axis are semicircles.
x  4  y  A y 
V

4
 4  y  dy

2
0

4
y 
 4 y  
2
2 0


2
2
 8  4
 4  y
2
Volume of a Sphere: Vertical Cross Sections Compute the volume of
R
a sphere of radius R.
V  2   R  x dx
2
2
0
R
 2
x 
 2  R x  
3 0

3
R
V
 A  x dx
R
 3 R3 
 2  R  
3 

2
4
3
3

2R    R

3
3
CONCEPTUAL INSIGHT Cavalieri’s principle states: Solids
with equal cross-sectional areas have equal volume. It is often
illustrated convincingly with two stacks of coins
b
Our formula V   A  y  dy includes Cavalieri’s principle,
a
because the volumes V are certainly equal if the crosssectional areas A(y) are equal.
Density
Next, we show that the total mass of an object can be expressed as the
integral of its mass density. Consider a rod of length ℓ. The rod’s
linear mass density ρ is defined as the mass per unit length. If ρ is
constant, then by definition,
Total mass  linear mass density  length    l
For example, if ℓ = 10 cm and ρ = 9 g/cm, then the total mass is
ρℓ = 9 · 10 = 90 g.
The symbol ρ (lowercase
Greek letter rho) is used often
to denote density.
Density
Now consider a rod extending along the x-axis from x = a to x = b
whose density ρ(x) is a continuous function of x. To compute the
total mass M, we break up the rod into N small segments of length
Δx = (b − a)/N. Then
N
M   Mi ,
i 1
where Mi is the mass of the ith segment.
Density
Eq. (2) Total mass  linear mass density  length    l
We cannot use Eq. (2) because ρ(x) is not constant, but we can argue
that if Δx is small, then ρ(x) is nearly constant along the ith segment.
If the ith segment extends from xi−1 to xi and ci is any sample point in
[xi−1, xi], then Mi ≈ ρ(ci)Δx and
N
N
i 1
i 1
Total mass M   M i    ci  x
Density
As N  , the accuracy of the approximation improves. However,
b
N
   c  x
i 1
i
   x dx,
is a Riemann sum whose value approaches
a
and thus it makes sense to define the total mass of a rod as the
integral of its linear mass density:
b
Total Mass     x dx
a
Note the similarity in the way we use thin slices to compute volume
and small pieces to compute total mass.
N
N
i 1
i 1
Total mass M   M i    ci  x
Total Mass Find the total mass M of a 2-m rod of linear density
ρ(x) = 1 + x(2 − x) kg/m, where x is the distance from one end of the
rod.
b
Total Mass     x dx
a
2

x  10
2
M   1  2 x  x dx   x  x   
kg
3 0 3

0
2
3
2
In some situations, density is a function of distance to the origin. For
example, in the study of urban populations, it might be assumed that
the population density ρ(r) (in people per square kilometer) depends
only on the distance r from the center of a city. Such a density
function is called a radial density function.
R
Population P within a radius R  2  r   r dr
0
Computing Total Population The population in a certain city has
radial density function ρ(r) = 15(1 + r2)−1/2, where r is the distance
from the city center in kilometers and ρ has units of thousands per
square kilometer. How many people live in the ring between 10 and 30
km from the city center?
R
Population P within a radius R  2  r   r dr
30

P  2  r 15 1  r
10
30
 30  r 1  r


2 1/ 2
2 1/ 2
 dr
0
dr u  1  r 2  du  2rdr
10
 15
901
u
1/ 2
1/ 2 901
1/ 2 901
du  15  2u   30 u 
101
101
101
 30  901  101   1881.825286 thousand  1,881,825.286 people
Average Value
DEFINITION Average Value The average value of an integrable
function f (x) on [a, b] is the quantity
b
1
Average Value 
f  x dx

ba a
GRAPHICAL INSIGHT
Think of the average value M of a
function as the average height of its
graph. The region under the graph
has the same signed area as the
rectangle of height M, because
b
 f  x dx  M  b  a  .
a
Average Value
DEFINITION Average Value The average value of an integrable
function f (x) on [a, b] is the quantity
b
1
Average Value 
f  x dx

ba a
Find the average value of f (x) = sin x on [0, π].
1

1
sin xdx    cos x 




0
0

1

 1  1 
2

Vertical Jump of a Bushbaby
To set up our bounds, we need roots!
The bushbaby is a small primate with remarkable jumping ability.
Find the average speed during a jump if the initial vertical velocity is
υ0 = 600 cm/s. Use Galileo’s formula for the height
1 2
h  t   v0t  gt (in centimeters, where g = 980 cm/s2).
2
 600  60
b
Axis  t 
 h  0 @ t  0 & t  2 
s
 
2a
 2  490   49
4.9
6
6 / 4.9

0
4.9 
600  980t dt 

6 
6 / 9.8

0

 600  980t dt    600  980t dt 
6 / 9.8

6 / 4.9
4.9   6 
  6 
 6  

F 
  F  0   F 
F
 
6   9.8 
4.9
9.8


 
 
4.9   6 
 6   4.9

2F 
 2 183.6735   0 
F
 

6   9.8 
 4.9   6
 300 cm/s
 F  x   600t  490t 2 
There is an important difference between the average of a list of
numbers and the average value of a continuous function. If the average
score on an exam is 84, then 84 lies between the highest and lowest
scores, but it is possible that no student received a score of 84. By
contrast, the Mean Value Theorem (MVT) for Integrals asserts that a
continuous function always takes on its average value somewhere in
the interval.
For example, the average of f (x) = sin x on [0, π] is 2/π. We have
f (c) = 2/π for c = sin−1(2/π) ≈ 0.69. Since 0.69 lies in [0, π],
f (x) = sin x indeed takes on its average value at a point in the interval
Theorem 1 Mean Value Theorem for Integrals
If f (x) is continuous on [a, b], then there exists a value c ε [a, b]
such that
b
1
f c 
f  x dx

ba a