Density and Center of Mass
Objectives
• Integrate a density function to find Total Population or Total Mass
• Integrate or use Summation to find Center of Mass of a Substance with
Certain Density
Density
Density of population is measured in quantity per unit area or per unit
volume. For example, people per square mile, bacteria per cubic centimeter
(cc). Density of a substance is measured as mass per unit volume. For
example, grams per cubic centimeter.
Finding Total Quantity from A Density Function
Example 1 Suppose Cities A, B, and C are as follows:
a) City A is a 2 mile square region bounded on the east by a river. The
population density, P , is a function of the distance r miles from the
river. That is, P = f (r) people/square mile.
b) City B is a circular city of radius 2 miles, where the population density,
P = g(x) people/square mile is a function of the distance, x, from the
city center.
c) City C is a very narrow canyon along a highway of length 6km. The
population density P = h(q) people/km, is a function of the distance q
km from the mouth of the canyon.
Express each of the populations as a definite integral.
Solution.
1
a) City A
1. Since density is constant along a vertical strip, we divide the region
into vertical rectangular strips of width ∆r, and height 2. So the
area of a strip is approximately 2∆r square miles.
2. The population per rectangular strip is approximately: (area of
slice in square miles)(density in people per square mile)
= 2∆r · f (r)
3. The total population of City A is
Z 2
2f (r)dr people
0
b) City B
1. Since the density is constant along a ring around the center of the
city, we divide the region into rings of width ∆x, and circumference 2πx.
2. The area of each ring is approximately 2πx∆x square miles.
3. The population per ring is approximately (area of slice in square
miles)(density in people per square mile)
= 2πxg(x)∆x
4. Total population of City B is
Z 2
2πxg(x)dx people
0
c) City C
1. Divide the line into subintervals of length ∆q km.
2. The population per segment is approximately (length of segment)(density
in people per km)
= h(q)∆q
2
3. The total population of City C is
Z 6
h(q)dq people
0
Example 2 A rod has length 2 meters. At a distance x meters from the left
endpoint of the rod, the density is given by δ(x) = 2 + 6x gm/m. Find the
total mass of the rod.
Solution. The total mass of the rod is given by
Z 2
2 + 6xdx
0
= 2x + 3x2 |20
= 4 + 12
= 16gm
Center of Mass
Finite Systems
Finding the center of mass of a finite system of n masses mi , along the x axis,
each located at coordinate xi can be thought of as finding where to put the
fulcrum of a see-saw to balance n people of weights mi located at position
xi .
Center of Mass = x̄ =
Sum of the Moments of the Mass
Total Mass of the System
Example 3 Find the center of mass of a system containing three point
masses of 5gm, 3gm, and 1gm located at x = −10, x = 1, and x = 2 respectively.
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Solution.
x̄ =
5(−10) + 3(1) + 1(2)
5+3+1
−45
=
9
= −5
Continuous Systems
Let δ(x) be the mass density (mass per unit length) at a point x. Then
Rb
xδ(x)dx
Center of Mass = x̄ = Ra b
δ(x)dx
a
Example 4 A rod of length 3 meters has density δ(x) = 1+x2 grams/meter,
and is positioned along the x-axis with left endpoint at 0. Find the center of
mass of the rod.
Solution.
R3
x(1 + x2 )dx
R3
1 + x2 dx
0
R3
x + x3 dx
0
= R3
1 + x2 dx
0
0
x̄ =
=
x2
2
+
x+
x4 3
|
4 0
x3 3
|
3 0
+ 81
4
3+9
99
=
4(12)
=
=
9
2
33
meters
16
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Mass may not always be distributed along a straight line. What if we
have a 2-D or 3-D object, and we want to determine its center of mass? Our
text only deals with objects of constant density here-if you would like to
know how this differs when the density is not constant throughout the entire
object, see me!
For the following, let Ax (x) be the area of a slice of the object, perpendicular to the x-axis. Let Ay (y) and Az (z) be defined in the same manner.
Note that for a 2-D figure, you will not have Az (z), and Ax (x) and Ay (y) are
interpreted as the height of a slice.
Theorem 1 For a region R, of constant density, δ, the center of mass of R
is given by the point (x̄, ȳ, z̄), where x̄, ȳ, and z̄ are defined as follows:
R
R
R
yδAy (y)dy
zδAz (z)dz
xδAx (x)dx
ȳ =
z̄ =
x̄ =
Total Mass
Total Mass
Total Mass
Example 5 Find the coordinates of the center of mass of the isosceles triangle, whose vertices are at (0, − 12 ), (0, 12 ), and (1, 0). The triangle has constant
density, δ, and mass m.
Solution. Since density is mass per unit area, we must have δ =
m
1/2
= 2m.
It will also be beneficial to know the equations of the lines which define
the sides of our triangle. Clearly, x = 0 is one of those lines. The other two
lines are y = 12 x − 12 and y = 12 − 12 x.
We now find x̄.
R1
x̄ =
0
δx ( 21 − 12 x) − ( 12 x − 12 ) dx
m
R1
2mx(1 − x)dx
= 0
m
R1
2m 0 x − x2 dx
=
m
5
Z
=2
1
x − x2 dx
0
x2 x3 1
− )|0
2
3
1 1
= 2( − )
2 3
1
= 2( )
6
1
=
3
= 2(
It may be obvious to you that ȳ = 0, but let’s show that this is truly the
case anyway. We convert our lines into functions of y rather than functions
of x, to obtain x = 2y + 1 and x = 1 − 2y. It should be evident from the
picture, that we will need two integrals to compute the numerator of our ȳ.
R0
ȳ =
δy(1 − 2y)dy +
−1/2
R 1/2
0
δy(1 + 2y)dy
m
2m
=
hR
0
−1/2
2
y − 2y dy +
R 1/2
0
2
y + 2y dy
m
2y 0
y
y 2 2y 3 1/2
)|−1/2 + ( +
)|
=2 ( −
2
3
2
3 0
= −2(
i
2
3
(−1/2)2 2(−1/2)3
(1/2)2 2(1/2)3
−
) + 2(
+
)
2
3
2
3
=0
That is, the center of mass for the given object is at (1/3, 0)!
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