Part 2: Mass of a Solid
In many situations, we are given the mass-density of a
solid, which is a function m( x,y,z) measured in units of
mass per unit volume. Typically, if dM is the approximate mass of a small
``chunk'' of a solid with volume dV and if ( x,y,z) is a
point in that ``chunk'', then
which is to say that m( x,y,z) dV is the mass of a small
sample of a given solid.
In the same manner illustrated in section 3, the Riemann definition of the
triple integral then implies that the total mass of the solid S with mass
density m( x,y,z) is given by
|
M = |
 |
dM = |
|
m( x,y,z) dV |
|
That is, the mass of a solid depends on how dense it is.
EXAMPLE 3 Find the mass of the bar corresponding to [0,10] ×[ 10,50] ×[ 0,10] when the
mass density is m( x,y,z) = 0.1y grams per centimeter.
Solution: This bar is ``more dense'' on the right end than it is on
the left end. We might say that the right end ``feels heavier'' than the
left end. It's total mass is given by
|
M = |
|
0.1y dV |
|
This integral can then be reduced to a triple iterated integral
|
M = |
ó
õ
|
10
0
|
|
ó
õ
|
50
10
|
|
ó
õ
|
10
0
|
0.1ydzdydx = 12,000 grams |
|
Thus, the mass of the bar is 12 kg, which is about 26.4 pounds at sea
level.
The key is that the density m( x, y, z) is a tool for relating the physical quantity of
mass to the mathematical quantity of volume.
Alternatively, densities allow us to imagine that a geometric structure has a
physical manifestation.
EXAMPLE 4 What is the
mass of a tetrahedron with vertices at ( 0,0,0) , ( 1,0,0,) , ( 0,1,0) , and (
0,0,1) if it has a uniform mass density of m(x,y,z)
= 18 kg per cubic meter.
Solution: We begin by identifying the two surfaces that bound the
tetrahedron above and below. Since the upper face is flat, it corresponds to the
plane through ( 1,0,0) , ( 0,1,0), and ( 0,0,1) . Since u = á0-1,1-0,0-0
ñ = á -1,1,0
ñ and v = á 0-1,0-0,1-0
ñ = á -1,0,1
ñ are parallel to the plane, their cross product n = u×v
= á 1,1,1 ñ is normal to
the plane, thus implying that the equation of the plane is z = 1-x-y.
Thus, the tetrahedron is between z = 1-x-y
and z = 0 over the region R between x = 0, x = 1, y
= 0, and y = 1-x.
As a result, the mass of the tetrahedron T is
| M = |
ó
õ |
|
ó
õ |
|
ó
õ |
T
|
m( x,y,z)
dA = |
ó
õ |
|
ó
õ |
R
|
|
ó
õ |
1-x-y
0
|
18 dzdA |
|
Evaluating the first integral yields
| M = |
ó
õ |
|
ó
õ |
R
|
18( 1-x-y)
dA = |
ó
õ |
1
0
|
|
ó
õ |
1-x
0
|
18(1-x-y)
dydx = 3 |
|
Thus, the tetrahedron has a mass of 3 kg.
Check Your Reading: Geometrically, what are the faces of the tetrahedron?