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

dM = m( x,y,z) dV

(2)

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    A box corresponding to [ 0,1] ×[ 0,1] ×[ 0,2] in xyz-coordinates is filled with a mixture of small machine parts (heavy) and styrofoam peanuts (light), which via settling is heavier at the bottom than the top. If the mixture has a density of

m( x,y,z) = ( 9-z3)    kg m3

then what is the mass of the machine parts, styrofoam peanuts combination.

Solution: The mixture occupies the solid W corresponding to the interior of the box.  Thus, the mass of the mixture is

M =    dM =      m( x,y,z)   dV

This integral can then be reduced to a triple iterated integral

M = ó õ 1 0  ó õ 1 0  ó õ 2 0  ( 9-z3) dz dydx = ó õ 1 0  ó õ 1 0  9z-  z4 4 ê ê 2 0   dydx

which reduces to M = 14 kg, which is about 31 lbs at sea level.

The settling means that near the bottom of the box (where z » 0 ), the mass density is about 9 kg per m³, whereas near the top ( z » 2) the density is about m(x,y,2) = 9-8 = 1 kg per m³.  Physically, this will means that the center of mass should be lower than the point (0.5,0.5,1) we would have expected for a uniform mixture, as we will explore in part 4.  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.

LiveGraphics3d Applet

 

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.

LiveGraphics3d Applet

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.

LiveGraphics3d Applet

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.