Michael Corral: Vector Calculus

126 CHAPTER 3. MULTIPLE INTEGRALS
The formulas for the center of mass of a region in 2 can be generalized to a solid S
in 3 . Let S be a solid with a continuous mass density functionδ(x,y,z) at any point
(x,y,z) in S. Then the center of mass of S has coordinates (¯x,ȳ,¯z), where
where
M yz =
S
xδ(x,y,z)dV, M xz =
¯x= M yz
M , ȳ= M xz
M , ¯z= M xy
M , (3.30)
S
M=
S
yδ(x,y,z)dV, M xy =
S
zδ(x,y,z)dV, (3.31)
δ(x,y,z)dV. (3.32)
In this case, M yz , M xz and M xy are called the moments (or first moments) of S around
the yz-plane, xz-plane and xy-plane, respectively. Also, M is the mass of S.
Example 3.14. Find the center of mass of the solid S={(x,y,z) : z≥0, x 2 +y 2 +z 2 ≤ a 2 },
if the density function at (x,y,z) isδ(x,y,z)=1.
Solution: The solid S is just the upper hemisphere inside the
sphere of radius a centered at the origin (see Figure 3.6.3). So
since the density function is a constant and S is symmetric
about the z-axis, then it is clear that ¯x=0 and ȳ=0, so we
need only find ¯z. We have
M= δ(x,y,z)dV= 1dV= Volume(S).
S
S
But since the volume of S is half the volume of the sphere of
radius a, which we know by Example 3.12 is 4πa3 2πa3
3
, then M=
3 . And
M xy = zδ(x,y,z)dV
S
=
=
=
=
zdV , which in spherical coordinates is
S
∫ 2π ∫ π/2 ∫ a
0 0
∫ 2π ∫ π/2
0 0
∫ 2π ∫ π/2
0
0
(ρ cosφ)ρ 2 sinφdρdφdθ
0
(∫ a
)
sinφ cosφ ρ 3 dρ dφdθ
a 4
4
0
sinφ cosφdφdθ
a
z
(¯x,ȳ,¯z)
0 a
x
Figure 3.6.3
y