Page 393 - Advanced engineering mathematics

12.1 Line Integrals 373

of length t = (b − a)/n, where n is a positive integer. We can make t as small as we want
by choosing n large, so that values of δ(x, y, z) are approximated as closely as we want on
[t j−1 ,t j ] by δ(P j ), where P j = (x(t j ), y(t j ), z(t j )). The length of wire between P j−1 and P j is
s = s(P j ) − s(P j−1 ) ≈ ds j . The density of this piece of wire is approximately δ(P j )ds j , and
n δ(P j )ds j approximates the mass of the wire. In the limit as n →∞, this gives
j=1

mass of the wire = δ(x, y, z)ds.
C
A similar argument gives the coordinates (˜, ˜y, ˜z) of the center of mass of the wire as
x
1 1 1
˜ x = xδ(x, y, z)ds, ˜y = yδ(x, y, z)ds, ˜z = zδ(x, y, z)ds,
m C m C m C
in which m is the mass.

EXAMPLE 12.7

A wire is bent into the shape of the quarter circle C given by x = 2cos(t), y = 2sin(t), z = 3for
2
0 ≤ t ≤ π/2. The density function is δ(x, y, z) = xy . We want the mass and center of mass of
the wire.
The mass is
π/2

2 2 2 2
m = xy ds = 2cos(t)[2sin(t)] 4sin (t) + 4cos dt
C 0
π/2 16
2
= 16cos(t)sin (t)dt = .
3
0
Now compute the coordinates of the center of mass. First,
1
˜ x = xδ(x, y, z)ds
m C
3 π/2
2
2
2
= [2cos(t)] [2sin(t)] 2 4sin (t) + 4cos dt
16 0
π/2 3π
2
2
= 6 cos (t)sin (t)dt = .
8
0
Next,
1
˜ y = yδ(x, y, z)ds
m C
3 π/2 3
2
2
= [2cos(t)][2sin(t)] 4sin (t) + 4cos dt
16 0
π/2 3
3
= 6 cos(t)sin (t)dt = .
0 2
And
3
2
˜ z = zxy ds
16 C
3 16 2
2
2
= 3[2cos(t)][2sin(t)] 4sin (t) + 4cos dt
16 0
π/2
2
= 9 sin (t)cos(t)dt = 3.
0
It should not be surprising that ˜z = 3 because the wire is in the z = 3 plane.
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October 14, 2010 14:53 THM/NEIL Page-373 27410_12_ch12_p367-424