James Stewart-Calculus_ Early Transcendentals-Cengage Learning (2015)

1018 Chapter 15 Multiple Integrals
5 The coordinates sx, yd of the center of mass of a lamina occupying the
region D and having density function sx, yd are
x − M y
m − 1 y x sx, yd dA
m
D
y − M x
m − 1 y y sx, yd dA
m
D
where the mass m is given by
m − y sx, yd dA
D
y
(0, 2)
y=2-2x
3 11
” , ’
8 16
D
0
(1, 0)
x
FIGURE 5
ExamplE 2 Find the mass and center of mass of a triangular lamina with vertices
s0, 0d, s1, 0d, and s0, 2d if the density function is sx, yd − 1 1 3x 1 y.
SOLUTION The triangle is shown in Figure 5. (Note that the equation of the upper
boundary is y − 2 2 2x.) The mass of the lamina is
m − yy sx, yd dA − y 1
0 y222x s1 1 3x 1 yd dy dx
0
D
− y
1
0
Fy 1 3xy 1 y 2
2
Gy−0
y−222x
dx
− 4 y 1
s1 2 x 2 d dx − 4Fx 2 x 3
0
1
3G0
− 8 3
Then the formulas in (5) give
x − 1 y xsx, yd dA − 3 8 y 1
m 0 y222x sx 1 3x 2 1 xyd dy dx
0
D
− 3 8 y 1
0
Fxy 1 3x 2 y 1 x y 2
2Gy−0
y−222x
dx
− 3 2 y1 sx 2 x 3 d dx − 3
0
2F x 2
2 2 x 4 1
− 3 8 4G0
y − 1 y ysx, yd dA − 3 8 y 1
m 0 y222x sy 1 3xy 1 y 2 d dy dx
0
D
− 3 8 y 1
0
F
y 2
2 1 3x y 2
− 1 4
F7x 2 9 x 2
2 1 y 3
3Gy−0
2 2 x 3 1 5 x 4
y−222x
1
4G0
dx − 1 4 y 1
s7 2 9x 2 3x 2 1 5x 3 d dx
− 11
16
The center of mass is at the point ( 3 8 , 11
16). ■
ExamplE 3 The density at any point on a semicircular lamina is proportional to the
distance from the center of the circle. Find the center of mass of the lamina.
0
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