Thomas Calculus 13th [Solutions]

Section 6.6 Moments and Centers of Mass 483
10. (a) Since the plate is symmetric about the line x y and its
density is constant, the distribution of mass is symmetric
about this line. This means that x y The typical vertical
strip has center of mass:
9 x
( x, y) x, ,
2
2
length:
mass:
x -axis is
2
9 x width: dx, area:
2
dA 9 x dx,
2
dm dA 9 x dx.
The moment about the
2
9 x 2 2
y dm 9 x dx 9 x dx
2 2
3 2
Thus, M
0 2 9 2 9 x
x y dm x dx x
3
0
2
(27 9) 9 ; M dm dA dA
(Area of a quarter of a circle of radius 3) 9 9
4 4 . Therefore, y Mx (9 ) 4 4
M 9
( x, y ) 4 , 4 is the center of mass.
3 3
(b) Applying the symmetry argument analogous to the
one used in Exercise 1, we find that x 0. The
typical vertical strip has the same parameters as in
3 2
part (a). Thus, M x y dm 9 x dx
3 2
2 3 2
0 2
9 x dx 2(9 ) 18 ; M dm dA
dA (Area of a semi-circle of radius 3) 9 9
2 2 . Therefore, y (18 ) 2 4 , the
M
9
same y as in part (a) ( x, y ) 0, 4 is the center of mass.
M x
11. Since the plate is symmetric about the x-axis and its density is
constant, the distribution of mass is symmetric about this line.
This means that y 0. The typical vertical strip has center of
mass: ( y, y) ( x , 0), length: 1 1 ,
2 2
1 x 1 x
width: dx, area: dA 2 dx , mass: dm dA 2 dx.
2
2
1 x
1 x
The moment about the y-axis is
x dm x 2 dx 2x
dx 2 x dx . Thus,
2 2 2
1 x 1 x 1 x
1 2 1
M 2 x
y dx [ln(1 x )]
0
2
0 ln 2.
1 x
1 1
M dm 2 dx 2 [arctan x ] 2
0
2
0 2 (arctan1) . Therefore,
1 x
4 2
M
x
y ln 2 2ln 2 ln 4 ( x, y ) ln 4 , 0 is the center of mass.
M /2
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