Michael Corral: Vector Calculus

3.6 Application: Center of Mass 125
Solution: The region R is shown in Figure 3.6.2. We have
M= δ(x,y)dA
=
=
=
R
∫ 1 ∫ 2x 2
0
∫ 1
0
∫ 1
0
0
⎛
y2
⎜⎝ xy+
2
= x4
2 + 2x5
5
(x+y)dydx
⎞
y=2x 2
∣
⎟⎠ dx
y=0
(2x 3 +2x 4 )dx
∣
1
0
= 9 10
y
y=2x 2 R
x
0 1
Figure 3.6.2
and
M x =
=
=
=
yδ(x,y)dA M y =
R
∫ 1 ∫ 2x 2
0
∫ 1
0
∫ 1
0
R
∫ 1 ∫ 2x 2
xδ(x,y)dA
y(x+y)dydx = x(x+y)dydx
0
0 0
⎛
⎞
xy 2
⎜⎝ 2 + y3
y=2x 2 ∫ ⎛
1
⎞
3∣
⎟⎠ dx = ⎜⎝ x2 y+ xy2
y=2x 2
y=0
0 2 ∣
⎟⎠ dx
y=0
∫ 1
(2x 5 + 8x6
3 )dx = (2x 4 +2x 5 )dx
0
1
= 5 = 2x5
∣ 7
5 + x6
1
= 11
3∣
15 ,
= x6
3 + 8x7
21
so the center of mass (¯x,ȳ) is given by
0
¯x= M y
M = 11/15
9/10 = 22
27 , ȳ= M x
M = 5/7
9/10 = 50
63 .
Note how this center of mass is a little further towards the upper corner of the region
Rthanwhenthedensityisuniform(itiseasytousetheformulasin(3.27)toshowthat
(¯x,ȳ)= ( 3
4 , 5) 3 in that case). This makes sense since the density functionδ(x,y)= x+y
increases as (x,y) approaches that upper corner, where there is quite a bit of area.
In the special case where the density functionδ(x,y) is a constant function on the
region R, the center of mass (¯x,ȳ) is called the centroid of R.
0