Chapter 6 | Applications of Integration
709
and right by the lines x = a and x = b , respectively. Let ρ denote the density of the associated lamina. Then we can make the following statements: i. The mass of the lamina is (6.18) m = ρ ∫ f ( x ) dx . ii. The moments M x and M y of the lamina with respect to the x - and y -axes, respectively, are a b
(6.19)
b ⎡
dx and M y = ρ ∫ a b
⎤ ⎦ 2
⎣ f ( x )
M x = ρ ∫
xf ( x ) dx .
2
a
⎝ x – , y –
iii. The coordinates of the center of mass ⎛
⎞ ⎠ are
M y m and y
(6.20)
– = M x m .
x – =
In the next example, we use this theorem to find the center of mass of a lamina.
Example 6.31 Finding the Center of Mass of a Lamina
Let R be the region bounded above by the graph of the function f ( x ) = x and below by the x -axis over the interval [0, 4]. Find the centroid of the region.
Solution The region is depicted in the following figure.
Figure 6.67 Finding the center of mass of a lamina.
Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant ρ cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume ρ =1. First, we need to calculate the total mass: m = ρ ∫ f ( x ) dx = ∫ 4 xdx
a b
0
x 3/2 |
0 4
16
= 2 3
= 2 3 [8−0] =
3 .
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