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Chapter 6 | Applications of Integration
⎛ ⎝ x i * ,
⎞ ⎠ , as shown in the following figure.
⎛ ⎝ f ( x i * ) ⎞
⎠ /2
Figure 6.66 A representative rectangle of the lamina.
Next, we need to find the total mass of the rectangle. Let ρ represent the density of the lamina (note that ρ is a constant). In this case, ρ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area of the rectangle by ρ . Then, the mass of the rectangle is given by ρf ( x i * )Δ x . To get the approximate mass of the lamina, we add the masses of all the rectangles to get m ≈ ∑ i =1 n ρf ( x i * )Δ x . This is a Riemann sum. Taking the limit as n →∞ gives the exact mass of the lamina: m = lim n →∞ ∑ i =1 n ρf ( x i * )Δ x = ρ ∫ a b f ( x ) dx . Next, we calculate the moment of the lamina with respect to the x -axis. Returning to the representative rectangle, recall its center of mass is ⎛ ⎝ x i * , ⎛ ⎝ f ( x i * ) ⎞ ⎠ /2 ⎞ ⎠ . Recall also that treating the rectangle as if it is a point mass located at the center of mass does not change the moment. Thus, the moment of the rectangle with respect to the x -axis is given by the mass of the rectangle, ρf ( x i * )Δ x , multiplied by the distance from the center of mass to the x -axis: ⎛ ⎝ f ( x i * ) ⎞ ⎠ /2. Therefore, the moment with respect to the x -axis of the rectangle is ρ ⎞ ⎠ Δ x . Adding the moments of the rectangles and taking the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the x -axis is ⎛ ⎝ ⎡ ⎣ f ( x i * ) ⎤ ⎦ 2 /2
⎡ ⎣ f ( x i * ) ⎤
→∞ ∑ i =1 n
⎦ 2 2 Δ
b ⎡
⎤ ⎦ 2
⎣ f ( x )
x = ρ ∫
M x = lim n
ρ
dx .
2
a
We derive the moment with respect to the y -axis similarly, noting that the distance from the center of mass of the rectangle to the y -axis is x i * . Then the moment of the lamina with respect to the y -axis is given by
→∞ ∑ i =1 n
ρx i * f ( x i * )Δ x = ρ ∫ a b
M y = lim n
xf ( x ) dx .
We find the coordinates of the center of mass by dividing the moments by the total mass to give x – = M y / m and y – = M x / m . If we look closely at the expressions for M x , M y , and m , we notice that the constant ρ cancels out when x – and y – are calculated. We summarize these findings in the following theorem.
Theorem 6.12: Center of Mass of a Thin Plate in the xy -Plane Let R denote a region bounded above by the graph of a continuous function f ( x ), below by the x -axis, and on the left
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