710
Chapter 6 | Applications of Integration
Next, we compute the moments:
b ⎡
⎤ ⎦ 2
⎣ f ( x )
M x = ρ ∫
dx
2
a
x 2 |
4
0 4
= ∫
x 2
dx = 1 4
=4
0
and
b
M y = ρ ∫
xf ( x ) dx
a
4 x xdx = ∫ 0 4
= ∫
x 3/2 dx
0
x 5/2 |
0 4
[32−0] = 64 5 .
= 2 5
= 2 5
Thus, we have
M y m = 64/5 16/3
M x
x – =
and y – =
= 64 5
· 3
= 12 5
y = 4
=4· 3 16
= 3 4 .
16
16/3
The centroid of the region is (12/5, 3/4).
6.31 Let R be the region bounded above by the graph of the function f ( x ) = x 2 and below by the x -axis over the interval [0, 2]. Find the centroid of the region.
We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function f ( x ), as before, but now, instead of having the lower bound for the region be the x -axis, suppose the region is bounded below by the graph of a second continuous function, g ( x ), as shown in the following figure.
Figure 6.68 A region between two functions.
Again, we partition the interval ⎡
⎤ ⎦ and construct rectangles. A representative rectangle is shown in the following figure.
⎣ a , b
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