Chapter 6 | Applications of Integration
713
M y m = − 9 4 ·
M x m = − 27 10 ·
x – =
y – =
3 5
2 9 = −
1 2 and
2 9 = −
.
The centroid of the region is ⎛
⎝ −(1/2), −(3/5) ⎞ ⎠ .
6.32 Let R be the region bounded above by the graph of the function f ( x ) =6− x 2 and below by the graph of the function g ( x ) =3−2 x . Find the centroid of the region.
The Symmetry Principle We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example. Example 6.33 Finding the Centroid of a Symmetric Region
Let R be the region bounded above by the graph of the function f ( x ) =4− x 2 and below by the x -axis. Find the centroid of the region.
Solution The region is depicted in the following figure.
Figure 6.71 We can use the symmetry principle to help find the centroid of a symmetric region.
The region is symmetric with respect to the y -axis. Therefore, the x -coordinate of the centroid is zero. We need only calculate y – . Once again, for the sake of convenience, assume ρ =1. First, we calculate the total mass:
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