# Weighted Average

advertisement ```Weighted Average
The centroid or center of mass of a planar region is the point at which that
region balances perfectly, like a plate on the end of a stick. The coordinates of
the centroid are given by weighted averages.
�
x dA
The x coordinate of the centroid is x̄ = �
, where dA is an inﬁntessimal
dA
portion of area; the weighting function in this average� is just x.
y dA
Similarly, the y coordinate of the centroid is ȳ = �
.
dA
Find the centroid (x̄, ȳ) of the parabolic region bounded by x = −1, x = 3,
y = (x − 1)2 and y = 4.
Solution
y
4
−1
3
x
We might guess from the symmetry of the region that x̄ = 1. We’ll do the
computation for practice. Since we’re integrating with respect to x, our area
dA will be a vertical rectangle with base at y = (x − 1)2 , top at y = 4 and width
dx.
�3
x̄ = �−31
�
3
−1
2
x(4 − (x − 1) ) dx
x(4 − (x − 1)2 ) dx
−1
(4 − (x − 1)2 ) dx
�
3
=
−1
�
=
3x − x3 + 2x2 dx
3 2 1 4 2 3
x − x + x
2
4
3
1
�3
−1
�
� �
�
3
1
2
3 1 2
&middot; 9 − &middot; 81 + &middot; 27 −
− −
2
4
3
2 4 3
2
10
3
=
=
�
3
−1
(4 − (x − 1)2 ) dx
�
3
=
−1
3 − x2 + 2x dx
�3
1 3
2
=
3x − x + x
3
−1
�
� �
�
1
1
=
3 &middot; 3 − &middot; 27 + 9 − −3 + + 1
3
3
2
= 10
3
�
10 23
= 1.
10 23
To ﬁnd ȳ we will be integrating with respect to y. Our area dA will have
width equal to the width of the region and height dy. The left side of the region
√
√
has the equation x = 1 − y and the right side has equation x = 1 + y, so the
√
√
√
width is (1 + y) − (1 − y) = 2 y.
We conclude that x̄ =
√
y(2 y) dy
�4 √
2 y dy
0
�4
ȳ =
4
�
√
0
y(2 y) dy
4
�
=
y 3/2 dy
2
0
0
�4
2 5/2 ��
2&middot; y �
5
0
4 5
(2 − 0)
5
27
5
=
=
=
�
4
√
2 y dy
4
�
=
0
y 1/2 dy
2
0
=
=
=
2
�4
2 3/2 ��
2&middot; y �
3
0
4
3
&middot; (2 − 0)
3
25
3
27
5
25
3
12
= 2.4.
5
The center of mass of the parabolic region is at (1, 2.4); this seems plausible
given that the top of the parabola (y = 4) will weigh more than the bottom
(y = 0).
Hence, ȳ =
=
3
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18.01SC Single Variable Calculus��
Fall 2010 ��
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