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
· 9 − · 81 + · 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 · 3 − · 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· 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· y �
3
0
4
3
· (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
MIT OpenCourseWare
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18.01SC Single Variable Calculus��
Fall 2010 ��
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