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Unformatted text preview: Centroids
Consider a region in the plane of area A. We can think of the region as a thin plate with uniform thickness
and density. The centroid of the region has coordinates (x, y). It can be found using
x= 1
A xe dA and y = A 1
A ye dA
A where (xe , ye ) is the coordinates of the centroid of the diﬀerential element of area dA.
Example 1. Find the centroid of the region bounded by the curve y = x3 and the lines y = 0 and x = 2.
Solution. We will use diﬀerential elements consisting of rectangular vertical slices of height y and width dx.
This means that variable x will be the variable of integration.
y
y = x3 8 (xe , ye )
y
0 dx x 2 x
We ﬁrst ﬁnd the area A.
2 A= dA =
A 2 x3 dx = y dx =
0 0 2 x4
4 =
0 dA 24
=4
4 Now, observe that the centroid of the diﬀerential element has coordinates (xe , ye ) where
y
xe = x and ye = .
2
Therefore,
2
2
2
2
x5
25
32
xe dA =
x y dx =
x(x3 ) dx =
x4 dx =
=
=
5 0
5
5
A
0
0
0
dA and 2 ye dA =
A 0 y
y dx =
2 2
0 y2
dx =
2 2
0 (x3 )2
dx =
2 2
0 x6
x7
dx =
2
14 dA We can now ﬁnd the coordinates of the centroid.
1
32/5
8
x=
xe dA =
=
A A
4
5 and y= We conclude that the centroid is located at the following point.
(x, y) =
Gilles Cazelais. Typeset with L T X on May 22, 2008.
A E 8 16
,
5 7 1
A ye dA =
A 2 =
0 27
64
=
.
14
7 64/7
16
=
4
7 Example 2. Find the centroid of the region bounded by the curve y = √ x and the lines x = 0 and y = 3. Solution. We will use diﬀerential elements consisting of rectangular horizontal slices of height x and width dy.
This means that variable y will be the variable of integration.
y
(xe , ye )
3 y= dy x y 0 y= √ x implies that x = y 2 . Let’s ﬁnd the area A. 3 dA =
A x 9 x Observe that A= √ 3 y 2 dy = x dy =
0 0 3 y3
3 =
0 dA 33
=9
3 Now, observe that the centroid of the diﬀerential element has coordinates (xe , ye ) where
x
2 xe = and ye = y. Therefore,
3 xe dA =
A 0 x
x dy =
2 3
0 x2
dy =
2 3
0 (y 2 )2
dy =
2 3
0 y4
y5
dy =
2
10 dA and 3 ye dA =
A 3 0 3 y(y 2 ) dy = y x dy =
0 y 3 dy =
0 y4
4 dA 3 =
0 3 =
0 35
243
=
10
10 34
81
=
.
4
4 We can now ﬁnd the coordinates of the centroid.
x= 1
A xe dA =
A 243/10
27
=
9
10 and y= We conclude that the centroid is located at the following point.
27 9
,
10 4 (x, y) = 2 1
A ye dA =
A 81/4
9
=
9
4 Let’s now consider the area between two curves. First consider the area bounded by the curves y = f (x) and
y = g(x) over a ≤ x ≤ b.
y
(xe , ye ) y = f (x) x y = g(x) a x b
dx In this case, we have
xe = x and ye = f (x) + g(x)
.
2 Therefore,
b A= (f (x) − g(x)) dx dA =
a A dA
b
A b x (f (x) − g(x)) dx xe dA = and f (x) + g(x)
2 ye dA = a A a (f (x) − g(x)) dx . dA dA Now consider the area bounded by the curves x = f (y) and x = g(y) over c ≤ y ≤ d.
y
x = g(y) x = f (y)
d
(xe , ye ) dy y
x c
In this case, we have
xe = f (y) + g(y)
2 and ye = y. Therefore,
d A= (f (y) − g(y)) dy dA =
A c
dA d xe dA =
A c f (y) + g(y)
2 d (f (y) − g(y)) dy and
A dA y (f (y) − g(y)) dy ye dA =
c dA 3 Example 3. Find the centroid of the region between the curves y = 2x and y = x2 over the interval 0 ≤ x ≤ 2.
Solution. We will use diﬀerential elements consisting of rectangular vertical slices of width dx. This means
that variable x will be the variable of integration.
y
4
y = 2x
(xe , ye )
x
y = x2 0 dx x 2 Let’s ﬁnd the area A.
2 A= (2x − x2 ) dx = dA =
A x2 − 0 2 x3
3 =4−
0 8
4
=
3
3 dA Now, observe that
2x + x2
.
2 xe = x and ye =
Therefore,
2
A 2 x (2x − x2 ) dx = xe dA = (2x2 − x3 ) dx = 0 0 2x3
x4
−
3
4 2 =
0 16
4
−4=
3
3 dA and
2 ye dA =
A 0 2x + x2
2 2 (2x − x2 ) dx =
0 4x2 − x4
dx =
2 2x3
x5
−
3
10 2 =
0 16 16
32
−
=
.
3
5
15 dA We can now ﬁnd the coordinates of the centroid.
x= 1
A xe dA =
A 4/3
=1
4/3 and y= We conclude that the centroid is located at the following point.
(x, y) = 4 1, 8
5 1
A ye dA =
A 32/15
8
=
4/3
5 Example 4. Find the centroid of the region bounded by the curves y = 3 − x, y = √ x − 1, and y = 0. Solution. We will use diﬀerential elements consisting of horizontal vertical slices of width dy. This means that
variable y will be the variable of integration.
y
y =3−x
2
y= (xe , ye ) √ x−1 1
dy
y
0 1 x 3 √
Let’s ﬁnd the point of intersection of y = 3 − x and y = x − 1.
√
3 − x = x − 1 =⇒ 9 − 6x + x2 = x − 1 =⇒ x2 − 7x + 10 = 0 =⇒ (x − 2)(x − 5) = 0
We conclude that x = 2 is the solution. Then, the point of intersection is (2, 1). Now, observe that
√
y = x − 1 =⇒ x = y 2 + 1 and y = 3 − x =⇒ x = 3 − y.
We can describe the region as the region between the curves x = y 2 + 1 and x = 3 − y over 0 ≤ y ≤ 1.
Let’s ﬁnd the area A.
1 A= 1 (3 − y) − (y 2 + 1) dy = dA =
A (2 − y − y 2 ) dy = 0 2y − 0 1 y3
y2
−
2
3 =
0 7
6 dA Now, observe that
xe = (3 − y) + (y 2 + 1)
2 and ye = y. Therefore,
1 xe dA =
A 0 1 (3 − y) + (y 2 + 1)
2 (3 − y)2 − (y 2 + 1)2
dy
2 (3 − y) − (y 2 + 1) dy =
0
dA 1 =
0 and 8 − 6y − y 2 − y 4
dy =
2 4y − 1
A =
0 67
30 1 y (3 − y) − (y 2 + 1) dy ye dA = 1 3y 2
y3
y5
−
−
2
6
10 0 (2y − y 2 − y 3 ) dy = y2 − 0 y3
y4
−
3
4 dA We can now ﬁnd the coordinates of the centroid.
x= 1
A xe dA =
A 67/30
67
=
7/6
35 and y= We conclude that the centroid is located at the following point.
67 5
,
35 14 (x, y) = 5 1
A ye dA =
A 5/12
5
=
7/6
14 1 =
0 5
12 ...
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 Spring '09
 sivaprasad

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