9.1 Centroids by Integration
9.1 Centroids by Integration Example 1, page 1 of 4
1. Locate the centroid of the plane area shown. Use a
differential element of thickness dx.
y
1
Definition of centroid coordinates
xc =
y = 3x2
yc =
dA
yel dA
dA
where (xel, yel) are the coordinates of the
centroid of the differential area
element dA.
12 ft
x
2 ft
xel dA
(1)
(2)
9.1 Centroids by Integration Example 1, page 2 of 4
2
Locate the differential element so that it extends
from an arbitrary point (x, y) on the curve to an
opposite boundary of the crosshatched region.
y
y = 3x2
(x, y)
3
Express the element area in terms of the coordinates of a
point (x, y) on the curve:
dA = height width (of rectangle)
y = height
or,
dA = y dx
x
dx = width
(3)
By choosing an element of width dx, we have also
implicitly chosen x to be the variable of integration.
x
9.1 Centroids by Integration Example 1, page 3 of 4
4
y
y = 3x2
Express the coordinates of the element
centroid in terms of the coordinates of the
point (x, y) on the curve. The x
coordinate of the element centroid is the
same as the x coordinate of the point on
the curve:
xel = x
5
Since the variable of integration is x, we
now have to express dA and yel in terms of
x (As can be seen from Eq. 4, xel already is
a function of x). The point (x, y) on the
curve satisfies
y = 3x2
(4)
Substituting the expression for y in Eq. 6
into the equations for dA (Eq. 3) and for yel
(Eq. 5) gives
Since the centroid of the differential
element is located in the center of the
element, the y coordinate of the element
centroid is
(x, y)
y
yel =
2
dA = y dx
(5)
y
yel = 2
3x2
= 2
y
x
x
2 ft
(Eq. 3 repeated)
= 3x2 dx
(xel, yel)
dx
(6)
6
(7)
(8)
Evaluate the integral in the denominator
of the equation for xc over the range (from
0 to 2):
2
dA = 3x2 dx = 8 ft2
0
(9)
9.1 Centroids by Integration Example 1, page 4 of 4
7
Evaluate the integral in the numerator of the equation for xc
over the range from 0 to 2:
2
xel dA =
0
x(3x2) dx = 12 ft3
(10)
Evaluate the integral in the numerator of the equation for yc over
the range from 0 to 2:
2
yel dA =
0
2
[ 3x ](3x2) dx = 28.8 ft3
2
(11)
Substitute the results given in Eqs. 9, 10, and 11 into the
definitions for xc and yc:
xc =
yc =
xel dA
dA
yel dA
dA
= 12
8 = 1.5 ft
Ans.
= 28.8
8 = 3.6 ft
Ans.
9.1 Centroids by Integration Example 2, page 1 of 3
2. Locate the centroid of the plane area shown, if a = 3 m
and b = 1 m. Use a differential element of thickness dy.
2
y
1
a
y = a sin(
x)
2b
3
Definition of centroid coordinates
xc =
xel dA
yc =
yel dA
Locate the differential element so that it
extends from an arbitrary point (x, y)
on the curve to an opposite boundary
of the crosshatched region.
y
(1)
dA
dA
x
(2)
dy
where (xel, yel) are the coordinates of the
centroid of the differential area
element dA.
Express the element area in terms of the
coordinates of a point (x, y) on the curve:
dA = width height (of rectangle)
(x, y)
y
y = a sin(
x)
2b
x
b
or,
dA = x dy
(3)
By choosing an element of width dy, we have
also implicitly chosen y to be the variable of
integration.
x
9.1 Centroids by Integration Example 2, page 2 of 3
4
y
x
(xel, yel)
dy
(x, y)
x
xel = 2
y = a sin( x )
2b
yel = y
5
Since the variable of integration is y, we now
have to express dA and xel in terms of y (As
can be seen from Eq. 5, yel already is expressed
as a function of y). The point (x, y) on the
curve satisfies
y = a sin (
x)
2b
Solving this equation for x gives
(4)
y
x = ( 2b ) sin-1( a )
The y coordinate of the element
centroid is the same as the y
coordinate of the point on the
curve:
a
y
Express the coordinates of the
element centroid in terms of the
coordinates of the point (x, y) on
the curve. Since the centroid of
the differential element is
located in the center of the
element, the x coordinate of the
element centroid is:
(6)
Substituting the expression for x in Eq. 6 into
the equations for dA (Eq. 3) and for xel (Eq. 4)
gives
dA = x dy
(5)
(Eq. 3 repeated)
y
= [( 2b ) sin-1( a )] dy
(7)
x
x
xel = 2
6
Substitute a = 3 m and b = 1 m, and evaluate the integral in
the denominator of the equation for xc over the range from 0
to 3 (Use the integral function on your calculator):
3
dA =
0
(
y
) sin-1( ) dy = 1.090 m2
3
(9)
(Eq. 4 repeated)
y
2b
= 12 ( ) sin-1( a )
y
= ( b ) sin-1( a )
(8)
9.1 Centroids by Integration Example 2, page 3 of 3
7
Similarly, evaluate the integral in the numerator of the equation for xc over the range
from 0 to 3 (Use the integral function on your calculator):
3
xel dA =
[
0
sin-1 (y/3)
] [(
sin-1(
y
)] dy = 0.2841 m3
3
(10)
And similarly evaluate the integral in the numerator of the equation for yc over the
range from 0 to 3 (Use the integral function on your calculator):
3
yel dA =
0
y [(
sin-1(
y
)] dy = 2.2500 m3
3
(11)
Substitute the results given in Eqs. 9, 10, and 11 into the definitions
for xc and yc:
xc =
xel dA
yc =
yel dA
dA
dA
= 0.2841
1.090 = 0.261 m
Ans.
= 2.2500
1.090 = 2.06 m
Ans.
9.1 Centroids by Integration Example 3, page 1 of 5
3. Locate the centroid of the plane area shown.
y
1
y = 4x5
y
Definition of centroid coordinates
1 in
3x2 + 12x + 1
dx
5
y = 4x
2
3x + 12x + 1
xc =
xel dA
yc =
yel dA
dA
(1)
(2)
dA
14
y
where (xel, yel) are the coordinates of the
centroid of the differential area
element dA.
2
13 in.
3
Locate the differential element so that it extends from
an arbitrary point (x, y) on the curve to an opposite
boundary of the crosshatched region.
14 in.
(x, y)
Express the element area in terms of the coordinates
of a point (x, y) on the curve:
y
dA = height width (of rectangle)
or,
1 in.
dA = (14
y) dx
(3)
x
x
By choosing an element of width dx, we have also
implicitly chosen x to be the variable of integration.
x
9.1 Centroids by Integration Example 3, page 2 of 5
4
Express the coordinates of the
element centroid in terms of the
coordinates of the point (x, y) on
the curve. The x coordinate of the
element centroid is the same as the
x coordinate of the point on the
curve:
5
y
dx
y = 4x5
(xel, yel)
xel = x
(4)
14
Since the centroid of the differential
element is located in the center of
the element, the y coordinate of the
element centroid is:
yel = y +
14
14
dA = (14
y
2
= [14
14 in
2
(x, y)
y = 4x
3x + 12x + 1
x
x
(4x5
= ( 4x5 + 3x2
5
= 4x
y
2
y) dx
y
yel = 7 + 2
4x5
=7+
(5)
5
3x2 + 12x + 1
(6)
Substituting the expression for y from Eq. 6 into the
equations for dA (Eq. 3) and for yel (Eq. 5) gives
y
y
y
=7+ 2
Since the variable of integration is x, we now have
to express dA and yel in terms of x (As can be seen
from Eq. 4, xel already is expressed as a function of
x). The point (x, y) on the curve satisfies
(Eq. 3 repeated)
3x2 + 12x + 1)] dx
12x + 13) dx
(7)
(Eq. 5 repeated)
3x2 + 12x + 1
2
2
3x + 12x + 15
2
(8)
9.1 Centroids by Integration Example 3, page 3 of 5
6
y
What if we had chosen a differential element dy wide instead of
dx? The figure shows that now the element area is
dA = x dy
Since dy is the variable of integration, we must express x as a
function of y. But x and y are related by
y = 4x5
3x2 + 12x + 1
and this equation is difficult to invert, that is, to solve for x as a
function of y. So it is much easier to use a differential element
dx wide, because then we don't have to solve for x as a function
of y.
7
dy
y = 4x5
The general conclusion to draw is that whether we should choose
a differential element dx or dy wide depends on whether the
equation defining the boundary of the region can be more easily
written as a function of x or as a function of y.
x
x
3x2 + 12x +1
9.1 Centroids by Integration Example 3, page 4 of 5
y
8
dx
Return now to a differential element dx wide.
9 Evaluate the integral in the denominator of the equation for xc over
the range from 0 to 1 (Use the integral function on your calculator):
1
5
y = 4x
dA =
2
3x + 12x + 1
0
( 4x5 + 3x2
12x + 13) dx = 7.333 in2
(9)
10 Similarly, evaluate the integral in the numerator of the equation for xc
over the range from 0 to 1 (Use the integral function on your calculator):
1
xel dA =
0
x ( 4x5 + 3x2
12x + 13) dx = 2.679 in3
(x, y)
11 And similarly evaluate the integral in the numerator of the equation for yc
over the range from 0 to 1 (Use the integral function on your calculator):
1
yel dA =
x
=
1 in.
0
5
[ 4x
69.849 in3
3x2 + 12x + 15 ]( 4x5 + 3x2
2
12x + 13) dx
(11)
(10)
9.1 Centroids by Integration Example 3, page 5 of 5
12 Substitute the results given in Eqs. 9, 10, and 11 into the
definitions for xc and yc:
xc =
xel dA
yc =
yel dA
dA
dA
= 2.679
7.333 = 0.37 in.
Ans.
= 69.849
7.333 = 9.52 in.
Ans.
9.1 Centroids by Integration Example 4, page 1 of 4
4. Locate the centroid of the plane area shown.
1
Definition of centroid coordinates
y
xc =
xel dA
yc =
yel dA
(1)
dA
0.5 m
xy = 1
(2)
dA
where (xel, yel) are the coordinates of the centroid of the
differential area element dA.
y
2m
xy = 1
0.5 m
x
dy
2m
(x, y)
2m
2
Locate the differential element so that it
extends from an arbitrary point (x, y)
on the curve to an opposite boundary.
y
0.5 m
x
x
9.1 Centroids by Integration Example 4, page 2 of 4
y
3
Express the element area in terms of the coordinates of a point
(x, y) on the curve:
xy = 1
dA = height width (of rectangle)
or,
(xel, yel)
dA = x dy
(3)
(x, y)
By choosing an element of width dy, we have also implicitly
chosen y to be the variable of integration.
dy
y
4
x
x
Express the coordinates of the element centroid in terms of the
coordinates of the point (x, y) on the curve. Since the centroid
of the differential element is located in the center of the element,
the x coordinate of the element centroid is
xel = x2
(4)
The y coordinate of the element is the same as the y coordinate
of the point on the curve:
yel = y
(5)
9.1 Centroids by Integration Example 4, page 3 of 4
5
Since the variable of integration is y, we now have to
express dA and xel in terms of y (As can be seen from
Eq. 5, yel already is a function of y). The point (x, y) on
the curve satisfies
y
xy = 1
xy =1
Solving for x gives
x = y1
dy
(6)
2m
Substituting the expression for x in Eq. 6 into the
equations for dA (Eq. 3) and for xel (Eq. 4) gives
dA = x dy
= 1y dy
(Eq. 3 repeated)
0.5 m
(7)
xel = x2
1/y
= 2
(Eq. 4 repeated)
= 1
2y
(8)
x
6
Evaluate the integral in the denominator of the equation for
xc over the range from 0.5 to 2:
2
dA =
0.5
1 dy = 1.3863 m2
y
(9)
9.1 Centroids by Integration Example 4, page 4 of 4
7
Evaluate the integral in the numerator of the equation for xc
over the range from 0.5 to 2:
2
xel dA =
( 1 )( 1y ) dy = 0.7500 m3
2y
0.5
(10)
8 Evaluate the integral in the numerator of the equation for yc
over the range from 0.5 to 2:
2
yel dA =
0.5
y ( 1y ) dy = 1.5 m3
10
What if we had chosen a vertical element rather
than a horizontal element? The figure below
shows that we would have to define the element
areas in two equations, one for the region where y
is constant (y = 2 m), and one for the region where
y varies. We would also have to evaluate two
integrals, one over each region. Thus using a
vertical element rather than a horizontal element
would almost double the amount of work required.
Nevertheless, it would give us the correct answer.
y
(11)
(x, 2)
xy = 1
9
Substitute the results given in Eqs. 9, 10, and 11 into the
definitions for xc and yc:
xc =
xel dA
yc =
yel dA
dA
dA
(x, y)
2m
0.7500 = 0.541 m
= 1.3863
Ans.
1.5 = 1.082 m
= 1.3863
Ans.
x
9.1 Centroids by Integration Example 5, page 1 of 3
5. Locate the centroid of the plane area shown.
y
1
x(13
y=
6
Definition of centroid coordinates
xc =
x)
yc =
xel dA
dA
yel dA
2
(1)
Locate the differential element so that it
extends from an arbitrary point on the
lower curve to a point directly above on
the upper curve.
y
(2)
dA
6m
dx
y = x2 +
2m
14
11x where (xel, yel) are the coordinates of the
centroid of the differential area element dA.
3
(x, y2)
y2 y1
x
3
1m
Express the element area in terms of the
coordinates of a point (x, y) on the curve:
(x, y1)
4m
y1
dA = height width (of rectangle)
x
x
or,
dA = (y2 y1) dx
(3)
By choosing an element of width dx, we have
also implicitly chosen x to be the variable of
integration.
9.1 Centroids by Integration Example 5, page 2 of 3
4
Express the coordinates of the element centroid in terms
of the coordinates of the point (x, y) on the curve. The
x coordinate of the element is the same as the x
coordinate of the point on the curve:
xel = x
(4)
Since the centroid of the differential element is located
in the center of the element, the y coordinate of the
element centroid is average of the y coordinates of the
points on the top and bottom curves.
yel =
y2 + y1
2
(5)
y
5
Since the variable of integration is x, we now have to express
dA and yel in terms of x (As can be seen from Eq. 4, xel already
is a function of x). The y coordinates of the top and bottom of
the element satisfy
x(13 x)
6
14 11x
y1 = x2 +
3
x(13 x)
14 11x
y2 y1 = [
] [x2 +
]
3
6
2
= 7x + 35x 28
(6)
6
y2 =
Substituting the expression for y in Eq. 6 into the equations for
dA (Eq. 3)
dx
(x, y2)
dA = (y2 y1) dx
=
(xel, yel)
y2 y1
2
y2 y1
y1
x
x = xel
28
dx
(7)
and for yel (Eq. 5) gives
y2 + y1
2
[x(13 x)/6] + [x2 + (14
=
2
(5x2 9x + 28)
=
12
yel =
(x, y1)
7x2 + 35x
6
(Eq. 3 repeated)
(Eq. 5 repeated)
11x)/3]
(8)
9.1 Centroids by Integration Example 5, page 3 of 3
6
4
dA =
7
y
Evaluate the integral in the denominator of the equation for xc over
the range from 1 to 4:
1
7x2 + 35x
6
28
y=
dx = 5.2500 m2
(9)
x(13
6
x)
Evaluate the integral in the numerator of the equation for xc over the
range from 1 to 4:
4
xel dA =
1
x[
7x2 + 35x
6
28
] dx = 13.1250 m3
y = x2 + 14 3 11x
dx
(10)
x
1m
4m
8
Evaluate the integral in the numerator of the equation for yc
over the range from 1 to 4:
4
yel dA =
1
[
(5x2
= 17.0625 m3
2
9x + 28)
][ 7x + 35x
12
6
9
Substitute the results given in Eqs. 9, 10, and 11 into the
definitions for xc and yc:
28 ] dx
xc =
xel dA
yc =
yel dA
= 13.125
5.25 = 2.50 m
dA
Ans.
(11)
dA
= 17.0625
5.25 = 3.25 m
Ans.
9.1 Centroids by Integration Example 6, page 1 of 3
6. Locate the centroid of the plane area shown.
1
Definition of centroid coordinates
y
xc =
x=4
y2
x = 3y
yc =
1m
xel dA
(1)
dA
yel dA
(2)
dA
x
3m
2
where (xel, yel) are the coordinates of
the centroid of the differential area
element dA.
1m
Locate the differential element so that it
extends from an arbitrary point (x, y)
on the curve to an opposite boundary.
3
Express the element area in terms of the
coordinates of a point (x, y) on the curve:
dA = width
y
x=4
(x1, y)
y2
height (of rectangle)
or,
(x2, y)
dA = (x2
x1) dy
(3)
x = 3y
y
dy
x
x1
x2
x1
By choosing an element of height dy, we
have also implicitly chosen y to be the
variable of integration.
9.1 Centroids by Integration Example 6, page 2 of 3
4
5
Express the coordinates of the element centroid in terms
of the coordinates of the point (x, y) on the curve. Since
the centroid of the differential element is located in the
center of the element, the x coordinate of the element
centroid is
xel =
=
x2
x2
x1
2
Since the variable of integration is y, we now have to
express dA and xel in terms of y (As can be seen from
Eq. 5, yel already is a function of y). The point (x, y)
on the curve satisfies
x1 = 3y
+ x1
x1
x2 = 4
(4)
2
x2
The y coordinate of the element is the same as the y
coordinate of the point on the curve:
yel = y
y2
= y2
(5)
y
x=4
(xel, yel)
(x1, y)
y2
xel =
dy
y
x
x1
=
=
x2
x1
3y + 4) dy
(6)
x1 in Eq. 6 into the
(Eq. 3 repeated)
(7)
and for xel (Eq. 4) gives
x = 3y
x1
3y + 4
x1) dy
= ( y2
(x2, y)
2
(3y)
Substituting the expression for x2
equations for dA (Eq. 3) gives
dA = (x2
x2
y2)
x1 = (4
x2
x1
2
y2) + (3y)
2
y2 + 3y + 4
2
(Eq. 4 repeated)
(4
(8)
9.1 Centroids by Integration Example 6, page 3 of 3
6
Evaluate the integral in the denominator of the equation for
xc over the range from 0 to 1 (Use the integral function on
your calculator):
1
dA =
0
( y2
3y + 4) dy = 2.1667 m2
y
x=4
y2
x = 3y
(9)
1m
x
7
Evaluate the integral in the numerator of the equation for xc over the range
from 0 to 1 (Use the integral function on your calculator):
9
xel dA =
1
0
[
y2 + 3y + 4
] [ y2
2
3y + 4] dy = 5.2667 m3
Substitute the results given in Eqs. 9, 10, and 11
into the definitions for xc and yc:
(10)
xc =
8
dA
Evaluate the integral in the numerator of the equation for yc over
the range from 0 to 1 (Use the integral function on your calculator):
yc =
1
yel dA =
0
xel dA
yel dA
dA
y( y2
3y + 4) dy = 0.75 m3
(11)
5.2667 = 2.43 m
= 2.1667
Ans.
0.75 = 0.346
= 2.1667
Ans.
9.1 Centroids by Integration Example 7, page 1 of 3
7. Locate the centroid of the plane area shown. Use a
differential element of thickness dx.
y
1 Definition of centroid coordinates
y = hb x
xc =
yc =
h
x
b
3
xel dA
dA
yel dA
dA
(1)
(2)
2
Locate the differential element so that it
extends from an arbitrary point (x, y)
on the curve to an opposite boundary.
where (xel, yel) are the coordinates of
the centroid of the differential area
element dA
y
Express the element area in terms of the
coordinates of a point (x, y) on the curve:
dA = height width (of rectangle)
(x, y)
y = hb x
or,
dA = y dx
y
(3)
By choosing an element of width dx, we
have also implicitly chosen x to be the
variable of integration.
x
dx
x
9.1 Centroids by Integration Example 7, page 2 of 3
4
Express the coordinates of the element centroid in terms of the
coordinates of the point (x, y) on the curve. The x coordinate
of the element is the same as the x coordinate of the point on
the curve:
xel = x
(4)
y
y=
Since the centroid of the differential element is located in the
center of the element, the y coordinate of the element centroid
is
y
yel = 2
h
x
b
(x, y)
(5)
(xel, yel)
y
5
Since the variable of integration is x, we now have to express dA and
yel in terms of x (As can be seen from Eq. 4, xel already is a function
of x). The point (x, y) on the curve satisfies
h
y=
x
b
(6)
Substituting the expression for y in Eq. 6 into the equations for dA
(Eq. 3) and for yel (Eq. 5) gives
h
x) dx
b
h
yel =
x
2b
dA = (
(7)
(8)
x
dx
x
y
2
9.1 Centroids by Integration Example 7, page 3 of 3
6
b
dA =
7
0
h x dx = bh
2
b
dx
(9)
Evaluate the integral in the numerator of the equation for
xc over the range from 0 to b:
b
xel dA =
8
y
Evaluate the integral in the denominator of the equation for
xc over the range from 0 to b:
0
2
x( hb x) dx = b 3h
b
(10)
Evaluate the integral in the numerator of the equation for
yc over the range from 0 to b:
b
yel dA =
0
2
h x)( h x) dx = bh
( 2b
6
b
x
(11)
9
Substitute the results given in Eqs. 9, 10, and 11
into the definitions for xc and yc:
xc =
xel dA
yc =
yel dA
dA
dA
=
b2h/3
= 2b
3
bh/2
Ans.
=
bh2/6 h
= 3
bh/2
Ans.
9.1 Centroids by Integration Example 8, page 1 of 3
8. Locate the centroid of the plane area shown. Use a
differential element of thickness dy.
y
1
b
x = a[1 (
Definition of centroid coordinates
y 2
)]
b
xc =
xel dA
yc =
yel dA
dA
(1)
(2)
dA
x
Locate the differential element so that it
extends from an arbitrary point (x, y) on
the curve to an opposite boundary of the
crosshatched region.
2
y
(a, y)
where (xel, yel) are the coordinates of
the centroid of the differential area
element dA.
a
(x, y)
3
Express the element area in terms of the
coordinates of a point (x, y) on the curve:
dA = width
y
x = a[1 (
dy
y 2
)]
b
height (of rectangle)
or,
x
dA = (a
x) dy
(3)
By choosing an element of width dy, we have
also implicitly chosen y to be the variable of
integration.
a
x
a
x
9.1 Centroids by Integration Example 8, page 2 of 3
4
y
Express the coordinates of the element centroid in terms of the
coordinates of the point (x, y) on the curve. Since the centroid
of the differential element is located in the center of the
element, the x coordinate of the element centroid is
xel =
(a
=
(xel, yel)
(a, y)
x)
2 +x
a+x
2
(4)
y
The y coordinate of the element is the same as the y coordinate
of the point on the curve:
yel = y
dy
(x, y)
x = a[1 (
y 2
)]
b
(5)
Since the variable of integration is y, we now have to express dA and
xel in terms of y (As can be seen from Eq. 5, yel already is a function
of y). The point (x, y) is on the curve so it satisfies
x = a[1 (
y 2
)]
b
(6)
Substituting the expression for x in Eq. 6 into the equation for dA
(Eq. 3) gives
dA = (a
x dy
= (a
a[1 (
y
= a( )2 dy
b
(Eq. 3 repeated)
y 2
) ] dy
b
(7)
(a x)
2
a
x
x
x
a
5
Substituting the expression for x in Eq. 6 into
the equation for xel (Eq. 4) gives
a+x
2
a + a[1 (y/b)2]
=
2
a[2 (y/b)2]
=
2
xel =
(Eq. 4 repeated)
(8)
9.1 Centroids by Integration Example 8, page 3 of 3
6 Evaluate the integral in the denominator of the equation for x
c over the range from 0 to b:
dA =
7
b
0
a(
y 2
ab
) dy = 3
b
y
(9)
dy
Evaluate the integral in the numerator of the equation for xc
over the range from 0 to b:
xel dA =
b
0
[
[2a
b
y
(a/b2) y2]
][a( )2] dy
b
2
x
= 7 a2b
30
8
(10)
Evaluate the integral in the numerator of the equation for yc over
the range from 0 to b:
b
yel dA =
0
y
ab2
y[a b )2] dy = 4
9
Substitute the results given in Eqs. 9, 10, and 11
into the definitions for xc and yc:
xc =
yc =
(11)
xel dA
dA
yel dA
dA
=
7a
(7/30)a2b
=
10
ab/3
Ans.
=
ab2/4
3b
=
ab/3
4
Ans.
9.1 Centroids by Integration Example 9, page 1 of 3
9. A sign is made of 0.5 in. thick steel plate in the shape shown.
Determine the reactions at supports B and C.
1
Free-body diagram of plate
y
Specific weight of steel = 490 lb/ft3
2
By
B
x = 50 + (10) sin y
24
Bx
The weight acts through the
centroid of the plate.
B
x = 50 + (10) sin
y
24
72 in.
72 in.
x
C
50 in.
3
Cx
Equilibrium equations for the plate
+
Fx = 0: Bx + Cx = 0
Fy = 0: By
+
+
Weight, W
xc
MB = 0: (72 in.)Cx
4
(1)
W=0
(2)
xcW = 0
(3)
x
C
Calculate the weight of the plate
by calculating the area. Also
calculate the distance xc to the
center of gravity.
5
Definition of centroid coordinate
xc =
xel dA
dA
(1)
where xel is the coordinate of the
differential element dA.
9.1 Centroids by Integration Example 9, page 2 of 3
6
Locate the differential element so that it
extends from an arbitrary point (x, y)
on the curve to an opposite boundary.
y
y
(xel, yel)
x = 50 + (10) sin 24
dy
7
Express the element area in terms of the coordinates of a
point (x, y) on the curve:
dA = width height (of rectangle)
or,
(x, y)
dA = x dy
(4)
By choosing an element of width dy, we have also implicitly
chosen y to be the variable of integration.
y
9
x
2
8
x
x
dA = x dy
Express the coordinate of the x-element centroid
in terms of the coordinates of the point (x, y) on
the curve. Since the centroid of the differential
element is located in the center of the element, the
x coordinate of the element centroid is
xel = x2
Since the variable of integration is y, we now have to express
dA and xel in terms of y (As can be seen from the equation of
the curve, x already is a function of y). Therefore
(5)
= [50 + (10) sin
y
dy
24
xel = x2
50 + (10) sin( y/24)
=
2
y
= 25 + 5 sin
24
(7)
(8)
9.1 Centroids by Integration Example 9, page 3 of 3
10 Evaluate the integral in the denominator of the equation
for xc over the range from 0 to 72 (Use the integral
function on your calculator):
dA =
(9)
y
72
0
[50 + (10) sin
y
dy = 3753 in2
24
12 Substitute the results given in Eqs. 9 and 10 into the
definition for xc:
xc =
72 in.
x = 50 + (10) sin y
24
xel dA
dA
=
99,439
3753 = 26.5 in.
13 Thus the weight is
W = area
thickness
specific weight
1 ft )3
= (3753 in2) (0.5 in.) (490 lb/ft3) ( 12
in.
x
11 Evaluate the integral in the numerator of the equation for
xc over the range from 0 to 72 (Use the integral function
on your calculator):
72
xel dA =
0
[25 + 5 sin y ][50 + (10) sin y ]dy
24
24
= 99,439 in3
(10)
= 532.1 lb
Substituting xc = 26.5 in. and W = 532.1 lb in the
equilibrium equations, Eqs. 1, 2, and 3, and solving gives
Bx = 196 lb
Ans.
By = 532 lb
Ans.
Cx = 196 lb
Ans.
9.1 Centroids by Integration Example 10, page 1 of 3
10. Locate the centroid of the wire shown.
y
1
y
Definition of centroid coordinates
3m
xc =
y = 2x2
yc =
xel dL
dL
yel dL
y = 2x2
(xel, yel)
(1)
(2)
dL
18 m
dL
y
where (xel, yel) are the coordinates of
the centroid of the differential length
element dL.
x
x
3
x
Since dy/dx is slightly easier to compute than
dx/dy, express dL in terms of dy/dx and dx:
dL =
[(dx)2 + (dy)2]
2
dL
dx
=
2
2
[1 + ( dy
dx ) ] (dx)
=
2
[1 + ( dy
dx ) ] dx
dy
Locate the differential element at an
arbitrary point (x, y) on the curve and
express the centroidal coordinates of
the element in terms of x and y.
(5)
By expressing the length dL in terms of dx, we
have also implicitly chosen x to be the variable
of integration.
xel = x
(3)
yel = y
(4)
9.1 Centroids by Integration Example 10, page 2 of 3
4
Since the variable of integration is x, we now have to express
dL and yel in terms of x (As shown in Eq. 3, xel already is
expressed as a function of x, since xel = x). The point (x, y) on
the curve satisfies:
y = 2x2
y
(6)
y = 2x2
(x, y)
Thus
dy = 4x
dx
(7)
dL
Substituting these expressions for y and dy/dx into the
equations for yel (Eq. 4) and dL (Eq. 5) gives
yel = y
x
= 2x2
dL =
=
3m
(8)
2
[1 + ( dy
dx ) ] dx
[1 + (4x)2] dx
(9)
5
Evaluate the integral in the denominator of the equation for
xc over the range from 0 to 3:
3
dL =
0
[1 + (4x)2] dx = 18.46 m
(10)
9.1 Centroids by Integration Example 10, page 3 of 3
6
Evaluate the integral in the numerator of the equation for xc over the range
from 0 to 3:
xel dL =
7
3
0
(11)
Evaluate the integral in the numerator of the equation for yc over
the range from 0 to 3:
3
yel dL =
8
x [1 + (4x)2] dx = 36.36 m2
0
(2x2) [1 + (4x)2] dx = 163.11 m2
(12)
Substitute the results given in Eqs. 10, 11, and 12 into the definitions for xc
and yc:
xc =
yc =
xel dL
dL
yel dL
dL
36.36 = 1.97 m
= 18.46
Ans.
= 163.11
18.46 = 8.84 m
Ans.
9.1 Centroids by Integration Example 11, page 1 of 3
11. Locate the centroid of the wire shown.
1 Definition of centroid coordinates
y
x = 300[1
y 4
(
)]
200
xc =
yc =
xel dL
(1)
dL
yel dL
(2)
dL
200 mm
where (xel, yel) are the coordinates of the
centroid of the differential length element dL.
x
y
300 mm
dL
x = 300[1
2
Locate the differential element at an arbitrary
point (x, y) on the curve and express the
centroidal coordinates of the element in terms
of x and y.
xel = x
(3)
yel = y
(4)
(
y 4
)]
200
(xel, yel)
y
x
x
9.1 Centroids by Integration Example 11, page 2 of 3
y
4
dL
x = 300[1
(
y 4
)]
200
Since the variable of integration is y, we now have to
express dL and xel in terms of y (As shown in Eq. 4, yel
already is expressed as a function of y, since yel = y). The
coordinates of the point (x, y) on the curve satisfy
x = 300[1
(xel, yel)
(
y 4
)]
200
(6)
Thus
y
dx
= 300[
dy
x
x
3
Since dx/dy is slightly easier to compute than dy/dx, express
dL in terms of dx/dy and dy:
dL
dy
dL = [(dx)2 + (dy)2]
dx
dx 2
2
= [(
) + 1] (dy)
dy
= [( dx )2 + 1] dy
(5)
dy
By expressing the length dL in terms of dy, we have also
implicitly chosen y to be the variable of integration.
= 7.5
4y3
2004
]
10-7 y3
(7)
Substituting these expressions for x and dx/dy into the
equations for xel (Eq. 3) and dL (Eq. 5) gives
xel = x
= 300[1
dL =
=
(
y 4
)]
200
(8)
[( dx )2 + 1] dy
dy
[(
.5
10-7 y3)2 + 1] dy
(9)
9.1 Centroids by Integration Example 11, page 3 of 3
5
Evaluate the integral in the denominator of the equation for xc over the
range from 0 to 200 (Use the integral function on your calculator):
200
dL =
[(
0
.5
10-7 y3)2 + 1] dy = 407.4 mm
y
dL
x = 300[1
(10)
(
y 4
)]
200
200 mm
6
200
xel dL =
0
y
300[1 ( 200 )4] [(
= 75 209.6 m2
.5
x
10-7 y3)2 + 1] dy
(11)
8
7
(x, y)
Evaluate the integral in the numerator of the equation for xc over the
range from 0 to 200 (Use the integral function on your calculator):
Evaluate the integral in the numerator of the equation for yc over
the range from 0 to 200 (Use the integral function on your
calculator):
200
yel dL =
0
y [(
.5
10-7 y3)2 + 1] dy = 54 861.7 m2 (12)
Substitute the results given in Eqs. 10, 11, and 12 into
the definitions for xc and yc:
xc =
yc =
xel dL
dL
yel dL
dL
209.6
= 75
407.4 = 184.6 mm
Ans.
861.7 = 134.7 mm
= 54407.4
Ans.
9.1 Centroids by Integration Example 12, page 1 of 3
12. The rod is bent into the shape of a circular arc.
Determine the reactions at the support A.
4
0.2 lb/ft
Equations of equilibrium for the rod.
+
Fx = 0: Ax = 0
(1)
3 ft
Free-body diagram of rod
2
W
y
xc
If L represents the length
of the rod, the weight W
is (0.2 lb/ft)L.
5
3 ft
20°
(0.2)L = 0
x
3 The weight of the rod acts
through the centroid (center of
gravity), located by xc.
(3)
The length L of the rod can be calculated without
using an integral:
(radius)
20 )( /180°)
= 8.378 ft
6
(2)
(0.2 lb/ft)(L)(xc) = 0
= (180
MA
Ay
MA = 0: MA
Ans.
L = (angle in radians)
0.2 lb/ft
Ax
+
1
Fy = 0: Ay
+
Therefore Ax = 0
20°
A
3 ft
(4)
Substituting L = 8.378 ft into Eq. 2 and solving
gives
Ay = 1.68 lb
Ans.
9.1 Centroids by Integration Example 12, page 2 of 3
7
To solve for the moment MA in Eq. 3 we have to
calculate the horizontal coordinate of the centroid:
xel dL
xc =
9
(5)
dL
where xel is the horizontal coordinate of the centroid
of the differential length element dL.
Express the element length in terms of the polar
coordinate angle
dL = (3 ft) d
(6)
10 Express the horizontal coordinate of the element in
terms of .
8
y
dL
(xel, yel)
Locate the differential
element at an arbitrary point
on the curve.
xel = 3 ft + (3 ft) cos
(7)
11 We do not need to evaluate the integral in the
denominator of Eq. 5 for xc, since we already know
that
d
3 ft
20°
A
(3 ft) cos
3 ft
xe1
dL = L = 8.378 ft
x
by Eq. 4.
(8)
9.1 Centroids by Integration Example 12, page 3 of 3
12 To evaluate the integral in the numerator in the
expression for xc, we note the limits of
integration.
y
radians
13
x
20°
xel dL =
(3 + 3 cos )(3)d = 22.055 ft2
(9)
20
180
Substitute the results given in Eq. 8 and 9 into the
definition of xC:
180°
xc =
xel dL
dL
= 22.055
8.378 = 2.632 ft
(10)
Substituting this result in Eq. 3 and solving for MA gives
MA = 4.41 lb·ft
Ans.
9.1 Centroids by Integration Example 13, page 1 of 3
13. a) Locate the centroid of the Gateway Arch in St.
Louis, Missouri, USA. b) During the pre-dawn hours of
September 14, 1992, John C. Vincent of New Orleans,
Louisiana, USA, climbed up the outside of the Arch to the
top by using suction cups and then parachuted to the
ground. Estimate the length of his climb.
1
xel dL
xc =
(1)
dL
yel dL
yc =
Approximate equation of centerline:
y = 639.9 ft
Definition of centroid coordinates
(2)
dL
(68.78 ft) cosh[(0.01003 ft-1)x]
where (xel, yel) are the coordinates of the centroid of the
differential length element dL.
y
Because of symmetry about the y-axis, the centroid must lie
on the y-axis, that is,
xc = 0
2
625 ft
x
299 ft
299 ft
Ans.
To avoid cluttering the equations with so many digits, define
a = 639.9 ft
(3)
b = 68.78 ft
(4)
c = 0.01003 ft-1
(5)
Then the equation of the arch becomes
y = 639.9 ft
=a
(68.78 ft) cosh[(0.01003 ft-1)x]
b cosh(cx)
(6)
9.1 Centroids by Integration Example 13, page 2 of 3
y
3
dL
Locate the differential element at a general point (x, y)
on the curve and express the y-centroidal coordinate of
the element in terms of the point on the curve.
yel = y
y=a
b cosh (cx)
y
5
(7)
Since the variable of integration is x, we now have to
express dL and yel in terms of x. The point (x, y) on the
curve satisfies
y=a
b cosh (cx)
(9)
x
Thus
4 Since dy/dx is easier to compute than dx/dy, express dL in
terms of dy/dx and dx:
dL
dy
dL = [(dx)2 + (dy)2]
dx
= [1+(
dy 2
) ] (dx)2
dx
dy 2
) ] dx
(8)
dx
By expressing the length dL in terms of dx, we have also
implicitly chosen x to be the variable of integration.
= [1 + (
dy
= bc sinh (cx)
dx
(10)
dy
Substituting these expressions for y and
into Eq. 7 for
dx
yel and Eq. 8 for dL gives
yel = y
=a
dL =
=
b cosh (cx)
[1+(
(11)
dy 2
) ] dx
dx
[1 + ( bc sinh (cx))2] dx
(12)
9.1 Centroids by Integration Example 13, page 3 of 3
y=a
b cosh (cx)
y
6
Evaluate the integral in the denominator of the equation for yc over the
range from 299 ft to +299 ft (Use the integral function on a calculator):
299
dL =
dL
-299
[1 + ( bc sinh (cx))2] dx = 1,476 ft
(13)
Similarly evaluating the numerator of the equation for yc gives
299
yel dL =
299 ft
b cosh (cx)) [1 + ( bc sinh (cx))2] dx
-299
= 439,685 ft2
x
299 ft
(a
(14)
Substitute the results given in Eqs. 13 and 14 into the definitions for yc.
yc =
yel dL
dL
=
439,685
= 298 ft
1,476
Ans.
Since Mr. Vincent only climbed half of the total arc length, we must
divide the length given in Eq. 13 by 2:
Mr. Vincent's climb =
1,476
= 738 ft
2
Ans.
(Actually he climbed a little more than 738 ft, since he climbed on the
outside surface, not on the centerline of the cross section.)
9.1 Centroids by Integration Example 14, page 1 of 4
14. Locate the centroid of the cone shown.
y
Radius = 2 m
1
Definition of centroid coordinates
xc =
3m
yc =
xel dV
dV
yel dV
(1)
(2)
dV
O
z
x
zc =
zel dV
dV
(3)
where (xel, yel) are the coordinates of the centroid of
the differential element dV.
By symmetry, the centroid must lie on the y-axis, so
xc = 0
Ans.
zc = 0
Ans.
The centroidal coordinate yc remains to be calculated.
9.1 Centroids by Integration Example 14, page 2 of 4
y
2 Use a differential element in the form of a
disk of thickness dy. The volume of the
disk is
3
B
dV = (area of circular base)
thickness of disk
= r2 dy
P(x, y, 0)
dy
(4)
By choosing an element of width dy, we
have also implicitly chosen y to be the
variable of integration.
A
The boundary of the disk
intersects the xy plane at an
arbitrary point P(x, y, 0).
r
y
(0, yel, 0)
O
z
x
4 The y coordinate of the
element centroid equals the
y coordinate of the point P
in the xy plane.
yel = y
(5)
9.1 Centroids by Integration Example 14, page 3 of 4
5 The distance from the y axis to
the boundary of the disk is x, so
r=x
6
2m
B
(6)
A
Since the variable of integration is y, we now have to
express dV and x in terms of y.
By similar triangles OCP and OBA, we have
y
x
C
B
P
Thus
y
P (x, y, 0)
x
C
3m
A
2m
2y
x= 3
dy
y
Thus the volume of the differential element in Eq. 4
becomes
x
O
(7)
2y
Substituting x = 3 into Eq. 6 gives
2y
r= 3
(8)
O
3m
r
y
x
2 = 3
dV = r2 dy
z
=
7
Evaluate the integral in the denominator of the
equation for yc over the range from 0 to 3:
3
dV =
0
2y 2
) dy = 4 m3
3
(10)
2y 2
) dy
3
(9)
9.1 Centroids by Integration Example 14, page 4 of 4
8
Evaluate the integral in the numerator of the equation for yc:
3
yel dV =
0
y[
2y 2
) ] dy = 9 m4
3
(11)
Using the results given by Eqs. 10 and 11 in the definition of yc yields
yc =
yel dV
dV
=
9
= 2.25 m
4
Ans.
9.1 Centroids by Integration Example 15, page 1 of 4
15. Locate the centroid of the volume shown.
y
One-eighth of a
sphere of radius "a"
1
Definition of centroidal coordinates
xc =
a
yc =
xel dV
dV
yel dV
(1)
(2)
dV
x
z
zc =
yel dV
dV
(3)
where (xel, yel, zel) are the coordinates of the centroid
of the differential volume-element dV.
Because of symmetry, xc = yc = zc, so we only have
to compute one centroidal distance. Let's arbitrarily
choose to compute zc.
9.1 Centroids by Integration Example 15, page 2 of 4
2
The differential element is one-quarter of a disk of thickness dz.
The volume of the disk is
dV = (one-fourth of the area of a circle)
thickness of disk
2
= r dz
4
(4)
Note that by choosing an element of width dz, we have also
implicitly chosen z to be the variable of integration.
4
y
The z coordinate of the element
centroid equals the z-coordinate of
the point P in the yz-plane.
zel = z
z
P(0, y, z)
3
The boundary of the quarter disk
intersects the yz plane at an
arbitrary point P(0, y, z).
(xel, yel, zel)
r
x
dz
z
(5)
9.1 Centroids by Integration Example 15, page 3 of 4
5
The distance from the z axis to the
boundary of the disk is y so
r=y
6
(6)
Since the variable of integration is z, we now have to express y
and dV in terms of z. Because point P lies on a circle of radius
"a" in the yz plane, y and z must satisfy the equation of a
circle
y2 + z2 = a2
y
Solving this for y gives
z
a2
y=
P (0, y, z)
z2
Substituting y =
a2
(7)
z2 into Eq. 6 gives
r=y
a
y
a2
=
x
r
(8)
Thus the volume of the differential element in Eq. 4 becomes
dV =
dz
z
z2
r2
dz
4
=
[ a2 z2 ]2
dz
4
=
a2 z2)
dz
4
(9)
9.1 Centroids by Integration Example 15, page 4 of 4
7 Evaluate the integral in the denominator of the
equation for yc over the range from 0 to a:
a2 z2)
a3
dz =
4
6
a
dV =
0
y
(10)
8 Evaluate the integral in the numerator of the equation
for zc over the range from 0 to a:
a
zel dV =
0
z[
4
a2 z2)
] dz = 16a
4
x
(11)
a
dz
9 Using the results given by Eqs. 10 and 11 in the
definition of zc yields
zc =
zel dV
dV
=
a4/16
= 3a
8
a3/6
z
Ans.
Thus from symmetry,
xc = yc = 3a
8
Ans.
9.1 Centroids by Integration Example 16, page 1 of 3
16. Determine the x coordinate of the centroid of the solid
shown. The solid consists of the portion of the solid of
revolution bounded by the xz and yz planes.
Definition of centroidal coordinates
1
xc =
y
a
xel dV
(1)
dV
where xel is the coordinate of the centroid
of the differential volume-element dV.
y
x
b
z
x = a[1
( bz )2]
dx
r
(This curve is rotated about the
x-axis to generate the solid.)
2
The differential element is one-half of a disk of thickness dx.
The volume of the disk is
dV = (one-half of the area of a circle)
=
r2
dx
2
x
x = a[1
x
z
P(x, 0, z)
( bz )2]
thickness of disk
(2)
By choosing an element of width dx, we have also implicitly
chosen x to be the variable of integration.
3
The boundary of the half disk intersects the xz
plane at an arbitrary point P(x, 0, z) on the
generating curve.
9.1 Centroids by Integration Example 16, page 2 of 3
4
The x coordinate of the element centroid
equals the x coordinate of the point P in
the xy plane.
xel = x
6
(3)
Since the variable of integration is x, we now have to
express z and dV in terms of x. Because point P lies
on the generating curve, x and z must satisfy the
equation of that curve
( bz )2]
x = a[1
y
(xel, yel, 0)
Solving this for z gives
z=b 1
( xa )
r
x
z
z
x
Substituting z = b [1
(5)
( ax )] into Eq. 4 gives
r=z
=b 1
P (x, 0, z)
( xa )
(6)
Thus the volume of the differential element in Eq. 2
becomes
5
The distance from the x-axis to the
boundary of the half-disk is z so
r=z
(4)
r2
dV = 2 dx
b 1 (x/a) ]2
=
dx
2
=
b2
x/a)
2
dx
(7)
9.1 Centroids by Integration Example 16, page 3 of 3
y
7
Evaluate the integral in the denominator of the equation for xc
over the range from 0 to a:
a
P(x, 0, z)
dV =
b2
x/a)
2
0
ab2
4
dx =
(8)
x
dx
a
8
z
Evaluate the integral in the numerator of the equation for xc over
the range from 0 to a:
a
xel dV =
9
0
x[
b2
x/a)
2
] dx =
a2b2
12
(9)
Using the results given by Eqs. 8 and 9 in the definition of xc
yields
xc =
xel dV
dV
a2b2/12
=
= a3
2
ab /4
Ans.
9.1 Centroids by Integration Example 17, page 1 of 4
17. Locate the centroid of the pyramid shown.
1
y
Definition of centroid coordinates
xc =
yc =
h
xel dV
(1)
dV
yel dV
(2)
dV
b
zc =
x
zel dV
(3)
dV
b
a
z
a
where ( xel, yel) are the coordinates of the
centroid of the differential volume-element
dV.
By symmetry, the centroid must lie on the
y-axis, so
xc = 0
Ans.
zc = 0
Ans.
yc remains to be calculated.
9.1 Centroids by Integration Example 17, page 2 of 4
Because of symmetry, we need consider
only one-fourth of the pyramid.
2
3
The differential element is a rectangular box of
thickness dy. The volume of the box is
y
dVo = (area of base)
A
= (xz) dy
(4)
By choosing an element of width dy, we have also
implicitly chosen y to be the variable of integration.
(xel, yel, zel)
h
P(x, y, 0)
Q(0, y, z)
y
dy
C
x
O
b
4
B
z
thickness
a
5
The boundary of the differential element
intersects the xy and xz planes at arbitrary points
P and Q on the sloping lines in those planes.
The y coordinate of the element centroid equals
the y coordinate of the point P in the xy plane.
yel = y
(5)
9.1 Centroids by Integration Example 17, page 3 of 4
6
Because the variable of integration is y, we now have
to dexpress dV in terms of y. Because point P lies on
the line AC in the xy plane, similar triangles can be
used to derive an equation relating x and y:
x= h y
a
h
7
Because point Q lies on the line AB in the yz plane,
similar triangles can be used to derive an equation
relating z and y:
z =h y
h
b
Solving for x gives
Solving for x gives
z = b(1
y
)
h
x = a(1
(6)
y
)
h
(7)
y
y
A
A
h
h
y
y
h
h
P(x, y, 0)
z
Q(0, y, z)
x
y
y
C
O
O
x
a
B
z
z
b
x
9.1 Centroids by Integration Example 17, page 4 of 4
Substituting Eqs. 6 and 7 in Eq. 4 and multiplying by
4 to get the volume of the whole pyramid (not just one
fourth) gives the volume element in terms of y.
8
dV = 4(dVo)
9
Evaluate the integral in the denominator of the equation for yc
over the range from 0 to h:
= 4(xz)dy
y
)] [b(1
h
y 2
) dy
h
= 4[a(1
= 4ab(1
y
] dy
h
dV = 4
h
ab(1
0
y 2
4abh
)
dy
=
3
h
(9)
(8)
Evaluate the integral in the numerator of the equation for yc:
y
yel dV = 4
y 2
) ]dy =
h
abh2
3
(10)
Using the results given by Eqs. 9 and 10 in the definition of yc
yields
h
yc =
x
z
h
y[ab(1
0
yel dV
dV
abh2/3
= 4abh/3 = h4
Ans.