Section 32–3
◆
Centroids
943
6. y2 24 4x from x 3 to 6
7. x2/3 y2/3 1 in the first quadrant
x3
1
8. y from x 1 to 3
2x
6
Find the area of the surface generated by rotating each curve about the y axis.
9. y 3x2 from x 0 to 5
10. y 6x2 from x 2 to 4
11. y 4 x2 from x 0 to 2
12. y 24 x2 from x 2 to 4
Geometric Figures
13. Find the surface area of a sphere by rotating the curve x2 y2 r2 about a diameter.
14. Find the area of the curved surface of a cone by rotating about the x axis the line connecting the origin and the point (a, b).
Applications
15. Find the surface area of the nose cone shown in Fig. 31–31 (in Chapter 31).
16. Find the cost of copper plating 10 000 bullets (Fig. 31–33 in Chapter 31) at the rate of $45
per square metre.
32–3 Centroids
Centre of Gravity and Centroid
The centre of gravity (or centre of mass) of a body is the point where all of the mass can be
thought to be concentrated, without altering the effect that the earth’s gravity has upon it. For
simple shapes such as a sphere, cube, or cylinder, the centre of gravity is exactly where you
would expect it to be, at the centre of the object. Figure 32–10 shows how the centre of gravity
of an irregular figure would be located.
A plane area has no thickness and hence has no weight or mass. Since it makes no sense
to speak of centre of mass or centre of gravity for a weightless figure, we use instead the word
centroid.
First Moment
To calculate the position of the centroid in figures of various shapes, we need the idea of the first
moment, often referred to as the moment. It is not new to us. We know from our previous work
that the moment of a force about some point a is the product of the force F and the perpendicular
distance d from the point to the line of action of the force. In a similar way, we speak of the
moment of an area about some axis [Fig. 32–11(a)]:
moment of an area area distance to the centroid
or moment of a volume [Fig. 32–11(b)]:
moment of a volume volume distance to the centroid
or moment of a mass [Fig. 32–11(c)]:
moment of a mass mass distance to the centroid
FIGURE 32–10 Any object
hung from a point will swing
to where its centre of gravity
is directly below the point of
suspension.
944
Chapter 32
◆
More Applications of the Integral
Centroid
Axis
d
Moment = Area × d
Area
(a)
Centroid
Axis
Volume
d
Moment = Volume × d
(b)
Centroid
Axis
d
Moment = Mass × d
Mass
(c)
FIGURE 32–11
We will learn about second
moments (moments of inertia) in
Sec. 32–6.
Moment of (a) an area, (b) a volume, and (c) a mass.
In each case, the distance is that from the axis about which we take the moment, measured to
some point on the area, volume, or mass. But to which point on the figure do we measure? To
the centroid or centre of gravity.
Centroids of Simple Shapes
If a shape has an axis of symmetry, the centroid will be located on that axis. If the centroid is on
_
_
the x axis, its coordinates will be (x, 0); if on the y axis, (0, y). If there are two or more axes of
symmetry, the centroid is found at the intersection of those axes.
We can easily find the centroid of an area that can be subdivided into simple regions, each
of whose centroid location is known by symmetry. We first find the moment of each region
about some convenient axis by multiplying the area of that region by the distance of its centroid from the axis. We then use the following fact, which we give without proof: For an area
subdivided into smaller regions, the moment of that area about a given axis is equal to the sum
of the moments of the individual regions about that same axis.
◆◆◆
Example 5: Find the location of the centroid of the shape in Fig. 32–12(a).
4 cm
4 cm
2 cm
y
4 cm
40 cm2
(5, 8)
4 cm
16 cm2
2 cm
(6, 4)
10 cm
20
cm2
(5, 1)
x
0
(a)
FIGURE 32–12
(b)
◆
Section 32–3
945
Centroids
Solution: We subdivide the area into rectangles as shown in Fig. 32–12(b), locate the centroid,
and compute the area of each. We also choose axes from which we will measure the coordinates
_
_
of the centroid, x and y. The total area is
40 16 20 76 cm2
The moment My about the y axis of the entire region is the sum of the moments of the individual
regions.
My 40(5) 16(6) 20(5) 396 cm3
Since
area distance to centroid moment
then
moment about y axis, My
x area
_
396 cm3
5.21 cm
76 cm2
The moment Mx about the x axis is
Mx 40(8) 16(4) 20(1) 404 cm3
So
moment about x axis, Mx
y area
_
404 cm3
5.32 cm
76 cm2
Centroids by Integration
If an area does not have axes of symmetry whose intersection gives us the location of the centroid, we can often find it by integration. We subdivide the area into thin strips, compute the first
moment of each, sum these moments by integration, and then divide by the total area to get the
distance to the centroid.
Consider the area bounded by the curves y1 f1(x) and y2 f2(x) and the lines x a and
x b (Fig. 32–13). We draw a vertical element of area of width dx and height (y2 y1). Since
the strip is narrow, all points on it may be considered to be the same distance x from the y axis.
The moment dMy of that strip about the y axis is then
dMy x dA x(y2 y1) dx
y
y2 = f2(x)
(x, y2)
dA
y2 − y1
A
y1 = f1(x)
(x, y1)
dx
0
a
FIGURE 32–13
b
x
Centroid of irregular area found by integration.
946
Chapter 32
◆
More Applications of the Integral
since dA (y2 y1) dx. We get the total moment My by integrating.
b
My a
3
x(y2 y1) dx
_
_
But since the moment My is equal to the area A times the distance x to the centroid, we get x by
dividing the moment by the area. Thus,
Horizontal
Distance to
Centroid
b
_
1
x
x(y2 y1) dx
A3
a
400
_
To find x, we must have the area A. It can be found by integration as in Chapter 31.
We now find the moment about the x axis. The centroid of the vertical element is at its
midpoint, which is at a distance of (y1 y2)/2 from the x axis. Thus, the moment of the element
about the x axis is
y1 y2
dMx (y2 y1) dx
2
_
Integrating and dividing by the area gives us y.
Equations 400 and 401 apply
only for vertical elements. When
using horizontal elements,
interchange x and y in these
equations.
Vertical
Distance to
Centroid
b
_
1
y (y1 y2)(y2 y1) dx
2A 3
a
401
Our first example is for an area bounded by one curve and the x axis.
Example 6: Find the centroid of the area bounded by the parabola y2 4x, the x axis, and
the line x 1 (Fig. 32–14).
◆◆◆
y (m)
y2 = 4x
2
(x, y)
1
y
0
1
x (m)
dx
FIGURE 32–14
7 8
Solution: We need the area, so we find that first. We draw a vertical strip having an area y dx
and integrate.
1
x3/2 1 4
A ydx 2
x1/2 dx 2 m2
3/2 0 3
0
3
3
Section 32–3
◆
947
Centroids
Then, by Eq. 400,
1
1
x
x(2x1/2 0) dx
A3
0
1
1
3
3 x5/2
3
3/2
2x dx · m
43
2 (5/2) 0 5
0
_
and, by Eq. 401,
1
1
y (2x1/2 0)(2x1/2 0) dx
2A 3
0
_
1
1
3
3 4x2
3
4xdx · m
0
83
8
2
4
0
Check: Does the answer seem reasonable? The given area extends from x 0 to 1, with more
_
area lying to the right of x 12 than to the left. Thus, we would expect x to be between x 12
and 1, which it is. Similarly, we would expect the centroid to be located between y 0 and 1,
◆◆◆
which it is.
For areas that are not bounded by a curve and a coordinate axis, but are instead bounded
by two curves, the work is only slightly more complicated, as shown in the following example.
Example 7: Find the coordinates of the centroid of the area bounded by the curves 6y x2 4x 4 and 3y 16 x2.
◆◆◆
Solution: We plot the curves as shown in Fig. 32–15 and find their points of intersection by
solving simultaneously. Multiplying the second equation by 2 and adding the resulting equation to the first gives
3x2 4x 28 0
Solving by quadratic formula (work not shown), we find that the points of intersection are at
x 2.46 and x 3.79. We take a vertical strip whose width is dx and whose height is
y2 y1, where
16 x2 x2 4x 4 28 4x 3x2
y2 y1 3
3
6
6
6
6
The area is then
3.79
A
(y2 y1) dx
2.46
3
3.79
1
(28 4x 3x2) dx 20.4
63
2.46
y
6
1
2
6y = x 2 − 4x + 4
4
3y = 16 − x2
2
−5
−2.46
5
0
3.79
FIGURE 32–15
x
948
Chapter 32
◆
More Applications of the Integral
Then, from Eq. 400,
3.79
_
M y Ax 2.46
3
x(y2 y1) dx
3.79
1
(28x 4x2 3x3) dx 13.6
63
2.46
My 13.6
_
x 0.667
A
20.4
We now substitute into Eq. 401, with
36 4x x2
y1 y2 6
Thus,
3.79
_
1
Mx Ay (36 4x x2)(28 4x 3x2) dx
72 3
2.46
7
3.79
8
1
(3x4 8x3 152x2 32x 1008) dx
72 3
2.46
1 3x5 8x4 152x3 32x2
1008x
72 5
4
3
2
Mx 52.5
y 2.57
A
20.4
3.79
2.46
52.5
Check by Physical Model: Here our check will be to make a model of the given figure. Make a
graph as in Fig. 32–15, being sure that you use the same scale for the x and y axes, and paste the
graph to cardboard or thin metal. Cut out the shaded area and locate the centroid by suspending
it as in Fig. 32–10 or by balancing it on the point of a tack. Compare your experimental result
◆◆◆
with the calculated one.
Centroids of Solids of Revolution by Integration
A solid of revolution is, of course, symmetrical about the axis of revolution, so the centroid
must be on that axis. We only have to find the position of the centroid along that axis. The procedure is similar to that for an area. We think of the solid as being subdivided into many small
elements of volume, find the sum of the moments for each element by integration, and set this
equal to the moment of the entire solid (the product of its total volume and the distance to the
centroid). We then divide by the volume to obtain the distance to the centroid.
◆◆◆
Example 8: Find the centroid of a hemisphere of radius r.
Solution: We place the hemisphere on coordinate axes as shown in Fig. 32–16 and consider it
as the solid obtained by rotating the first-quadrant portion of the curve x2 y2 r2 about the x
axis. Through the point (x, y) we draw an element of volume of radius y and thickness dx, at a
distance x from the base of the hemisphere. Its volume is thus
dV y2 dx
Of course, these methods
work only for a solid that is
homogeneous, that is, one
whose density is the same
throughout.
and its moment about the base of the hemisphere (the y axis) is
dMy xy2 dx x(r2 x2) dx
Integrating gives us the total moment.
My 7
r
8
x(r2 x2) dx 0
0
3
3
r
r2x2 x4
r4
2
4 0
4
r
(r2x x3) dx
◆
Section 32–3
949
Centroids
y
x2 + y2 = r2
r
(x, y)
Axis of
rotation
y
r
0
x
dx
dV
−r
FIGURE 32–16
x
Finding the centroid of a hemisphere.
_
2
The total moment also equals the volume (r3 for a hemisphere) times the distance x to the
3
centroid, so
r4/4
3r
x 2r3/3
8
_
◆◆◆
Thus, the centroid is located at (3r/8, 0).
As with centroids of areas, we can write a formula for finding the centroid of a solid of
revolution. For the volume V (Fig. 32–17) formed by rotating the curve y f(x) about the x axis,
the distance to the centroid is the following:
y
x−
y = f(x)
Axis of
rotation
0
a
FIGURE 32–17
b
x
Centroid of a solid of revolution.
Distance to the
Centroid of Volume
of Revolution about
x Axis
b
_
x
xy2 dx
V3
a
402
950
Chapter 32
◆
More Applications of the Integral
For volumes formed by rotation about the y axis, we simply interchange x and y in Eq. 402
and get
Distance to the
Centrold of Volume of
Revolution about y Axis
d
_
y
yx2 dy
V3
c
403
These formulas work for solid figures only, not for those that have holes down their centre.
Exercise 3
◆
Centroids
Without Integration
1. Find the centroid of four particles of equal mass located at (0, 0), (4, 2), (3, 5), and
(2, 3).
2. Find the centroid in Fig. 32–18(a).
3. Find the centroid in Fig. 32–18(b).
4. Find the centroid in Fig. 32–18(c).
x−
1
2
6
1
12
3
8
C
6
3
−y
C
1
2
2
4
C
−y
1
32
4
(a)
−y
1
(b)
3
(c)
FIGURE 32–18
Centroids of Areas by Integration
Find the specified coordinate(s) of the centroid of each area.
_
_
5. bounded by y2 4x and x 4; find x and y.
_
_
6. bounded by y2 2x and x 5; find x and y.
_
7. bounded by y x2, the x axis, and x 3; find y.
_
_
8. bounded by x y 1 and the coordinate axes (area 61); find x and y.
Areas Bounded by Two Curves
Find the specified coordinate(s) of the centroid of each area.
_
9. bounded by y x3 and y 4x in the first quadrant; find x.
_
10. bounded by x 4y y2 and y x; find y.
_
_
11. bounded by y2 x and x2 y; find x and y.
_
12. bounded by y2 4x and y 2x 4; find y.
_
13. bounded by 2y x2 and y x3; find x.
_
_
14. bounded by y x2 2x 3 and y 6x x2 3 (area 21.33); find x and y.
_
15. bounded by y x2 and y 2x 3 in the first quadrant; find x.
3
8
Section 32–3
◆
951
Centroids
Centroids of Volumes of Revolution
Find the distance from the origin to the centroid of each volume.
16. formed by rotating the area bounded by x2 y2 4, x 0, x 1, and the x axis about the
x axis.
17. formed by rotating the area bounded by 6y x2, the line x 6, and the x axis about the
x axis.
18. formed by rotating the first-quadrant area under the curve y2 4x, from x 0 to 1, about
the x axis.
19. formed by rotating the area bounded by y2 4x, y 6, and the y axis about the y axis.
20. formed by rotating the first-quadrant portion of the ellipse x2/64 y2/36 1 about the x axis.
21. a right circular cone of height h, measured from its base.
Applications
22. A certain airplane rudder (Fig. 32–19) consists of one quadrant of an ellipse and a quadrant
of a circle. Find the coordinates of the centroid.
y
A
2.00 m
B
0
x
C
1.20 m
FIGURE 32–19
Airplane rudder.
23. The vane on a certain wind generator has the shape of a semicircle attached to a trapezoid
_
(Fig. 32–20). Find the distance x to the centroid.
0.400 m
x−
C
1.00 m
2.50 m
FIGURE 32–20
Wind vane.
24. A certain rocket (Fig. 32–21) consists of a cylinder attached to a paraboloid of revolution.
Find the distance from the nose to the centroid of the total volume of the rocket.
6.00 m
28.0 m
FIGURE 32–21
Rocket.
6.00 m
952
Chapter 32
◆
More Applications of the Integral
25. An optical instrument contains a mirror in the shape of a paraboloid of revolution
(Fig. 32–22) hollowed out of a cylindrical block of glass. Find the distance from the flat
bottom of the mirror to the centroid of the mirror.
Mirrored
surface
164 mm
90.0 mm
72.0 mm
190 mm
FIGURE 32–22
32–4
Paraboloidal mirror.
Fluid Pressure
The pressure at any point on a submerged surface varies directly with the depth of that point
below the surface. Thus, the pressure on a diver at a depth of 20 m will be twice that at 10 m.
The pressure on a submerged area is equal to the weight of the column of fluid above
that area. Thus, a square metre of area at a depth of 7 m supports a column of water having a
volume of 7 12 7 m3. Since the density of water is 1000 kg/m3, the weight of this column
is 7 1000 7000 kg, so the pressure is 7000 kg/m2 at a depth of 7 m. Further, Pascal’s law
says that the pressure is the same in all directions, so the same 7000 kg/m2 will be felt by a
surface that is horizontal, vertical, or at any angle.
The force exerted by the fluid (fluid pressure) can be found by multiplying the pressure per
unit area at a given depth by the total area.
Recall that
weight volume density
(Eq. A43).
Total Force
on a Surface
Surface of fluid
x
0
force pressure area
A46
A complication arises from an area that has points at various depths and hence has different
pressures over its surface. To compute the force on such a surface, we first compute the force
on a narrow horizontal strip of area, assuming that the pressure is the same everywhere on that
strip, and then add up the forces on all such strips by integration.
◆◆◆ Example 9: Find an expression for the force on the vertical area A submerged in a fluid of
density (Fig. 32–23).
y
dA
Solution: Let us take our origin at the surface of the fluid, with the y axis downward. We draw
a horizontal strip whose area is dA, located at a depth y below the surface. The pressure at depth
y is y, so the force dF on the strip is, by Eq. A46,
dF y dA (y dA)
A
y
FIGURE 32–23 Force on a
submerged vertical surface.
Sometimes it is more
convenient to take the origin at
the surface of the liquid.
Integrating, we get the following equation:
Force of
Pressure
F y dA
3
A47