PROBLEM 5.128*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the x-axis.
SOLUTION
First, note that symmetry implies
0y
0z
Choose as the element of volume a disk of radius r and thickness dx.
Then
2
,
EL
dV r dx x x
Now
sin 2
x
rb a
PROBLEM 5.128* (Continued)
Then
2
2
2
22
sin sin
22
a
xx
a
aa
EL a
aa
a
xx
xdV b x dx
PROBLEM 5.129*
Locate the centroid of the volume generated by revolving the
portion of the sine curve shown about the y-axis. (Hint: Use a thin
cylindrical shell of radius r and thickness dr as the element of
volume.)
SOLUTION
First note that symmetry implies
0x
0z
Choose as the element of volume a cylindrical shell of radius r and thickness dr.
Then 1
(2 )( )( ), 2
EL
dV r y dr y y
PROBLEM 5.129* (Continued)
Also
21
2
2
22
sin 2 sin
22
sin 2
a
EL a
a
a
rr
ydV b br dr
aa
r
br dr
a
Use integration by parts with
2
sin 2
r
ur dv dr
a
PROBLEM 5.130*
Show that for a regular pyramid of height h and n sides
(3,4,)n
the centroid of the volume of the
pyramid is located at a distance h/4 above the base.
SOLUTION
Choose as the element of a horizontal slice of thickness dy. For any number N of sides, the area of the
base of the pyramid is given by
2
base
Akb
where
();kkN
see note below. Using similar triangles, we have
shy
bh
PROBLEM 5.131
Determine by direct integration the location of the centroid of one-half of
a thin, uniform hemispherical shell of radius R.
SOLUTION
First note that symmetry implies
0x
The element of area dA of the shell shown is obtained by cutting the shell with two planes parallel to the
xy plane. Now
()( )
2
EL
dA r Rd
r
y
where sinrR
PROBLEM 5.132
The sides and the base of a punch bowl are of
uniform thickness t. If
tR
and R 250 mm,
determine the location of the center of gravity of
(a) the bowl, (b) the punch.
SOLUTION
(a) Bowl:
First note that symmetry implies
0x
0z
for the coordinate axes shown below. Now assume that the bowl may be treated as a shell; the
center of gravity of the bowl will coincide with the centroid of the shell. For the walls of the bowl,
an element of area is obtained by rotating the arc ds about the y-axis. Then
wall
(2 sin )( )dA R R d
PROBLEM 5.132 (Continued)
(b) Punch:
First note that symmetry implies 0x
0z
and that because the punch is homogeneous, its center of gravity will coincide with the centroid of
the corresponding volume. Choose as the element of volume a disk of radius x and thickness dy.
Then
2
,
EL
dV x dy y y
Now
22 2
xyR
so that
22
()dV R y dy
PROBLEM 5.133
Locate the centroid of the section shown, which was cut from a thin circular
pipe by two oblique planes.
SOLUTION
First note that symmetry implies 0x
Assume that the pipe has a uniform wall thickness t and choose as the element of volume a vertical strip
of width ad
and height
21
().yy
Then
PROBLEM 5.133 (Continued)
Then 0
0
2(1cos) [sin]
2
aht
V d aht
aht
and 0
2
2
0
2 (5 cos ) (1 cos )
12 2
(5 6cos cos )
12
EL
haht
ydV d
ah t d
PROBLEM 5.134*
Locate the centroid of the section shown, which was cut from an
elliptical cylinder by an oblique plane.
SOLUTION
First note that symmetry implies 0x
2
PROBLEM 5.134* (Continued)
and 22
2
22 2
3
1()2 ()
22 2
1()
4
b
EL b
b
b
hah
ydV b z b z b zdz
bb b
ah bz b zdz
b
Let sin coszb dzb d
Then
2/2 2
3/2
/2
22 222
/2
1[(1 sin )]( cos ) ( cos )
4
1(cos 2sin cos sin cos )
4
EL
ah
ydV b b b d
b
abh d
PROBLEM 5.134* (Continued)
Now 2
15
:232
EL
yV y dV y abh abh
or
5
16
yh
PROBLEM 5.135
Determine by direct integration the location of the centroid
of the volume between the xz plane and the portion shown
of the surface y 16h(ax x
2
)(bz z
2
)/a
2
b
2
.
SOLUTION
PROBLEM 5.135 (Continued)
and 22 22
22 22
00
2
22 3 4 22 3 4
44 00
116 16
()() ()()
2
128 (2 )(2 )
ba
EL
ba
hh
y dV ax x bz z ax x bz z dx dz
ab ab
ha x ax x b z bz z dx dz
ab
PROBLEM 5.136
After grading a lot, a builder places four stakes to
designate the corners of the slab for a house. To provide a
firm, level base for the slab, the builder places a minimum
of 3 in. of gravel beneath the slab. Determine the volume
of gravel needed and the x coordinate of the centroid of
the volume of the gravel. (Hint: The bottom surface of
the gravel is an oblique plane, which can be represented
by the equation y a bx cz.)
SOLUTION
The centroid can be found by integration. The equation for the bottom of the gravel is
,yabxcz
where the constants a, b, and c can be determined as follows:
For
0x
and
0,z
3 in., and therefore,y
31
ft , or ft
12 4
aa