Post on 02-Jun-2018
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1
1
CHAPTER 3
APPLICATIONS
OF THE DEFINITE
INTEGRAL
2
OUTLINE3.1 Area of a plane region
3.2 Volume of a solid of revolution3.3 Center of mass of a rod
3.4 Centroid of a plane region and a
Theorem of Pappus
3.5 Centroid of a solid of revolution
3.6 Length of an arc of a curve
3.7 Area of a surface of revolution
3.8 Work
3.9 Force due to Liquid Pressure
33
CHAPTER OBJECTIVES
At the end of the chapter, you should be able to use
the definite integral to find/compute for
1. the area of a plane region using vertical or horizontal
strips,
2. the volume of a solid of revolution using the
cylindrical or shell method,
3. the center of mass of a rod, plane region and of a solid
of revolution,
4. the length of an arc of a plane curve,
5. the area of a surface of revolution and
6. work and force due to liquid pressure. 4
3.1 Area of a plane region
Let f be a function of x which iscontinuous and non-negativeon the closedinterval [a, b].
Subdivide the closed interval [a, b] into
nsub-intervals by choosing (n- 1)
intermediate numbers
121 nx,...,x,x
.... bxxxxxa nn 1210
where
5
For example, if we want to subdivide
[a,b] into 2 sub-intervals, we only choose one
number x1between aand b.
a b| |
x1
|
If we want to subdivide [a,b] into 3 sub-
intervals, we choose two numbers x1 and x2
between aand b.
a b| || |x1 x2
66
x
y
y= f(x)
ba| |x1|
xi-1|
xi|
Denote the ith sub-
interval by Iiso that
101 x,xI
212 x,xI
323 x,xI
iii x,xI
1
nnn x,xI
1
For each ,
choose a number
n,...,,i 21
.Iii
xn-1|
1 i n
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7
Denote the length of the ith sub-interval
by
so that
xi
011 xxx
122 xxx
233 xxx
1
iii xxx
1
nnn xxx
88
Our objective is to
find a formula for
the area of the
region R enclosed
by the graph of f,
the x-axis and the
lines given by x = a
and x = b.x
y
y= f(x)
ba| |
R
9
x
y
y= f(x)
ba| |x1|
xi-1|
xi|
xn-1|
1 i n
1
f
if
nf On the ith sub-
interval, construct
a rectangle of length
and whose width is
.xi
if
10
The area of the ith rectangular region is
.xiif
Thus, the sum of areas of all nrectangular
regions is
xf 11
xf 22
...xf... ii
xf nn
or
.xfii
n
i
1
11
x
y
y= f(x)
ba| |
RThe area of the region
Ris given by
.xfl im ii
n
in
1
12
.dxxfxfl imb
aii
n
in
1
By definition,
Thus, the area of the region Ris given
by
.dxxfRAb
a
length of a
rectangularelement
width of a
rectangular
element
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Example 3.1.1 Find the area of the region
enclosed by the graphs of
.yx,x,xy 0and212
RA2
1
3
3
x
3
1
3
23
3
7
3
1
3
8
Thus, the area of the
region is 7/3 sq. units.2x1x
2xy
Solution:
dx 2x1
2
14
Example 3.1.2 Find the area of the region
enclosed by the graphs of
.yex,xl ny 0and
exxl nx 1
Thus, the area of the region Ris 1 sq. unit.
111 l neel ne1 0
11ee0y
xlnyex
Solution:
1x
RA dx xln1
e
15
Let g be a function of y which is
continuous and non-negative on the
closed interval [c, d].
Subdivide the closed interval [c, d]into
nsub-intervals by choosing (n- 1)
intermediate numbers
121 nyyy ,...,,
.... dyyyyyc nn 1210
where
1616
x
y
x= g(y)i
The area of the ith
rectangular region is
.yiig
The area of the
regionRis given
by
.ygl im ii
n
in
1
.dyygRAd
c
y= d
y= c
For each ,
choose a number
n,...,,i 21
.Iii
igx
1717
Example 3.1.3 Find the area of the region in
the first quadrant enclosed by the graphs of
.xy,y,xy 0and412
4
1
23
23
/
y /
Thus, the area of the region is 14/3 sq. units.
2xy
yx
41
23
3
2 /y
232314
3
2 //
183
2
3
14
0x
1y
4y
Solution:
RA dy y4
1
1818
Example 3.1.4 Find the area of the region in
the first quadrant enclosed by the graphs of
.ln 2 yxy and
20ye
Thus, the area of the region is (e2 - 1) sq. units.
xy ln 02 ee 2y
Solution:
RA dy ye2
0
yex
12 e
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1919
Theorem 3.1.1Let fand g be functions of
x which are continuous on [a, b] and
f(x) g(x)
for all xin [a, b].
Then the area of the region bounded by
the graph of the graphs of fand gand
the lines given by x = aand x= bis given
by
. b
a dxxgxfA
2020
xgy
xfy
ax bx
xgxf
b
adxxgxfA
21
Example 3.1.5 Set-up the definite integral that
gives the area of the region R bounded by the
graphs of and the y-axis.93 y,y x
Solution:
(2,9)f(x) = 9
g(x) = 3x
dxxgxfRA ba
dx 9(2
0)x
3
22
Solution:
Example 3.1.6 Set-up the definite integral that
gives the area of the region R bounded by the
graphs of and .12 xy 42 xy
4212 xx
0322 xx
013 xx
13 xx or
POI: (3,10) and (-1,2)
23
dxxgxfRA ba
dx 42 x3
-112 x
3-1
12 xxg
42 xxf
2424
Theorem 3.1.2Let fand g be functions of
y which are continuous on [c, d] and
f(y) g(y)
for all yin [c, d].
Then the area of the region bounded by
the graph of the graphs of fand gand
the lines given by y = cand y= dis given
by
.dyygyfA d
c
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2525
dy
cy
.dyygyfAd
c
ygyf
yfx ygx
26
Solution:
Example 3.1.7 Set-up the definite integral that
gives the area of the region bounded by the
graphs of and y= x3.12
xy
312 yy
022 yy
012 yy
12 yy or
POI: (5,2) and (2,-1)
27
y2= x - 1
y=x - 3
(2,-1)
(5,2)
x=y + 3
x = y2+1
dyygyfAd
c
dyyy 13 22
1
.dyyy 2
1-
22
28
3.2 Volume of a Solid of Revolution
Soli d of Revolution
A soli d of revoluti on is obtained when
a region in the xy-plane is rotated about a
fixed line in the plane.
The fixed line is called the axis of
symmetry or axis of revolution of the
resulting solid.
We shall only consider vertical or
horizontal line as axis of revolution.
29
Rotate a tri angular regionabout one of
its legs to obtain a conical soli d.
30
Rotate a rectangular regionabout one of its
sides to obtain a ri ght-cir cular cyli ndrical
solid.
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31
Rotate a parabolic regionabout its axis of
symmetry to obtain a parabolic cyli ndri cal
solid.32
33
Methods of finding the volume of a solid of
revolution
1. Circular Disk Method
- a rectangular element isperpendicular to the axis ofrevolution.
2. Cylindrical Shell
- a rectangular element is parallelto the axis of revolution.
3434
3.2.1 Circular Disk Method
We derive the formula for finding
the volume of a solid formed when a
plane region is revolved about the x-axis
using the circular disk method.
The method may be applied even
when the axis of revolution is no longer
the x-axis but with a slight modification.
35
We recall that
the volume of a solid
enclosed by a right
circular cylinder of
radius rand altitude
his
V=r2h.
r
h
36
Let the function f be continuous on
the closed interval [a,b]and assume that
f(x) 0 for all xin [a,b].
Let Rbe the
region bounded by
the graph of y = f(x),
the x-axisand the
lines given by x= a
and x= b.
xfy
ax bx
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3737
Suppose Ris revolved around the x-axis.
38
xfy
ax bx
38
i
When the ith rectangular
region is revolved about
the x-axis, a circular disk
is formed.Its volume is
Vi=r2h = xf ii
2
The sum of volumes of
the ndisks is
i
n
i
V1
.xfii
n
i
2
1
3939
Thus, the volume of the solid formed
when the region Ris revolved about
the x-axis is
xfl imV iin
in
2
1
or equivalently,
.dxxfV ba 2
40
Example 3.2.1
xxf
Consider the region bounded by the graphs of
, and the x-axis.xy 4x
If the region is
revolved about the x-
axis, a solid isformed.
Using the Disk
Method, the radius of
the disk is given by
xxf
Solution:
4141
Using the Disk
Method, the
volume of the
solid generated
is
dxxfVb
a 2
dxxV24
0
xdx4
0
4
0
2
2 x 8
242
cu. units
42
Example 3.2.2
Consider the region bounded by the graphs of
in the first quadrant.
2xy 4yand
Set up the definite integral
which gives the volume of the
solid formed when the given
region is revolved about
1. x= 0 2. x= 2 3. x= -1
4. y= 0 5. y= 4 6. y= -1
2xy
(2,4)
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43
2xy
1. x= 0
yxr
V
(2,4)
dyh
2y dy0
4
44
2xy
2. x= 2
yx
21 VVV
dyh
2r
dyh
yr 2
1V 2
2 dy0
4
xr 2
2V 2
2 y dy0
4
45
3. x= -1
2xy
yx
x= -1 x= -1 x= -1
dyh
xr 1
yr 1
dyh
1r
21 VVV
1V 2
1 y dy0
4
2V 2
1 dy0
4
46
2xy
4. y= 0
21 VVV
dxh
4r
yr
dxh
2xr
1V 2
4 dx0
2
2V 22
x0
2dx
4747
2xy
5. y= 4
yr 4
dxh
V dx0
2
24 x
224 x
48
2xy
6. y= -1
21 VVV
dxh
5r1 yr
dxh
12
xr
1V
25 dx
0
2
2
V
221x
0
2dx
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4949
3.2.2 Cylindrical Shell Method
A cylindrical shell is asolid contained between
two cylinders having the
same altitude and axis.
The Cylindrical Shell
method involves taking
rectangular elements
which are parallel to the
axis of revolution.
r1
r2
h
hrV 2
hrhrV2
1
2
2
hrrV 21
2
2
5050
Theorem 3.2.2Let the
function f be continuous
on the closed interval [a, b]
and assume that f(x) 0
for all xin [a, b].
Let Rbe the region
bounded by the graph of
y= f(x), the x-axis and the
lines given by x= aand
x= b.
Suppose Ris revolved
around the y-axis.
xfy
ax bx
5151
xfy
ax bxi
When the ith rectangular
region is revolved about
the y-axis, a circular shell
is formed.
Its volume is
hrrVi
2
1
2
2
iii f-xx 2
1
2
iiiii fxx- xx 11
iii fx 2Let .
2
1 iiixx
xf iii 2
5252
The sum of volumes of the nshells is
.xfxf iiin
i
iii
n
i
2211
Thus, the volume of the solid formed
when the region Ris revolved about the y-
axis is
or equivalently,
xflimV iiin
in
21
.dxxxfVb
a
2
Distance from the
axis of revolution
Length of a
rectangular element
Width of a
rectangular element
5353
Example 3.2.3
2xy
Consider the region bounded by the graphs of
, and they-axis in the first quadrant.2xy 4y
Using Shell Method, set up thedefinite integral which gives thevolume of the solid formedwhen the given region isrevolved about
1. x= 0 4. y= 0
2. x= -1 5. y= 53. x= 3 6. y = -2
yx
Solution:
54
24 xh
1. x= 0
xd
.dxxxV 2
0
242
x= 02xy
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2. x= -1
24 xh
1xd
.dxxxV 2
0
2412
x= -12
xy
5656
3. x= 3
24 xh
xd 3
.dxxxV 2
0
2432
x= 3
2
xy
57
yh
4. y= 0
yd
.dyyyV 4
02
y= 0
2xy
5858
5. y= 5
yd 5
y= 5
yh
.dyyyV 4
052
2xy
59
6. y= -2
2yd
y= -2
yh
.dyyyV 4
022
2xy
60
3.3 Center of mass of a rod
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0
Consider a horizontal rod, of
negligible weight and thickness, placedon the x-axis.
On the rod is a system of nparticles
located at points x1, x2, , xn.
m1
m2 mi mnx1 x2 xi xn
62
The total mass of the systemis
M= m1
+ m2
+ + mn
.mn
i i
1
The moment of mass of the ith
particle with respect to the originis
The total moment of mass of the
system with respect to the originis
.xmn
i
ii
1
.xmii
nnxm...xmxmM 22110
63
The center of mass of the systemis
at
.
m
xm
n
i
i
n
i
ii
1
1
.xmxmn
i
ii
n
i
i
11
M
Mx 0
64
Example 3.3.1 Find the center of mass of the
system of 5 particles with masses 3,4,2,1 and 6
kgms located at -2,-1,0,3 and 4 respectively.
Solution:
5
1
5
1
i
i
i
ii
m
xm
x 61243
4333031323
.4
3
16
12
The center of mass of the system is 0.75 to the right
of the origin.
65
A rod is said to be homogeneous if the
mass of a part of it is proportional to the length
of that part.
im li
ii klm
Linear density
The center of mass of a homogeneous rod is at
its center.66
Consider a non-homogeneousrod, that is
the linear density varies along the rod. Let L
meters be the length of the rod and place the
rod on the x-axisso that an endpoint of the rod
coincides with the origin and the other
endpoint is at L .
0 L
Suppose the linear density at point x is(x) where is continuous on [0, L ].
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Consider a partition of [0, L ] into nsub-
intervals and let the ith subinterval be
.x,xI iii 1
Let i be a number in Ii .
The length of the ith subinterval is
.xmiii
.xi
An approximation to the mass of the part of the
rod in the ith subinterval is
68
0 Li
x1ix
i
.xmiii
An approximation to the mass of the rod is
i
n
i
m1
xii
n
i
1
The mass of the rod is
xl imMii
n
in
1
dxxL
0
69
An approximation to the moment
of mass of the part of the rod in the ith
subinterval with respect to the originis
iixm
iii x
An approximation to the momentof mass of the rod with respect to the
originis
ii
n
i
xm1
.xiii
n
i
1
.xiii
70
xl imMiii
n
in
1
0 .dxxxL
0
The moment of mass of the rod
with respect to the originis
The center of mass of the rod is at
.
dxx
dxxx
L
L
0
0
M
Mx 0
71
Example 3.3.2 Find the center of mass of a rod that
is 20 cm. long and the linear density on the rod at a
point xcentimeters from one end is (3x+ 2)
gm/cm.
Solution:
dxx
dxxxx
L
L
0
0
dxx
dxxx
23
23
2
2
0
0
0
0
dxx
dxxx
23
23
2
22
0
0
0
0
20
0
20
0
xx
xx
2
2
3 2
23
20220
2
3
2020
2
23
20220
2
3
2020202
32
420
8
105
72
Example 3.3.3 A rod is 12 cm. long and the linear
density at a point on the rod is a linear function of
the distance of the point from the left endpoint of
the rod. If the linear density at the left end is 3
gm/cm and at the right end is 4 gm/cm, find the
center of mass of the rod.
Solution:
Let xbe the distance from the left endpoint
of a point on the rod and (x)be the linear
density at x. Then for some constants aand b,
.baxx
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Since (0) = 3 and (12) = 4, and
(x) =ax+ b,
30 3 b
3 axx
412 4312 a
12
1a
312
1 xx
74
dxx
dxxxx
L
L
0
0
dxx
dxxx
312
1
312
1
1
1
2
0
2
0
dxx
dxxx
312
1
312
1
1
21
2
0
2
0
12
0
12
0
xx
xx
324
1
2
3
36
1
2
23
75
1231224
1
122
312
36
1
2
23
1231224
1
122
312
36
1 2
3
2
1
1223
31
2
7
184
2
7
22
7
44
76
3.4 Centroid of a Plane Region
Consider a system of nparticles located at
the points (x1, y1), (x2,y2) , (xn,yn) in the xy
plane and let their masses be m1, m2, .., mn,
respectively. Imagine the particles as supported
by a sheet of negligible weight and thickness.
(x1, y1) (x2, y2)
(x3, y3)
(x4, y4)
(x5, y5)
77
Let the ith particle, located at (xi,yi) have
mass mi.
The total mass of the system is
.mMn
i
i
1
The moment of mass of the system with
respect to the x-axisis
.xmMn
i
iiy
1
.ymMn
i
iix
1
The moment of mass of the system with
respect to the y-axisis
78
The center of mass of the system is at
y,xwhere
andM
xm
M
Mx
n
i
iiy
1
.M
ym
M
M
y
n
i
ii
x
1
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Example 3.4.1 Find the center of mass of the
system of 4 particles with masses 4,2,3,6 and which
are located at the points (-2,-3), (1,2), (3,-2) and
(4,-3).
Solution:
4
1i
imM 156324
4
1i
iix ymM 36232234 4
4
1i
iiy xmM 46331224 27
80
The center of mass of the system is
at
y,x
where
.M
My x
15
4
and5
9
15
27
M
Mx y
81
A lamina is said to be homogeneousif
the mass of a portion of it is proportional
to the area of the part. Thus, if M is the
mass of the region R and A is its area,
then there is a constant ksuch that
M= kA .
Area density
82
So if we have a homogeneous
rectangular region, its mass Mi is
given by
Mi=k(lw).
Also, its centr oid or center of mass
is at its center.
83
Let Rbe the region
bounded above and
below, respectively by
the graphs of
and the lines given by
xgy,xfy
ax .bxand
xgy
ax bx
xfy
84
The area of the region is given by
.dxxgxfAb
a
Its mass is given by
.dxxgxfkMb
a
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85
b,a
xgy
ax bx
xfy
i
As we did in the
first 2 sections,
we subdivide
[a, b] into n
subintervals and
we choose a
number i in
each subinterval.
This time, we choose ias the abscissa
of the midpoint of each sub-interval.8686
Then we construct a
rectangle whose width
is and whose
length is .
xi
ii gf
The area of the ith
rectangular region is
.xgfiii
Thus, the mass of the
ith rectangular region
is
. xgfk iii
b,a
xgy
ax bx
xfy
i
8787
b,a
xgy
ax bx
xfy
The centroid of the ith
rectangular region is
at
.
2
ii gf
i,
2
ii gf
i,
The moment of mass
mi,ywith respect to the
yaxis of the ith
rectangular region is
the product of its mass
and the distance of its
centroid from the y-axis. Thus,
iy,im xgfk iii
.xgfk iiii
8888
b,a
xgy
ax bx
xfy
2
ii gf
i,
The moment of mass
mi,xwith respect to the
x-axis of the ith
rectangular region is
the product of its
mass and the distance
of its centroid fromthe x-axis. Thus,
2
ii gf
x,im xgfk iii
.xgfk iii 2
22
89
The total moment of massMi,ywith respect to
the yaxisof the nrectangular regions is given
by
y,i
n
i
y,i mM
1
.xgfk iiiin
i
1
The moment of massMywith respect to the y
axisof region Ris given by
xgfklim iiiin
in
1
y,i
n
in
y,in
y mlimMlimM
1
xgflimk iiiin
in
1
dxxgxfxkb
a
90
.xgfk iiin
i
2
22
1
The total moment of massMi,xwith respect to
the xaxisof the nrectangular regions is given
by
x,i
n
i
x,i mM
1
The moment of massMxwith respect to the x
axisof region Ris given by
xgfklim iiin
in
2
22
1
x,i
n
in
x,in
x mlimMlimM
1
dxxgxfk ba
22
2
xgflimk
iii
n
in 2
22
1
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If is the centroid of the plane region R
whose mass is M, then
y,x
M
Mx y M
My xand .Since
,dxxgxfkMb
a
,dxxgxfxkMb
ay
dxxgxfkM ba
x
22
2
92
M
My x
dxxgxfk
dxxgxfk
b
a
b
a
22
2
dxxgxf
dxxgxf
b
a
b
a
2
22
M
M
x
y
dxxgxfk
dxxgxfxk
b
a
b
a
dxxgxf
dxxgxfx
b
a
b
a
9393
Summary:
xgy
ax bx
xfy
dxxgxf
dxxgxfxx
b
a
b
a
dxxgxf
dxxgxfy
b
a
b
a
2
22
y,x
94
xxf
Example 3.4.2
Find the centroid of the region bounded by the
graphs of , and thex-axis.xy 4x
4x 0xg
Solution:
95
dxxgxf
dxxgxfxx
b
a
b
a
dxx
dxxx
0
0
4
0
4
0
dxx
dxx
/
/
214
0
234
0
4
0
23
4
0
25
23
25
/
x
/
x
/
/
4
0
23
4
0
25
3
2
5
2
/
/
x
x
3
42
5
42
23
25
/
/
3
165
64
.5
12
16
3
5
64
9696
dxx
dxx
02
0
4
0
224
0
dxx
xdx
/214
0
4
0
2
3
162
2
4
0
2
x
dxxgxf
dxxgxfy
b
a
b
a
2
22
3
162
2
42
3
32
8 .
4
3
32
38
8/11/2019 Math 37 - Chapter 3
17/2817
9797
xxf
(2.4, 0.75)
The centroid of the region bounded by the
graphs of , and thex-axis is atxy 4x
(2.4, 0.75).
98
24 xxf
Solution:
Example 3.4.3
Find the centroid of the region bounded by the
graphs of and .2
4 xy 2xy
2xxg
022 xx
242 xx
012 xx
.xx 1or2
(-2,0)
(1,3)
9999
dxxgxf
dxxgxfxx
b
a
b
a
dxxx
dxxxx
24
24
21
2
21
2
dxxx
dxxxx
21
2
21
2
2
2
dxxx
dxxxx
21
2
231
2
2
2
1
2
23
1
2
342
232
34
xx
x
xxx
2
2
3
222
2
1
3
12
3
2
4
22
3
1
4
11
23
342
100
23
84
2
1
3
12
3
842
3
1
4
11
3
8
62
1
3
1
2
3
82
3
1
4
11
18
1
9
2
4
1
2
94
1
2
15
4
1
101
dxxx
dxxx
242
24
21
2
2221
2
dxxgxf
dxxgxfy
b
a
b
a
2
22
dxxx
dxxxxx
21
2
2421
2
22
44816
dxxx
dxxxx
21
2
421
2
22
4912
102102
dxxx
dxxxx
21
2
421
2
22
4912
1
2
23
1
2
25
3
2322
25
312
xx
x
xx
xx
2
2
3
222
2
1
3
122
225
2232122
5
1312
23
25
3
23
84
2
1
3
122
85
322424
5
17
425
12
9
5
108
2
152
5
3315
.
8/11/2019 Math 37 - Chapter 3
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103103
(-1/18, 2.4).
The centroid of the region bounded by the
graphs of and is at24 xy 2xy
24 xxf
2xxg
104104
ygxcy
dy
yfx
dyygyf
dxygyfx
d
c
d
c
2
22
dyygyf
dyygyfyy
d
c
d
c
y,x
105
12 xy
Solution:
Find the centroid of the region bounded by the
graphs of and .
Example 3.4.4
12 xy 5x
5x
(5,-2)
(5,2)
1
2
y
ygx 5yf
106106
dyygyf
dxygyfx
d
c
d
c
2
22 dyy
dxy
152
15
22
2
2222
2
dyygyf
dyygyfyy
c
c
c
c
dyy
dxyy
15
15
22
2
22
2
The centroid of the region
is at where,y,x
107
Theorem of Pappus
When a plane region R is revolved
about a line which does not intersect the
region except possibly at its boundaries,
the volume of the solid formed is equal to
the area of the region times the distance
traveled by the centroidof the region.
108
y,x
Let Abe the area of the
region.
d
The distance traveled by the
centroid of the region when
the region is revolved about
the given line is 2d.
The volume of the solid of
revolution is A*(2d).
8/11/2019 Math 37 - Chapter 3
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109
Use the Theorem of Pappusto find the volume of
the solid formed when the region bounded by the
graphs of , and thex-axis
is revolved about the indicated line.
a. y= 0 b. y= 6 c. x= 0 d. x= -2
Example 3.4.5
xy 4x
110
From Example 3.4.2,
the centroid of the
region is at (2.4, 0.75).
Solution:
xxf
4x
0xgAlso from Example
3.4.2, the area of the
region is 16/3 sq. units.
111
(2.4, 0.75) 750.yd
3
16V 7502 .
4
32
3
16
.unitscubic8
a. Abouty= 0.
112112
b. Abouty= 6.
4
216
4
3 yd
3
16V
4
212
.unitscubic56
(2.4, 0.75)
6y
113
(2.4, 0.75)
c. Aboutx= 0.
5
1242 .xd
3
16V
5
122
.unitscubic5
128
114114
d. Aboutx= -2.
5
222
5
122 xd
3
16V
5
222
.unitscubic15
704
(2.4, 0.75)
2x
8/11/2019 Math 37 - Chapter 3
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115
3.5 Center of Mass of a Solid of
Revolution
In this section, we only consider
finding the center of mass of a solid of
revolution formed when a region whose
boundary includes the x-axis is revolved
about the x-axis or when a region whose
boundary includes the y-axis is revolved
about the y-axis.
116116
Remarks:
In a solid of revolution, the center of mass
lies along the axis of revolution, that is,
i. if the solid of revolutionSis generated by
revolving a region Rabout the x-axis, then
the centroid of Sis located at . 00,,x
ii. if Sis generated by revolving a region R
about the y-axis, then the centroid of Sis
located at . 00 ,y,
117117
x
y
z
x
y
z
x
y
00,,x xfy
b
a
b
a
dxxf
dxxfxx
2
2
118118
x
y
00 ,y, ygx
d
c
d
c
dyyg
dyygyy
2
2
x
y
z x
y
z
119
Example 3.5.1 Find the center of mass of the
solid formed when the region bounded by the
graphs of
and thex-axis in the first quadrant is revolved
about thex-axis.
xl ny ex,
120
0y
xlnyex
Solution:
The centroid of the
solid is at
where,
00,,x
e
e
dxxln
dxxlnxx
1
2
1
2
8/11/2019 Math 37 - Chapter 3
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121
dxxlnx2
udv
vduuv
dxxxlnxx
xln1
222
222
xdxlnx
xlnx
2
22
xdxdv,xlnu 2
2
12
2xv,dx
xxlndu
122
xdxdv,xlnu
2
12x
v,dxxdu
xdxlnx
udv vduuv
dxxxx
xln1
22
22
xdxxlnx
2
1
2
2
Cxxlnx
22
1
2
22
Cxxlnx
42
22
123
dxxlnx2
xdxlnxxlnx
2
22
xdxlnx Cxxlnx
42
22
Cxxlnxxlnx
dxxlnx
4222222
2
Cxxlnxxlnx
dxxlnx 422
22222
124
e
e xxlnxxlnxdxxlnx
1
2222
1
2
422
4
1
2
1
2
1
422
22222 lnlneelneelne
0 01 1
4
1
4
1
4
22
ee
125
dxxln2
udv dxdv,xlnu 2
xv,dxx
xlndu 1
2 vduuv
dxxxlnxxxln
12
2
xdxlnxlnx 22
Cxxlnxxlnx 22
Cxxl nxxl nx 222
126126
ee xxl nxxl nxxl n 1221
22
212122 2221
l nl neel neel nexl ne
0 0
222221
eeeexl ne
1 1
The center of mass of the solid is at where, 00,,x
e
e
dxxln
dxxlnx
x
1
21
2
241
24
12
2
e
e
e
e
242.
8/11/2019 Math 37 - Chapter 3
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127
Example 3.5.2 Find the center of mass of
the solid formed when the region boundedby the graphs of
,
and they-axis in the first quadrant is
revolved about they-axis.
2xy 4y
128
2xy
Solution:
The centroid of the region
is at where,00 ,y,
d
c
d
c
dxyg
dxygyy
2
2
4
0
2
4
0
2
dxy
dxyy
4
0
4
0
2
ydy
dyy
.3
8
8
3
64
4
0
2
4
0
3
2
3
y
y
129
3.6 Length of Arc of a Curve
Let the function f and its derivative f
be continuous on the closed interval [a, b].
af,a
xfy
bf,b
130
af,a
xfy
bf,b
Subdivide the closed interval [a, b] into
nsub-intervals by choosing (n- 1)
intermediate numbers .121 nx,...,x,x
x1
|
xi-1
|
xi
|
xn-1
|
x0
|
xn
|
131
The distance between (xi-1, f(xi-1)) and
(xi, f(xi)) is
21
2
1 iiiii xxxfxfd
221 xxfxf iii
22
2
1 1 xx
xfxfi
i
ii
2
2
1 1 xx
xfxfi
i
ii
132
x
x
xfxfd
i
i
ii
i
2
11
By the Mean-Value Theorem, there exists
an iin [xi-1, xi] such that
1
1
ii
ii
ixx
xfxff
x
xfxf
fi
ii
i
1
8/11/2019 Math 37 - Chapter 3
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133
xfdiii
2
1
An approximation to the length of the curve
from (a, f(a)) to (b, f(b)) is
.xfdii
n
i
i
n
i
2
11
1
134
b
adxx'fL
21
The length of the curve from (a, f(a)) to
(b, f(b)) is
.xfl imdl imLii
n
in
i
n
in
2
11
1
135
Example 3.6.1 Find the length of the curve
given by from the point where to the
point where
xey 1x.x 2
Solution:
b
a dxx'fL2
1
2
1
2
1 dxex
2
1
21 dxe x
xx ex'fexf
136
dxe x2
1
dxy
ydy
12
xey2
1
xey22
1
dxeydy x222
dxeydy x2
dxyydy 12
y1
2 y
ydy
12
2
y
dyy
dyy 1
11
2
dy
yy 12
1
12
11
Cyl nyl ny 12
11
2
1
137
Cy
yl nydxe
x
1
1
2
11
2
Ce
el nydxe
x
x
x
11
11
2
11
2
22
2
1
21 dxe x
2
1
2
2
11
11
2
1
x
x
e
el ny
138
11
11
2
11
11
11
2
12
2
2
4
4
e
el n
e
el n
11
11
2
1
11
11
2
11
2
2
4
4
e
el n
e
el n
11
11
11
11
2
11
2
2
4
4
e
e
e
el n
8/11/2019 Math 37 - Chapter 3
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139
Solution:
2
2
44
x
xx'fxxf
Example 3.6.2 Find the length of the curve
given by on the interval24 xy .,22
b
adxx'fL
21
2
2
2
24
1 dxx
x
2
2 2
2
41 dxx
x
140
2
2 24
12 dx
x
2
0 2
0
2 24
124
12 dxx
dxx
(Improper of type DB and DA)
b
baa
dxx
l imdxx
l im0 22
0
22 4
12
4
12
bxb
a
x
a
si nArcl imsinArcl im
022
0
22
22
141
2
2
02 aa
sinArcsi nArcl im
022
2
sinArcsinArcl im bb
1-1
x
y
2
2
0
0
2
2
2 aa
sinArcl im
2
2
2 bb
si nArcl im
2
2 2
2
2
142
Let the function g and its derivative gbe
continuous on the closed interval . Then the
length of the curve from the point
to the point is given by
d,c ygx c,cg
d,dg
d
cdyy'gL
21
143
Example 3.6.3 Find the length of the curve
given by from the point where to the
point where
1xy 1y
.y 2
Solution:
2
111
yy'g
yygxxy
d
cdyy'gL
21
2
1
2
2
11 dy
y
L
2
1
4
11 dy
y
The evaluation of the integral is left to you as an exercise.144
3.7 Area of a surface of
revolution
When a region
is revolved about a
line, a solid is
formed.
When an arc is
revolved about a
line, a surface is
formed.
8/11/2019 Math 37 - Chapter 3
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145
.rrlSA 21
l
1r
2r
From Euclidean Geometry, the surface
area of a frustum of a cone is
The frustum is generated by revolving
the red segment about the blue one.
146
Let the function f be continuous on
the closed interval [a, b], differentiable on
the open interval (a, b) and f(x) 0 foreach xin [a, b].
Let Cbe the arc of the graph of fon
[a, b]. We wish to find the area of the
surface formed when Cis revolved about
the x-axis.
147
x
y
y=f(x)
ba| | x
y
ba| |
x1|
xi-1|
xi|
xn-1|
148
The length of the ith segment on the curve is
xfdiii
21
11With ixfr and, 2 ixfr
,xfl ii 2
1
The area of the ith frustum is
iiiii
xfxfxfSA 12
1
21rrlSA
149
,xfxfiii 21
2
xff iii 2
12
2
1 ii
i
xfxfxf
xfxf iii 212
The area of the surface of revolution is
approximately equal to
.xff iiin
i
2
1
12 150
b
adxx'fxfSA
212
Distance of a point of the
curve from the axis of
revolution
Element of arclength
xffl imSA iiin
in
2
1
12
The area of the surface of revolution is equal to
b
adsldSA 2
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151
Example 3.7.a Find the area of the surface
formed when the arc of from the point
where to the point where is
revolved about the given indicated line.
xey
1x 2
x
a. y= 0 d. x= 0
b. y= 9 e. x= 5
c. y= -2 f. x= -1
Solution:
.dxeL x 2
1
21
From Example 3.6.1, the length of the
arc is given by
152
x
y
xey
.dxeL x 2
1
21
a. y= 0
b
adxx'fxfSA
212
2
1
212 dxee xx
153
x
y
xey
.dxeL x 2
1
21
b. y= 9
2
1
21?2 dxeSA x
2
12192 dxee xx
154
x
y
xey
.dxeL x 2
1
21
c. y= -2
2
1
21?2 dxeSA x
2
1
2122 dxee xx
155
x
y
xey
.dxeL x 2
1
21
d. x= 0
2
1
212 dxex x
2
1
21?2 dxeSA x
156
x
y
xey
.dxeL x 2
1
21
e. x= 5
2
1
21?2 dxeSA x
x = 5
2
1
2152 dxex x
8/11/2019 Math 37 - Chapter 3
27/2827
157157
x
y
xey
f. x= -1
.dxeL x 2
1
21
2
1
21?2 dxeSA x
x= -1
2
1
2112 dxex x
158
Example 3.7.b Find the area of the surface
formed when the arc of from the point
where to the point where is
revolved about the given indicated line.
xey1x 2x
a. y= 0 d. x= 0
b. y= 9 e. x= 5
c. y= -2 f. x= -1
Solution:
lnxy e y x
1dy dx
y
1dx
dy y
159
2
2
11
e
eL dy
y
The length of the arc is given by
2
2
11
e
eL dy
y
xey
,1when x ey
,2when x 2ey
160160
Example 3.7.c Find the area of the surface
formed when the arc of from the point
where to the point where is
revolved about the given indicated line.
1y
x1y 2y
a. x= 0 d. y= 0
b. x= 3 e. y= 3
c. x= -1 f. y= -1
Solution:1
xy
2
1dx
dy y
161161
b. x= 3
x
y
yx
1
a. x= 0
c. x= -1
2
1 4
11?2 dy
ySA
x= 3x= -1
2
1 4
1
1 dyyL
d
cdyy'gygSA
212
2
1 4
11
12 dy
yy
2
1 4
11
132 dy
yy
2
1 4
11?2 dy
ySA
2
1 4
11
112 dy
yy
162162
d. y= 3
x
y
yx
1
c. y= 0
e. y= -1
2
1 4
11?2 dy
ySA
y = 3
y= -1
2
1 4
1
1 dyyL
d
cdyy'gySA
212
2
1 4
112 dy
yy
2
1 4
1132 dy
yy
2
1 4
11?2 dy
ySA
2
1 4
1112 dy
yy
8/11/2019 Math 37 - Chapter 3
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163163
The End
of Chapter