36
Week 14 Week 14 Integration Integration Engineering Applications Engineering Applications
Week 14Week 14
IntegrationIntegration
Engineering ApplicationsEngineering Applications
Some common applications of integration are:
1. Volume of a solid of revolution
2. Centroid of a plane area
3. Centre of gravity of a solid of revolution
4. Arclength and surface area
5. Mean values
6. Root mean square values
Page 2
Applications of Integration
1. Volume of a Solid of a Revolution
Imagine rotating the plane area A under the graph of the function through a complete revolution about the x axis, as in Figure 14.1.
Page 3
Applications of Integration
],[),( baxxf
Figure 14.1
1. Volume of a Solid of a Revolution (continue)
The result would be to generate a solid having the x-axis as axis of symmetry as shown in Figure 14.2(a).
This is called a solid of revolution.
Page 4
Applications of Integration
Figure 14.2(a)
1. Volume of a Solid of a Revolution (continue)
To determine the volume of this solid, we subdivide the rotating area into n vertical strips.
When a vertical strip within the subinterval is rotated through a revolution about the x-axis, it will generate a thin disc of radius f(xr*) (xr-1< xr*< xr) and thickness ∆xr-1 as shown in Figure 14.2(b).
Page 5
Applications of Integration
Figure 14.2(b)
],[ 1 rr xx
1. Volume of a Solid of a Revolution (continue)
The volume of each disc is given by
Thus, the volume of the solid can be approximated by
This approximation is closer to the exact volume as the number of strips is increased. Thus, limiting case as n ∞ and ∆x0, ∆x=max ∆xr gives the volume as,
Page 6
Applications of Integration
1
2*)]([ rrr xxfV
1
2
11
*)]([
rr
n
rr
n
r
xxfVV
b
a
rr
n
rxn
dxxfxxfV 2
1
2
10
)]([*)]([lim (14.1)
2. Centroid of a Plane Area
Consider the plane region of Figure 14.3 bounded between the graphs of the two continuous functions f(x) and g(x) on the interval with g(x)≤f(x) on the interval:
Page 7
Applications of Integration
],[ bax
Figure 14.3
2. Centroid of a Plane Area (continue)
The area A of this region is clearly given by:
A = area under graph f(x) – area under graph g(x)
That is,
Page 8
Applications of Integration
dxxgdxxfb
a
b
a )()(
dxxgxfAb
a )]()([ (14.2)
2. Centroid of a Plane Area (continue)
The centroid of the area is given by coordinates
where,
and area A is given by equation (14.2).
Page 9
Applications of Integration
),( yx
dxxgxfA
y
dxxgxfxA
x
b
a
b
a
22 )]([)]([2
1
)]()([1
(14.3)
(14.4)
2. Centroid of a Plane Area (continue)
In the particular case when g(x) is the x-axis, the centroid of the plane area is bounded by
and the coordinates of the centroid is given by
dxxfA
ydxxxfA
xb
a
b
a 2)]([2
1,)(
1
Page 10
Applications of Integration
]),[)(( baxxf
(14.5)
3. Centre of Gravity of a Solid of Revolution
We can obtain the coordinates of the centre of gravity of the solid of revolution generated by
as in Figure 14.1.
By symmetry, it lies on the x-axis, so that
where volume V is given by equation (14.1).
dxxfxV
X
Y
b
a
2)]([
0
Page 11
Applications of Integration
]),[)(( baxxf
(14.6)
),( YX
Example 14.1:
The area enclosed between the curve and the ordinates x = 2 and x = 5 is rotated through 2π radians about the x-axis. Calculate
(a) The rotating area and the coordinated of its centroid
(b) The volume of the solid of revolution generated and the coordinates of its centre of gravity.
Page 12
Applications of Integration
)2( xy
Solution 14.1:
(a) The rotating area is the shaded region shown in Figure 14.4.
From Equation 14.2,
where
Thus,
Page 13
Applications of Integration
Figure 14.4
dxxgxfAb
a )]()([
5,2,0)(,)2()( baxgxxf
dxxA 5
2
2/1)2(
32)2(3
25
2
2/3
x square units
Solution 14.1:
(a) To find the coordinates of the centroid, note that g(x) is actually the x-axis. Thus, using formula 14.5:
known A =2√3
Thus,
Page 14
Applications of Integration
dxxxfA
xb
a )(1
dxxxA
5
2
2/1)2(1
dxxxA
5
2
2/12/3 )2(2)2(1
5
2
2/32/5 )2(3
4)2(
5
21
xx
A
2/32/5 )3(3
4)3(
5
21
A
35
381
A
5
19x
Solution 14.1:
(a) Similarly,
known A =2√3
Thus,
Hence, the coordinates of the centroid are
Page 15
Applications of Integration
dxxfA
yb
a 2)]([2
1dxx
A
5
2
22/1 ])2[(2
1
dxxA
5
2)2(
2
1
5
2
2)2(2
1
2
1
x
A
2
9
2
1
A
8
33y
8
33,
5
19
Solution 14.1:
(b) The volume V of the solid of revolution is calculated from equation 14.1:
Page 16
Applications of Integration
cubic units
b
a
dxxfV 2)]([
5
2
22/1 ])2[( dxx
5
2
)2( dxx
5
2
2)2(2
1
x
2
9
Solution 14.1:
(b) The coordinates of its centre of gravity is given by equation 14.6.
Since the solid is symmetrical on the x-axis,
Thus, coordinates of centre of gravity is (4, 0).
Page 17
Applications of Integration
0Y
dxxxV
X 5
2
22/1)2(
dxxxV
5
2)2(
dxxxV
5
2
2 2
5
2
23
3
x
x
V
V
18 Known
2
9V
4X
4. Arclength and Surface Area
In many practical problems, we are required to work out the length of a curve or the surface area generated by rotating a curve.
The formula for the length s of a curve with formula y = f(x) between 2 points corresponding to x =a and x = b is given by the equation:
Page 18
Applications of Integration
dxdx
dys
b
a
2
1 (14.7)
4. Arclength and Surface Area
The surface area S generated by s when it is rotated through 2π radians about the x-axis is given by
Page 19
Applications of Integration
dxdx
dyyS
b
a
2
12 (14.8)
Example 14.2:
A parabolic reflector (Figure 14.5) is formed by rotating the part of the curve between x = 0 and x = 1 about the x-axis. What is the surface area of the reflector?
Page 20
Applications of Integration
xy
Figure 14.5:
A parabolic reflector
Solution 14.2:
Given differentiating gives,
From equation (14.8) the surface area is given by
Page 21
Applications of Integration
2/1xy x
xdx
dy
2
1
2
1 2/1
dxdx
dyyS
1
0
2
12 dxx
x 1
0 4
112
dxx
x 1
0 4
112
dxx
xx
1
0 4
142
Solution 14.2 (continue):
Page 22
Applications of Integration
dxdx
dyyS
1
0
2
12 dxx
xx
1
0 4
142
dxx
xx
1
0 2
142
dxx
xx
1
0 2
142
dxx 1
014
1
0
2/3)14(3
2
4
1
x
)15(6
1 2/3 square units
Example 14.3:
The curve described by the cable of the suspension bridge shown in Figure 14.6 is given by
Where x is the distance measured from one end of the bridge. What is the length of the cable?
Page 23
Applications of Integration
hxl
h
l
hxy
22
2
Figure 14.6:
Suspension bridge
Solution 14.3:
Given, equation of the curve is
Differentiating gives
Page 24
Applications of Integration
hxl
h
l
hxy
22
2
1
22
2
l
x
l
xh
2
1
l
xh
1
2
l
x
l
h
dx
dy
Solution 14.3 (continue):
From equation (14.1), the length s of the cable is
To simplify the integral, let
Thus,
Page 25
Applications of Integration
dxl
x
l
hs
l
2
0
2
)1(2
1
dxl
x
l
hl
2
2
22
01
41
1
2
l
x
l
ht
dtth
ls
lh
lh /2
/2
22
)1(2
dtth
ls
lh
/2
0
22
)1( (from symmetry)
Solution 14.3 (continue):
Recall from Week 13 notes, the integral obtained can be further simplified using substitution of , giving
Page 26
Applications of Integration
duuh
ls
lh
/2sinh
0
22 1
cosh
ut sinh
duuh
l lh
/2sinh
0
2 1
12cosh
)/2(sinh
0
21
2sinh2
1lh
uuh
l
)/2(sinh
0
21
coshsinhlh
uuuh
l
l
h
l
h
l
h
h
l 2sinh
41
2 1
2
22
l
h
h
lhl
2sinh
24 1
222
5. Mean values
In many engineering applications we often need to know the mean value of a continuously varying quantity.
When dealing with a sequence of values, the mean value can be computed simply by adding the values together and then divide by the number of values taken.
When dealing with a continuously varying quantity, we cannot do the same directly.
Page 27
Applications of Integration
5. Mean values (continue)
In continuous varying quantity problems, we can compute the mean value by using integration.
Consider a function f(x) with graphical
representation as in
Figure 14.7:
Page 28
Applications of Integration
Figure 14.7
5. Mean values (continue)
The sum of the shaded areas above the line y = (mean value) is equal to the sum of the shaded areas below it, so that the area of the rectangle ABCD is the same as the area between the curve and the x-axis.
The mean value is given by
Page 29
Applications of Integration
Figure 14.7
dxxfab
xfvmb
a )(1
))(.(.
(14.9)
Example 14.4:
Find the mean value of function
between the interval of t = 1 and t = 3.
Page 30
Applications of Integration
2t)t(f
Solution 14.4:
Using equation (14.9), the mean value is given by,
Page 31
Applications of Integration
dx)x(fab
1v.m
b
a
dtt13
1v.m
3
1
2
3
1
3
3
t
2
1v.m
3
13
3
1
3
3
2
1v.m
33
6. Root Mean Square values
In some context, the computation of the mean value of a function is not useful.
For example, the mean of an alternating current is zero but that does not imply it is not dangerous!
In such situations, we use root mean square (r.m.s) of the function f(x).
Page 32
Applications of Integration
6. Root Mean Square values (continue)
Literally, this is the square root of the mean value of [f(x)]2, written as
Common applications of r.m.s. values are in electrical engineering.
Page 33
Applications of Integration
dxxfab
xfsmrb
a
22)(
1))(.(.. (14.10)
Example 14.5:
An electric current i is given by the expressions
where I is a constant. Find the root mean square value of the current over the interval .
Page 34
Applications of Integration
sinIi
20
Solution 14.5:
Using equation (14.10), the r.m.s. value of the given current is,
Thus, r.m.s current,
Page 35
Applications of Integration
dxxfab
xfsmrb
a
22)(
1))(.(..
dIismr
2
0
222sin
02
1...
dI 2
0
2
)2cos1(2
1
2
222
0
2
2
12
42sin
2
1
4I
II
22...
2 IIismr
Thank You!Thank You!
