Download Application of Definite Integrals

Lesson-14
Introduction to Calculus
Application of
Integrals (continuation)
1.
Volumes of revolution
The volume of revolution can be determined in the same way as the
area under the curve. The area is divided into a large number of strips. If
this area is made to rotate about the x-axis these strips all rotate and
produce figures, which are very nearly discs with uniform areas of cross
section, see in Fig.14.1 below.
Fig. 14.1
The summation of the volumes of the discs gives the volume of
revolution. As the thickness of the discs dx → 0 they can be regarded ad
discs with uniform area of cross-section.
Volume of disc = area of cross-section times thickness
V = y2x
Where y is the radius of the disc.
The volume of revolution between the limits x = a and x = b is the sum of
the volumes of these discs. When this summation is an integral; that is
V = ∫ a y2 dx
b
(14.1)
Example 1
Find the volume of revolution between the limits x = 2 and x = 5, when
the curve y = 2x2 is rotated about the x-axis.
Solution:
According to the formula 14.1
5
V = ∫ 2 y2 dx = ∫ 2 (2x2)2 dx = 4∫ 2 x4 dx = 4 x [ 2 = 4( 5
5
5
5
5
5
5
5
-
25
) = 2473 (cubic units)
5
2.
The concept of work
Fig. 14.2 shows a force F acting a body (box) moving to right on a
surface. If  is the angle between the force and the direction of motion,
then force F cos  is the component of F parallel to the direction of motion.
Fig. 14.2
For the work W done by F when the body has displacement x we can
write.
W = (F cos ) x
When n times x distances move the body, the work we can write down.
n
W = lim [  f ( x) cos a ]x
(14.2)
i 1
But the limit of the sum in equation above is simple a definite integral
between a and b.
b
W =∫ a f (x) cos  dx
(14.3)
Example 2
A force F = 3x2 - 2x acts on a body moving along the x-axis. Here
x is in meters and F is in Newtons (N). If the force is applied at an
angle of 300 with the direction of motion, find the work done by the
force in moving the object from x = 1m to x = 2m.
Solution:
The work is calculated from equation (14.3) by substitution of
f (x) = 3x2 – 2x
Data:
 = 300
S = 1m
b = 2m
W = ∫ 1 (3x2 - 2x) cos 300 dx = cos 300 ∫ 1 (3x2 – 2x) dx
2
2
3
2
3
2
= cos 300[3 x - 2 x ] 1 = cos 300[(2)3 – (2)2] 2
- [(1)3 – (1)2] = 0.8660 4 = 3.46N
Example 3
A cylindrical tank of radius 20cm and height 100cm is filled with water.
Give that the density () of water is 1000kgm-2; find the work done in
emptying the tank.
Solution:
At a height of y m from the base of the tank (see above), imagine an
element of water of radius 0.2m and height dy. Its volume is V = r2dy, and
thus its weight is:
F = Vyg where
g = 9.81ms-2 is acceleration due the gravity. This element
must be raised the remain distance of 1 - ym, and the work necessary to do
this is F (1- y) dy. We can look at this as the differential element of the
work, dW, that is
W = ∫ 0 F(1 - y) dy
1
Data:  for water = 1000kg/m3
r = 20cm = 0.2m
a=0
b = 100cm = 1m
F = Vyg = r2g dy = 10000.229.81dy = 392.4dy (N)
2
1
1
1
W = ∫ 0 F (1 – y) dy = 392.4∫ 0 (1 – y) dy = 392.4(y - y )[ 0
2
W = 392.4[(1 – ½) - (0 –0)] = 196.2 (Joules)
3.
The rectilinear motion.
By analogy to the work done, both the displacement and the velocity
maybe written as the definite integral with respect of time.
t2
Distance, d = ∫ t V dt
(14.4)
1
t2
Velocity, V = ∫ t a dt
(14.5)
1
Example 4
A body moves in a straight line so that its velocity V at time t is given
by the equation V = 48t – t3 (ms-1).
How far does it move between the two instants t = 0 and t =
4
seconds?
Data :
t1 = 0s
t2 = 4s
V = 48t - t3
t2
Distance = ∫ t (48t – t3) dt = ∫ o (48t – t3) dt =
4
1
t3
t2
2
3
4
= [48 2 - 3 ] 0 = [48 (4) - (4) ] - [(0) – (0)] = 384 – 64 =
2
3
= 320m
Example 5
A particle starts from rest at the origin and moves along the x-axis so
that its acceleration after t seconds is a = 30 - 6x. When and where will
the particle come to rest again?
Solution:
Data:
t1 = t s (after t)
t2 = 0
(at rest)
a = 30 – 6x
V=?
S=?
t2
2
t
V =∫ t (30 – 6t) dt = [30t -6 t ] 0 = [(30t - 3t2) - (0 - 0)] =
1
2
= 3t(10 – t)
The particle will be again at rest when Velocity V drops to 0.
therefore:
V = 3t(10 - t) = 0
then t = 10 seconds and of course at t = 0 (stated at the beginning).
How far the particle will travel during this 10s we find from the equation
(14.4).
Where t1 = 0 and t2 =10s
S = ∫ 0 (30x - 3x2) dx = [15x2 – x3] 0 =
10
10
= [15(10)2 - (10)3] - (0 - 0) = 1500 - 1000 = 500m
Answer
The particle will come to rest again after 10s and travelling 500m
EXERCISES:
In the problems 1- 4, find the work done by the given force in moving
an object along the x-axis from an initial position x = x1, to a final
position x = x2. Assume the force is applied in the direction of the
displacement (x = 0).
1.
F = 3N, x1 = 0, x2 = 2m
2.
F = 8N, x1 = - 2m, x2 = 3m
3.
F = (3x + 1)N, x1 = - 1m, x2 = 2m
4
F = 3ex N, x1 = 0, x2 = 2m
5.
One force acting on a body as a function of position is given by F =
3x2 + 2x (Newton), G = 100 (Newton) and T = 100 – (2x + 20) sin 600
(Newton). If the object moves from x1 = 1m to x2 = 3m, find the work
done by the force if the angle between the force and the direction of
the displacement is:
a) = 00
b) = 300
c) = 1200
d) = 1800
6.
Three force act on the crate (shown below). If F = (2x + 20)N, G =
100N and T = [- 100 - (2x + 20) sin 60O]N, find the total work done as
the crate moves horizontally from x = 1m to x = 3m.
7.
In the following problems, the V =f (t) equation of a moving body is
given. Compute the distance travelled in the indicated time interval.
(units are ms-1)
a)
V = 4t +3, during the third second.
b)
V = 3t2 - 6t + 4, during the first second.
c)
V = 6 - t - t2, between t = 2s and t =5 s.
d)
V = 3t2 + 4t + 3, during the fourth second.
e)
V = t3 + 2t, during the first fourth seconds.
f)
V = t + cos t, between t = 0 and t =¼(measured in the
Radians)
g)
8.
V = 2t - sin t; between t = 0 and t =1/3
The acceleration a of a body moving in relation to an origin is given by
a = 2 + 3t for t ≥ 0. If the initial displacement is + 5, and the body attain
a velocity 10ms-1 after 10 seconds, find the displacement after 1
second.
9.
The acceleration of a moving body at distance x from an origin is given
by a = 4x - 2. If the velocity is 1ms-1 at x = 1m, find the velocity at x =
2m.
10. A particle commences to move along the x-axis at the point x = 16,
with a velocity
acceleration
of
8ms-1 towards
the
origin
and
constant
of 2ms-1, away from the origin. Find its velocity as it
passes through the origin. Also find the V = f (t) and x = f (t) equations.
Answers:
1.
6J
2.
40J
3.
7.5J
4.
19.2J
5a.
34J b.
17 3 J
c.
-17J
d.
-3 4J
6.
24J
7a.
13m
b.
2m
c.
31.5m
d.
54m
7e.
80m
f.
0.5m
g.
(1/92 – 1/3)m
8.
13m 3ms-1 9.
a = 0, V = 2t – 8, x = t2 – 8t + 16