VECTOR INTEGRATION
The vector integration is similar in all respects to
The ardmany integration of algebraic functions.
As usu
r = xity; + zK
So (dr at = rat) + C
where c is the arbitrang u constant vector which is
independent of t. It is also the indefinte integral of
Also the definite integral of n between a andb,
r.
act «b, is given by
a
Example 1: Gwen r = (at -E) i + 3t] - 4K. Evaluate
a fralt as f'rolt.
Solution.
1 frat = : (F-1+4) +) (8+4) - K(44+6)
= (1-5) i + 1; - At K + G, i+ Sj-SK
= (1-1)i + t3 - 44k + C,
is a constant veclor which is independent
where
C
of t (C = Gi+ Sj - GK , G, S, g ane constants).
1- 61+ 63j - 12K.
=
Example a: If the acceleration of a body is goven by
a = 24 cos at i + 16 sinat] - 6tK,
find the velocity I and displacement o when both
time shes and it = 0 and t= Th. Find their respective
Vanishes and It =0 and t'= Tip. Find their respective
scalar components.
Solution
- dIV = 24 coset i + 16sinatj - 6tK
a=
dt
V = S (24cosat i + 16sinat; - 61 k) dt
= 12sinati - 8cos atj-st'K + G
For V=0 (Ine V vanishes) and t=0
0 = 12 sin (0) 1 - 8 Cos (0); - 300)K + G
= 0-8; -0+ C
= 4 =
Therefine
8j
V = 128 nati - 8 cosatj - 3t'k +8;
= 12 sin 2t i - 8(Cos at -1); - 3t'K
for t=7/4
V = 12i + 8j - 1.85K (show detwils).
Also 63
.:r = J [12sinati - 8(cosat -8) -31 K] dit
=- 6Cosati - (4 sinst -84) j-1'* + C
Fir i#O (tile + vanishes) and t=0
C2 = Gi
+ = - 6 (Cos2t -1) i - 4(sin2t-2t) j - t'K
Also
Fir r=0 and t= Th
+ = 61 +2-2883 - 0.484K (Show detub)
Hence
Ir/ a 6044 units
/V/ a 14.54 rint / sec
Exercise
Given acceleration
a = (25 cos 2t) i + (16sin 2t); + 9EK.
Repent the questions of the last example
LINE INTEGRAL
Let Po and Pr be any two paints janed by an arc C
which is assumed to be continuous at each of its pointe
and does not intersect rtself. If F is a vector finchen
such that
and displacement r = xit yg+ ZK is the position
Vector of P(x, y, 2) on C. Then, the line inter
the integral along the given path C und is defned
Fidr = Fidr = Pedae+ Pady+ Bade,
C P.
where F= Rij+ Paj+ fak and dr = dei+dyj+dzK.
Thes, live integral represent the workdone by the
force I in moving partice along the path O from to
to b,
If I is a closed curve, the integral around c
is aften denated by
§Fadr = S fide + I dy + Be de.
In aerodynamics and fluid dynamics, this integra
called the circulation of F about C, where
is called
represents the velocity of the fluid.
( a straight line from (0,1) to (2,4)
ught hme from (0,1) to (12) and then (,) to 123
ON) the Curves (or parabola) x= at and y = t'73.
SOLUTION
1) From (0,1) to (2,4).
The eguntion jorning (9,1) to (2, 4) is
y = 22 =3 dy = 3 3 9 = 3x+ C
and c = 1.
The inlygnal is J' (ai-2(37+9) de
+ [2 (38+1) + 3x]g dx
=
= 95
1) From (0,1) to (12): the equation joining (0,) to (12)
is ydy==X+1 =
I -s dy = de (y= 27C , and c=1)
The live integral =
= (32-2y) dx + (ay'+ 2x) dy
(322-2(471) de + (28 +1) + 32) doe
=
= (62-22-2)de + (1x8'+7x +2) de
(6x75x) de = 2% = 4%
Also from (11) to fi3); the equation of the line faring
(11) to (73) is y = 22-1 => dy = 2dx
it The line mtegral = J (192-12+6) de
= 8% = 29⅓
The required value of the line integral is
35 + 88 = 291 = 33%.
til) Gwen that X=2t at (0,1), x=t = 0
At (2,4): t=1 and x = 2t
die = adt and y-f'73 as dy = at lt
"(32-2y)de + (2y +3x) dy
: The line integral =
(+76t + 9t'+ 8t -3) dt
=4
6% = 2273.
=
625)
ExercISE
1 Find the tokal work done in moving a partice in
the force field given by F = z1 + y + *K along
the helix C given by
*= cost, y-sint, z=t
from t=0 to t= M
2. Suppose A = (32°+ 67)i-1447 j+ 20X2 K.
Evaluate fader firm (000) to (110) along the
fillaung piths C.
a x=t, y=t,7=t
) the straight line from (9,90) to (190) then to
(1,1,0) and then to (1,11).
@ the straight line joinng (90,0) and (111).
1) Suppose a force freld is guven by
F= (2x-y+2)i + (x+y-2)j + (32-29+42) k.
Find the work done in moving a particle one around
a circe C in the sey-plane with its centre at the
Origin and a radundaf 3.
4) Es Suppose F is a conservilure field. Prove that
Curl F = DXF = 0 (re F is irratational).
4) Conversely, if PXF = 0, prove that F is conservative
NoTE
Theorem.
Suppose 1 = St everywhere in region R of a space, where
R is defred by 9, 5X692, b, 6y sha, 4, 5266
and where 464, 9, 2) is single- value and has contemar
derwalives in R. Then
9) GAto = O avanal any clared come C in R.
In such a case, A is called a conservative vector field
and of is its scalar potential.
SOLuTION TO EXERCISE NO. 2
2 SArds = [(32 + 6)i -14yzj + 2022 K]. (dxirdy +dek)
](32'+6y) de - 1yz dy +20x2 de
=
• rat get a ten but done net, them
to,
(gELt - 28t It + 60t' It
=
= (94 - 28t'+ 601) lt
- 38-48+61= 5
§ the straight-line from (090) to (1,90), y-0, 2=0
du =0. dz=0. while a vanes from O to 1. Then the
integral over this part of the path is
('(32-60)de - 446080)(0) + 20340760) = 32th
220 25д
Along the strught line from (100) tr (1,10).
S' (368°767)0 - 19y oldy + 20411010) = 0
4-0
Along the straught line (1,10) to (4, 11): X=1, y=1,
dr =0, dy =0, while z vanes from o to 1 Then the
integnal over this part of the puth is
('(308°+660)0-14(1)z (0) + 2041)2° da
ZE0 1022 doc = 202°|"
20
=
Z=0
3
Alans Jail = 1101 39 = 28.
(6)
6 The straught lime joining (0,0,0) and (111) is gaven
in parametric form x=t, y=t, z-t.
Then
JAndr =
B£ +6t-1442+2073) at
=
= 1⅔.
