Chaper Vecor Calculus. Elemenar. Vecor Produc. Differeniaion of Vecors 4. Inegraion of Vecors 5. Del Operaor or Nabla (Smbol 6. Polar Coordinaes
Chaper Coninued 7. Line Inegral 8. Volume Inegral 9. Surface Inegral 0. Green s Theorem. Divergence Theorem (Gauss Theorem. Sokes Theorem
. Elemenar Vecor Analsis Definiion. (Scalar and vecor Scalar is a quani ha has magniude bu no direcion. For insance mass, volume, disance Vecor is a direced quani, one wih boh magniude and direcion. For insance acceleraion, veloci, force
We represen a vecor as an arrow from he origin O o a poin A. O OA A or O a A The lengh of he arrow is he magniude of he vecor wrien as or. OA a 4
.. Basic Vecor Ssem Uni vecors,, Perpendicular o each oher In he posiive direcions of he aes have magniude (lengh 5
.. Magniude of vecors Le P (,, z. Vecor wih magniude (lengh OP p i j z k [,, z] OP p OP p z is defined b 6
.. Calculaion of Vecors. Vecor Equaion Two vecors are equal if and onl if he corresponding componens are equals Le a ai a Then a b a j a k b, andb bi b a b, a b j bk. 7
8. Addiion and Subracion of Vecors. Muliplicaion of Vecors b Scalars k b a j b a i b a b a ( ( ( k b j b i b b ( ( ( hen isascalar, If
Eample. Given p 5 i j - k and q 4 i - j k. Find a p q b p - q c Magniude of vecor p d q - 0 p 9
. Vecor Producs If a a i a j a k and b bi b j b k, Scalar Produc (Do produc or a b ab ab ab a. b a b cos, is heangle beween aandb 0
Vecor Produc (Cross produc i j k a b a a a or a b a bsin n b b b ab -ab i- ab - ab j ab -ab k
Whaever direcion of a b is, i has o be perpendicular o boh a and b. Le c a b Swing a o b The direcion of c is poined ino he page. hps://www.ouube.com/wach?vh0njk4meiju Do produc and cross produc Visualizaion
Applicaion of Muliplicaion of Vecors a b a b a Given vecors and, projecion ono is defined b The vecor projecion of a on b is he uni vecor of b b he scalar projecion of a on b: Vecor projecion ab. proj a b b b ab. comp a b b ab. lengh ( l b Scalar projecion a comp b a b The scalar projecion of a on b is he magniude of he vecor projecion of a on b.
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b The area of riangle A ab. b a 8
9 c The area of parallelogram d The volume of erahedrone e The volume of parallelepiped a b a b A a b c 6 c c c b b b a a a 6 V a. b c a b c c c c b b b a a a V a. b c
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Eample. and. beween angle andhe, deermine., and Given b a b a a b k j i b k j i a -
. Vecor Differenial Calculus Le A be a vecor depending on parameer u, A( u a ( u i a ( u j a ( u k The derivaive of A(u is obained b differeniaing each componen separael, z da da du da du da du i j z k du 4
A ( u. The nh derivaive of vecor is given b dn A dna dna dna i j k du dun n dun dun z d n A The magniude of is n du d n n du A dna n n n d a d a n n z du du du 5
Eample. If A u hence i-u j 5k da du d A du 6
Eample.4 The posiion of a moving paricle a ime is given b 4,, z 5. Obain The veloci and acceleraion of he paricle. The magniude of boh veloci and acceleraion a. 7
Soluion The parameer is, and he posiion vecor is r ( (4 i ( j ( 5 The veloci is given b dr 4i ( j ( d The acceleraion is dr j d 0 k (6 0 k.. k. 8
A, he veloci of he paricle is dr ( 4i (( j (( d 4i 5j k. and he magniude of he veloci is 0( k dr ( 4 5 d 0. 9
A, he acceleraion of he paricle is d r( j (6( 0 k d j 6k. and he magniude of he acceleraion is d r( 6 d 65. 0
.. Differeniaion of Two Vecors A ( B( If boh and are vecors, hen a b c d u u d ( ca c da du du d ( A B da db du du du d db da ( A. B A.. B du du d da ( A B A db du B du du du
.. Parial Derivaives of a Vecor If vecor depends on more han one parameer, i.e A,,, (,,, (,,, (,,, ( k u u u a j u u u a i u u u a u u Au n z n n n
Parial derivaive of wih respec o is given b e..c. A u, k u u a j u u a i u u a u u A k u a j u a i u a u A z z
Eample.5 6, 6, 6 4, 6, 4 hen ( ( If vi u v F u v F k ui v F uk j u F vk j uvi v F k u u j i v u F k v u j v u uv i F - - 4
Eercise. - u v F u v F v F u F v F u F k v u j v u u vi F,,, hen ( ( If 5
.4 Vecor Inegral Calculus The concep of vecor inegral is he same as he inegral of real-valued funcions ecep ha he resul of vecor inegral is a vecor. If hen Au ( a b a ( u i a Au ( du b a b a ( u a j a ( u k a ( u dui z ( u duj b a a ( u duk. z 6
Eample.6. 80 4 ] [ 5] [ ] [ 4 5 ( 4 (. calculae, 4 5 ( 4 ( If 4 k j i k j i dk dj di Fd Fd k j i F - - - - Answer 7
Eercise.. 7 4 7 4 ( (. calculae, 4 ( ( If 0 0 0 0 0 k j i dk dj di Fd Fd k j i F - - - Answer 8
.5 Del Operaor Or Nabla (Smbol Operaor is called vecor differenial operaor, defined as i j k. z 9
.5. Grad (Gradien of Scalar Funcions If,,z is a scalar funcion of hree variables and is differeniable, he gradien of is defined as grad i * isascalar funcion * isa vecor funcion j k. z 40
Eample.7 z z z z z z z z z z z,hence Given (,,. P a grad,deermine If Soluion 4
. 7 84. ( (( (( (( ( (( ( (( ( ( ( ((( wehave (,,, P A. ( ( ( z Therefore, k j i k j i k z z j z z i z z k j i 4
Eercise. If z z, deermine grad a poin P (,,. 4
Soluion 0. 6 (,,, P A Grad z hen, Given k j i z z 44
.5.. Grad Properies If A and B are wo scalars, hen ( A B AB ( AB A( B B( A 45
.5. Direcional Derivaive Direciona lderivaive of d ds wherea a. grad dr dr, in hedirecion ofais which isauni veco r in hedirecion ofdr. 46
Eample.8 Compue hedireciona lderivaive of z a hepoin (,, -in hedirecion ofhevecor A i j-4k. z 47
Soluion Direcional derivaive of in he direcion of d a. grad ds wheregrad i j k z Given z z, hence (z i (4 z j ( anda z k. a A. A 48
9. 4 ( hen, 4 given Also,. 9 6. (( (( ( (4(( ( ((( (,,-, A - - - - - - A k j i A k j i k j i 49
A Therefore, a i j- 4 k. 9 9 9 A Then, dφ a. ds i j- 4 k.(6 i 9j-k 9 9 9 (6 (9-4 ( - 9 9 9 5 9.47046. 9 50
.5. Uni Normal Vecor Equaion (,, z consan is a surface equaion. Since (,, z consan, he derivaive of is zero; i.e. d dr cos 0 90. dr.grad grad cos 0 0 5
This shows ha when (,, z consan, grad dr. grad dr z Vecor grad is called normal vecor o he surface (,, z consan 5
Uni normal vecor is denoed b Eample.9 Calculae he uni normal vecor a (-,, for z z 0. n. 5
Soluion Given z z 0. Thus 6. 4 and ( ( ( (-,,, A. ( ( ( - - k j i k j i k j z i z ( 6 6 is vecor normal uni The k j i k j i n 54
.5.4 Divergence of a Vecor...(. as defined is of divergence he, If z a a a A diva k a j a i a k z j i A diva A k a j a i a A z z z 55
Eample.0. (( (( (( (,,, poin A.. (,,. poin a deermine, If - - - diva z z z a a a A diva diva k z zj i A z Answer 56
Eercise.4. 4 (,,, poin A. (,,. poin a deermine, If - diva z a a a A diva diva k z z j i A z Answer 57
Remarks Aisa vecor funcion, bu diva isascalar funcion. IfdivA 0, vecor Aiscalled solenoid vecor. 58
.5.5 Curl of a Vecor. ( b isdefined of curl he, If z z z a a a z k j i A A curl k a j a i a k z j i A A curl A k a j a i a A 59
Eample. If A ( 4 - z i ( deermine curla a (,, -. j - zk, 60
Soluion. 4 ( ( ( ( ( ( ( ( 4 4 4 k j z z zi k z j z z z i z z z z z k j i A A curl - - - - - - - - - - - - - - 6
Eercise.5 06. 8 4( (( ( ( ( (( ( ( ( (,,-, A k j i k j i A curl - - - - - - - - - 6 (,,. poin a deermine, ( ( If A curl k z j z i z A - -
Answer. 6 5 (,,, A. ( ( ( k j i A curl k z j z z i z z A curl - - - - - - 6 Remark funcion. isalsoa vecor and funcion isa vecor A curl A
.6 Polar Coordinaes Polar coordinae is used in calculus o calculae an area and volume of small elemens in eas wa. Les look a siuaions where des Cares Coordinae can be rewrien in he form of Polar coordinae. 64
.6. Polar Coordinae for Plane (r, θ d ds rcos rsin ds rdrd 65
.6. Polar Coordinae for Clinder (,, z z ds dv cos sin z z z ds d dz dv d d dz 66
.6. Polar Coordinae for Sphere (r,, z rsin cos rsin sin z rcos ds r sin d d dv r sin drd d r 67
Eample. (Volume Inegral Calculae FdVwhereF i z j k andv and V isaspace bounded b z 0, 9. z 4 - z 4 68
Soluion Since i is abou a clinder, i is easier if we use clindrical polar coordinaes, where cos, where 0 sin, z z, dv d,0,0z4. d dz 69
.7 Line Inegral Ordinar inegral f ( d, we inegrae along he -ais. Bu for line inegral, he inegraion is along a curve. f (s ds f (,, z ds A r O rdr B 70
.7. Vecor Field, FInegral Le a vecor field and r di The scalar produc F. dr ( F i F d Fd Fd Fdz. F F i F j F k z dj dzk. F. dr is wrien as j F k.( di dj dzk z z 7
Ifa vecor field isalong hecurve, C hen he lineinegral offalonghe curve C fromapoin A oanoher F. dr F d F d c F c poin Bisgiven b c c F dz. z 7
7 Eample.. if, 4, curve he along (4,, ob (0,0,0 froma. Calculae zk zj i F z dr F c -
74 Soluion. 4 4 And. 4 4 ( ( (4( ( (4 Given 5 4 4 dk dj di dzk dj di dr k j i k j i zk zj i F - - -
75. 6 (8 6 8 ( 4 ( (4 (4 (4 ( 4 (4 4 4 (. Then 7 5 4 7 5 4 5 4 4 5 4 4 d d d d d d d dk dj di k j i F dr - - - -.,, 4, 4 (4,,, and,a B 0. 0, 0, 0, 4 (0,0,0, A A
76. 0 6 8 5 8 8 5 8 6 (8. 0 8 6 5 0 7 5 4 - - - d F dr B A
77 Eercise.6. 68 6 7..,, curve on he (,, (0,0,0oB froma. calculae, If - F dr z dr F zk zj i F B A c Answer
78 * Double Inegral * 4uni., ( inegrals. order boh in, ( Find. and 0, line asraigh b region bounded in 4, Given ( - R R da f da f R f Answer.6 Eample
.8 Volume Inegral.8. Scalar Field, F Inegral If V is a closed region and F is a scalar field in region V, volume inegral F of V is FdV Fdddz V V 79
Eample.4 Scalar funcion F defeaed in one cubic ha has been buil b planes 0,, 0,, z 0 and z. Evaluae volume inegral F of he cubic. z O 80
Soluion FdV V z 0 0. z 0 0 z 0 0 0 dz [ z] z 0 z 0 ddz 0 0 dddz 0 [ ] dz ddz 6 8
.9 Surface Inegral.9. Vecor Field, Inegral If vecor field inegral F F F defeaed on surface S, surface of S is defined as where S S n S F. ds S F. nds 8
Eample.5 Vecor field F i j k defeaed on surface S: z 9 and bounded b 0, 0, z 0 in he firs ocan. Evaluae F. ds. S 8
Soluion Given S: z 9is bounded b 0, 0, z 0in he s ocan. This refer o sphere wih cener a (0,0,0 and radius, r, in he s ocan. z O 84
So, grad S is S S i S j S k z i j zk, and S ( ( ( z z 9 6. 85
Therefore, S n S S F. ds F. nds S i j zk 6 ( i j zk. ( i j k ( S i j zk ds ( z ds. S 86
Using polar coordinae of sphere, rsin cos sin cos rsin sin sin sin z rcos cos ds r sin d d 9sin d d where 0,. 87
F ds S. [(sin cos (sin sin 0 0 9 [sin sin cos 0 0 (sin sin cos ][9sin ] dd sin sin sin cos ] dd 9 4 88
89 Eercise.7 ocan. 0in he 0and 0, 4, b bounded ofheregion isasurface and where on, Evaluae s z z S k z j i F S F ds S :8 6 Answer
.0 Green s Theorem If c is a closed curve in couner-clockwise on plane-, and given wo funcions P(, and Q(,, - dd ( Pd S Q P c Qd where S is he area of c. 90
Eample.6 Prove Green's Theorem for [( d ( d ] c which has been evaluaed b boundar ha defined as 0, 0and 4 in he firs quarer. Soluion C C O C 9
c Given [( d ( d ] where P and Q.We defined curve c as c, c and c. i For c : 0, d 0and 0 c c d 0 8. 0 ( Pd Qd ( d ( d 9
ii For c : 4,in he firs quarer from (,0 o (0,. This curve acuall a par of a circle. Therefore, i's more easier if we inegrae b using polar coordinae of plane, cos, sin, 0 d - sin d, d cos d 9
94 4. 4 8 4sin sin 8cos cos 8sin cos 8sin ( cos 8sin 4cos 8sin ( ] cos ( (sin ((cos sin ( (sin [((cos ( ( ( 0 0 0 0 - - - - - d d d d d d Qd Pd c c
iii For c : 0, d 0, 0 Pd Qd d d c ( ( c ( 0 d 0-4. ( Pd Qd 8 ( 4 4 6. c - - - 95
b Now, we evaluae Q P S - dd where Q and P. Again, because his is a par of he circle, we shall inegrae b using polar coordinae of plane, rcos, rsin wh ere 0r, 0 and dd ds rdrd. 96
S Q - P dd S ( - dd 0 r 0 ( -r sin sin 0 r - r 6 sin 0 - d 6 cos 0-6. rdrd 0 d 97
Therefore, ( Pd Qd Q P dd C S - 6 -. LHS RHS Green's Theorem has been proved. 98
. Divergence Theorem (Gauss Theorem If S is a closed surface including region V in vecor field V F divfdv S F. ds. f f f divf z 99
Eample.7 Prove Gauss' Theorem for vecor field, F i j z k in he region bounded b planes z 0, z 4, 0, 0and 4 in he firs ocan. 00
Soluion 4 z S S 4 S O S S 5 0
For his problem, he region of inegraion is bounded b 5 planes : S : z 0 S : z 4 S : 0 S4 : 0 S5 : 4 To prove Gauss' Theorem, we evaluae boh and F. ds, he answer should be he same. S V divfdv 0
We evaluae divfdv. Given F i j z k. V So, divf ( ( ( z z z. Also, divfdv ( z dv. V V The region is a par of he clinder. So, we inegrae b using polar coordinae of clinder, cos ; sin ; z z dv dd dz where 0,0,0z4. 0
Therefore, V ( z dv ( 4 0 0 z 0 4 [ z z ] 0 0 0 0 0 0 0 40 0 0. divfdv 0. V [0 ] (0 dd 0 (40 d d z dzd d dd 04
Now, we evaluae F. ds F. nds. S S i S : z 0, n- k, ds rdrd F i j 0k F. n ( i j.( - k 0 F. nds 0. S 05
ii S : z 4, n k, ds rdrd F i j (4 k i j 6 k. 6 F. n ( i j 6 k.( k 6. Therefore for S, 0r, 0 F nds S 0 r 0 6. rdrd 06
iii S : 0, n- j, ds ddz F i j z k F n i j zk j. (.( - -. Therefore for S, 0, 0z4 4 F. nds (- dzd S 0 z 0-6. 07
iv S : 0, n- i, ds ddz 4 F. nds 0. S 4 F 0i j zk j zk F. n ( j zk.( i 0. - 08
v 5: 4, S ds d dz S i j and S 4 5 5 S i j 5 n S 5 4 ( i j. B using polar coordinae of clinder : cos, sin, z z where for S5 :, 0, 0z 4, ds d dz 09
F. n ( i j z k. i j ( cos ( sin cos sin ; kerana. (cos sin. S 5 F. nds 4 0 z 0 6 4. ((cos sin ( d dz 0
Finall, F. ds F. ds F. ds F. ds F. ds F. ds S S S S S S 4 5 0 6-6 0 6 4 0. F. ds 0. S LHS RHS Gauss' Theorem has been proved.
. Sokes Theorem If F is a vecor field on an open surface S and boundar of surface S is a closed curve c, herefore S curlf ds F dr c i j k curlf F z f f f z
Eample.8 Surface S is he combinaion of iapar of he clinder 9beween z 0 and z 4for 0. iiahalf of he circlewih radius a z 4,and iii plane 0 If F zi j zk,prove Sokes' Theorem for his case.
Soluion z S 4 S We can divide surface Sas S : 9for0 z4and 0 S : z 4, half of he circle wih radius S : 0 C C O S 4
We can also mark he pieces of curve C as C : Perimeer of a half circle wih radius. C : Sraigh line from (-,0,0 o (,0,0. Le sa, we choose o evaluae S curlf ds firs. Given F zi j zk 5
6 So, ( ( ( ( ( ( ( k j z k z j z z z i z z z z z k j i F curl - - - -
B inegraing each par of he surface, ( i Forsurface S : 9, S i j and S ( ( 6 7
Then, S i j n ( i S 6 j and curlf n ( -z j ( -z. k i j 8
B using polar coordinae of clinder ( because S : 9 is a par of he clinder, cos, sin, z z ds d dz where, 0 and 0z4. 9
Therefore, curl F n ( -z sin -z sin ( - z; because Also, dsd dz 0
curl F ds curlf nds S S 4 z 0 0 4 0 4 0 sin ( -z ddz ( -z -cos 0 ( -z( -(- dz -4 dz
(ii For surface surface is S : z 4 nk., normal vecor uni o he B using polar coordinae of plane, rsin, z 4 dan ds rdrd where 0r and 0.
S S 8 r 0 0 r r 0 0 curl F n ( z j k - k rsin curl F ds curl FndS ( rsin ( rdrd sin d dr
(iii For surface S : 0, normal vecor uni o he surface is ds d dz n -j The inegraion limis :. - and 0z4 So, curlf n (( -z j k ( -j z- 4
Then, curlf. ds curlf. nds S S 4 ( z- dzd - z 0 4. curlf. ds curlf. ds curlf. ds curlf. ds S S S S - 4 8 4 8. 5
Now, we evaluae F. dr for each pieces of he curve C. C i C is a half of he circle. Therefore, inegraion for C will be more easier if we use polar coordinae for plane wih radius r, ha is cos, sin dan z 0 where 0. 6
F zi j zk and (cos (sin j 9sin cos j dr di d j dzk - sin d i cos d j. 7
From here, F. dr 7sin cos d. F. dr 7sin cos C 0-9cos 8. 0 d 8
ii Curve C is a sraigh line defined as, 0 and z 0, where -. Therefore, F zi j zk 0 F. dr 0 C 9
F. dr F. dr F. dr C C C 8 0 8. We alread show ha curlf. ds F. dr S C Sokes' Theorem has been proved. 0