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Vector calculus
- 1. VECTOR CALCULUS 1.10 GRADIENTOF A SCALAR 1.11 DIVERGENCE OF A VECTOR 1.12 DIVERGENCE THEOREM 1.13 CURL OF A VECTOR 1.14 STOKES’S THEOREM 1.15 LAPLACIAN OF A SCALAR
- 2. 1.10 GRADIENT OFA SCALAR Suppose is the temperature at , and is the temperature at as shown. ( )zyxT ,,1 ( )zyxP ,,1 2P( )dzzdyydxxT +++ ,,2
- 3. The differential distancesare the components of the differential distance vector : dzdydx ,, zyx dzdydxd aaaL ++= Ld However, from differential calculus, the differential temperature: dz z T dy y T dx x T TTdT ∂ ∂ + ∂ ∂ + ∂ ∂ =−= 12 GRADIENT OF A SCALAR (Cont’d)
- 4. But, z y x ddz ddy ddx aL aL aL •= •= •= So, previous equationcan be rewritten as: Laaa LaLaLa d z T y T x T d z T d y T d x T dT zyx zyx • ∂ ∂ + ∂ ∂ + ∂ ∂ = • ∂ ∂ +• ∂ ∂ +• ∂ ∂ = GRADIENT OF A SCALAR (Cont’d)
- 5. The vector insidesquare brackets defines the change of temperature corresponding to a vector change in position . This vector is called Gradient of Scalar T. Ld dT GRADIENT OF A SCALAR (Cont’d) For Cartesian coordinate: zyx z V y V x V V aaa ∂ ∂ + ∂ ∂ + ∂ ∂ =∇
- 6. GRADIENT OF ASCALAR (Cont’d) For Circular cylindrical coordinate: z z VVV V aaa ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ φρ φρρ 1 For Spherical coordinate: φθ φθθ aaa ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ V r V rr V V r sin 11
- 7. EXAMPLE 10 Find gradient ofthese scalars: yxeV z cosh2sin− = φρ 2cos2 zU = φθ cossin10 2 rW = (a) (b) (c)
- 8. SOLUTION TO EXAMPLE10 (a) Use gradient for Cartesian coordinate: z z y z x z zyx yxe yxeyxe z V y V x V V a aa aaa cosh2sin sinh2sincosh2cos2 − −− − += ∂ ∂ + ∂ ∂ + ∂ ∂ =∇
- 9. SOLUTION TO EXAMPLE10 (Cont’d) (b) Use gradient for Circular cylindrical coordinate: z z zz z UUU U a aa aaa φρ φρφρ φρρ φρ φρ 2cos 2sin22cos2 1 2 + −= ∂ ∂ + ∂ ∂ + ∂ ∂ =∇
- 10. SOLUTION TO EXAMPLE10 (Cont’d) (c) Use gradient for Spherical coordinate: φ θ φθ φθ φθφθ φθθ a aa aaa sinsin10 cos2sin10cossin10 sin 11 2 − += ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ r r W r W rr W W
- 11. 1.11 DIVERGENCE OFA VECTOR Illustration of the divergence of a vector field at point P: Positive Divergence Negative Divergence Zero Divergence
- 12. DIVERGENCE OF AVECTOR (Cont’d) The divergence of A at a given point P is the outward flux per unit volume: v dS div s v ∆ • =•∇= ∫ →∆ A AA lim 0
- 13. DIVERGENCE OF AVECTOR (Cont’d) What is ??∫ • s dSA Vector field A at closed surface S
- 14. Where, dSdS bottomtoprightleftbackfronts • +++++=• ∫∫∫∫∫∫∫ AA And,v is volume enclosed by surface S DIVERGENCE OF A VECTOR (Cont’d)
- 15. For Cartesian coordinate: z A y A x Azyx ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ A For Circular cylindrical coordinate: ( ) z AA A z ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ φρ ρ ρρ φ ρ 11 A DIVERGENCE OF A VECTOR (Cont’d)
- 16. For Spherical coordinate: () ( ) φθθ θ θ φθ ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ A r A r Ar rr r sin 1sin sin 11 2 2 A DIVERGENCE OF A VECTOR (Cont’d)
- 17. EXAMPLE 11 Find divergence ofthese vectors: zx xzyzxP aa += 2 zzzQ aaa φρφρ φρ cossin 2 ++= φθ θφθθ aaa coscossincos 1 2 ++= r r W r (a) (b) (c)
- 18. 18 (a) Use divergencefor Cartesian coordinate: SOLUTION TO EXAMPLE 11 ( ) ( ) ( ) xxyz xz zy yzx x z P y P x P zyx += ∂ ∂ + ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ 2 02 P
- 19. (b) Use divergencefor Circular cylindrical coordinate: ( ) ( ) ( ) ( ) φφ φρ φρ φρ ρρ φρ ρ ρρ φ ρ cossin2 cos 1 sin 1 11 22 += ∂ ∂ + ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ Q z z z z QQ Q z SOLUTION TO EXAMPLE 11 (Cont’d)
- 20. SOLUTION TO EXAMPLE11 (Cont’d) (c) Use divergence for Spherical coordinate: ( ) ( ) ( ) ( ) ( ) φθ θ φθ φθ θθ θ φθθ θ θ φθ coscos2 cos sin 1 cossin sin 1 cos 1 sin 1sin sin 11 2 2 2 2 = ∂ ∂ + ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ + ∂ ∂ =•∇ W r r rrr W r W r Wr rr r
- 21. It states thatthe total outward flux of a vector field A at the closed surface S is the same as volume integral of divergence of A. ∫∫ •∇=• VV dVdS AA 1.12 DIVERGENCE THEOREM
- 22. EXAMPLE 12 A vector fieldexists in the region between two concentric cylindrical surfaces defined by ρ = 1 and ρ = 2, with both cylinders extending between z = 0 and z = 5. Verify the divergence theorem by evaluating: ρρ aD 3 = → ∫ • S dsD ∫ •∇ V DdV (a) (b)
- 23. SOLUTION TO EXAMPLE12 (a) For two concentric cylinder, the left side: topbottomouterinner S d DDDDSD +++=•∫ Where, πφρ φρρ π φ ρ ρρ π φ ρ ρρ 10)( )( 2 0 5 0 1 4 2 0 5 0 1 3 −=•−= −•= ∫ ∫ ∫ ∫ = = = = = = z z inner dzd dzdD aa aa
- 24. πφρ φρρ π φ ρ ρρ π φ ρ ρρ 160)( )( 2 0 5 0 2 4 2 0 5 0 2 3 =•= •= ∫ ∫ ∫ ∫ == = = = = z z outer dzd dzdD aa aa ∫ ∫ ∫ ∫ = = = = = = =•= =−•= 2 1 2 0 5 3 2 1 2 0 0 3 0)( 0)( ρ π φ ρ ρ π φ ρ ρφρρ ρφρρ z ztop z zbottom ddD ddD aa aa SOLUTION TO EXAMPLE 12 Cont’d)
- 25. Therefore π ππ 150 0016010 = +++−=•∫ SD S d SOLUTION TOEXAMPLE 12 Cont’d)
- 26. SOLUTION TO EXAMPLE12 Cont’d) (b) For the right side of Divergence Theorem, evaluate divergence of D ( ) 23 4 1 ρρρ ρρ = ∂ ∂ =•∇ D So, πρ φρρρ π φ π φ ρ 150 4 5 0 2 0 2 1 4 5 0 2 0 2 1 2 = = =•∇ = = = = = = ∫ ∫ ∫∫∫∫ z r z dzdddVD
- 27. 1.13 CURL OFA VECTOR The curl of vector A is an axial (rotational) vector whose magnitude is the maximum circulation of A per unit area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum.
- 28. maxlim 0 a A AA n s s s dl Curl ∆ • =×∇= ∫ →∆ Where, CURLOF A VECTOR (Cont’d) dldl dacdbcabs • +++=• ∫∫∫∫∫ AA
- 29. CURL OF AVECTOR (Cont’d) The curl of the vector field is concerned with rotation of the vector field. Rotation can be used to measure the uniformity of the field, the more non uniform the field, the larger value of curl.
- 30. For Cartesian coordinate: CURLOF A VECTOR (Cont’d) zyx zyx AAA zyx ∂ ∂ ∂ ∂ ∂ ∂ =×∇ aaa A z xy y xz x yz y A x A z A x A z A y A aaaA ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ =×∇
- 31. z z AAA z φρ φρ ρ φρ ρ ρ ∂ ∂ ∂ ∂ ∂ ∂ =×∇ aaa A 1 ( ) z zz AA z AA z AA a aaA ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ =×∇ φρ ρ ρ ρφρ ρφ φ ρ ρ φ 1 1 ForCircular cylindrical coordinate: CURL OF A VECTOR (Cont’d)
- 32. CURL OF AVECTOR (Cont’d) For Spherical coordinate: ( ) φθ φθ θ φθθ ArrAA rr r r sin sin 1 2 ∂ ∂ ∂ ∂ ∂ ∂ =×∇ aaa A ( ) ( ) ( ) φ θ θ φθφ θ φθφθ θ θ a aaA ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ =×∇ r r r A r rA r r rAA r AA r )(1 sin 11sin sin 1
- 33. EXAMPLE 13 zx xzyzxP aa+= 2 zzzQ aaa φρφρ φρ cossin 2 ++= φθ θφθθ aaa coscossincos 1 2 ++= r r W r (a) (b) (c) Find curl of these vectors:
- 34. SOLUTION TO EXAMPLE13 (a) Use curl for Cartesian coordinate: ( ) ( ) ( ) ( ) zy zyx z xy y xz x yz zxzyx zxzyx y P x P z P x P z P y P aa aaa aaaP 22 22 000 −−= −+−+−= ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ − ∂ ∂ − ∂ ∂ =×∇
- 35. (b) Use curlfor Circular cylindrical coordinate ( ) ( ) ( ) ( ) ( ) z z z zz zz z z y Q x QQ z Q z QQ aa a aa aaaQ φρρφ ρ φρρ ρ ρφ ρ ρ ρρφρ ρ φρ ρφ φ ρ ρ φ cos3sin 1 cos3 1 00sin 11 3 2 2 −++−= −+ −+ − − = ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ =×∇ SOLUTION TO EXAMPLE 13 (Cont’d)
- 36. (c) Use curlfor Spherical coordinate: ( ) ( ) ( ) ( ) ( ) ( ) φ θ φ θ θ φθφ θ θ φθ θ φ θ θφ φθ θ θθ θ θ φθφθ θ θ a aa a aaW ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ = ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ =×∇ 22 2 cos )cossin(1 cos cos sin 11cossinsincos sin 1 )(1 sin 11sin sin 1 r r r r r rr r r r W r rW r r rWW r WW r r r r r SOLUTION TO EXAMPLE 13 (Cont’d)
- 37. SOLUTION TO EXAMPLE13 (Cont’d) ( ) ( ) a aa a aa φ θ φ θ θφ θ φ θ θ θ φθ θφθθ θ sin 1 cos2 cos sin sin 2cos sin cossin2 1 cos0 1 sinsin2cos sin 1 3 2 ++ + += ++ −++= r rr r r r r r r r r
- 38. 1.14 STOKE’S THEOREM Thecirculation of a vector field A around a closed path L is equal to the surface integral of the curl of A over the open surface S bounded by L that A and curl of A are continuous on S. ( )∫∫ •×∇=• SL dSdl AA
- 39.
- 40. EXAMPLE 14 By using Stoke’sTheorem, evaluate for ∫ • dlA φρ φφρ aaA sincos += →
- 41.
- 42. SOLUTION TO EXAMPLE14 Stoke’s Theorem, ( )∫∫ •×∇=• SL dSdl AA where, andzddd aS ρφρ= Evaluate right side to get left side, ( ) zaA φρ ρ sin1 1 +=×∇
- 43. SOLUTION TO EXAMPLE14 (Cont’d) ( ) ( ) 941.4 sin1 1 0 0 60 30 5 2 = +=•×∇ ∫ ∫∫ = = aA z S dddS ρφφρρ ρφ ρ
- 44. EXAMPLE 15 Verify Stoke’s theoremfor the vector field for given figure by evaluating:φρ φφρ aaB sincos += → (a) over the semicircular contour. ∫ • LB d (b) over the surface of semicircular contour. ( )∫ •×∇ SB d
- 45. SOLUTION TO EXAMPLE15 (a) To find∫ • LB d ∫∫∫∫ •+•+•=• 321 LLL dddd LBLBLBLB Where, ( ) ( ) φφρρφρ φρρφφρ φρφρ dd dzddd z sincos sincos += ++•+=• aaaaaLB
- 46.
- 47. 20 2 1 sincos 0 2 2 00,0 0 23 =+ −= + =• = = = == = ∫∫∫ r zz L ddd ρ φφρρφρ π πφφρ LB SOLUTION TO EXAMPLE15 (Cont’d) Therefore the closed integral, 8242 =++=•∫ LB d
- 48. SOLUTION TO EXAMPLE15 (Cont’d) (b) To find ( )∫ •×∇ SB d ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) z z z zz a aaa a aa aaB += +++= ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ + ∂ ∂ − ∂ ∂ = +×∇=×∇ ρ φ φρφ ρ φρ φ φρ ρρ ρ φρφ φρ φφρ φρ φρ φρ 1 1sin sinsin 1 00 cossin 1 0cossin0 1 sincos
- 49. SOLUTION TO EXAMPLE15 (Cont’d) Therefore ( ) 8 2 1 cos 1sin 1 1sin 0 2 0 2 0 2 0 0 2 0 = +−= += • +=•×∇ = = = = = = ∫ ∫ ∫ ∫∫∫ π φ ρ π φ ρ π φ ρ ρρφ φρρφ φρρ ρ φ aaSB dd ddd zz
- 50. 1.15 LAPLACIAN OFA SCALAR The Laplacian of a scalar field, V written as: V2 ∇ Where, Laplacian V is: ∂ ∂ + ∂ ∂ + ∂ ∂ • ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇•∇=∇ zyxzyx z V y V x V zyx VV aaaaaa 2
- 51. For Cartesian coordinate: 2 2 2 2 2 2 2 z V y V x V V ∂ ∂ + ∂ ∂ + ∂ ∂ =∇ ForCircular cylindrical coordinate: 2 22 2 2 11 z VVV V ∂ ∂ + ∂ ∂ + ∂ ∂ ∂ ∂ =∇ φρρ ρ ρρ LAPLACIAN OF A SCALAR (Cont’d)
- 52. LAPLACIAN OF ASCALAR (Cont’d) For Spherical coordinate: 2 2 22 2 2 2 2 sin 1 sin sin 11 φθ θ θ θθ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ =∇ V r V rr V r rr V
- 53. EXAMPLE 16 Find Laplacian ofthese scalars: yxeV z cosh2sin− = φρ 2cos2 zU = φθ cossin10 2 rW = (a) (b) (c) You should try this!!
- 54. SOLUTION TO EXAMPLE16 yxeV z cosh2sin22 − −=∇ 02 =∇ U ( )θ φ 2cos21 cos102 +=∇ r W (a) (b) (c)
- 55. Yea, unto Godbelong all things in the heavens and on earth, and enough is God to carry through all affairs Quran:4:132 END