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Coordinate systems (and transformations) and vector calculus

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Coordinate systems (and transformations) and vector calculus

1. 1. Coordinate Systems andTransformations & Vector Calculus By: Hanish Garg 12105017 ECE Branch PEC University ofTechnology
2. 2. Coordinate Systems • Cartesian or Rectangular Coordinate System • Cylindrical Coordinate System • Spherical Coordinate System Choice of the system is based on the symmetry of the problem.
3. 3. Cartesian Or Rectangular Coordinates P (x, y, z) x y z P(x,y,z)    z y x A vector A in Cartesian coordinates can be written as ),,( zyx AAA or zzyyxx aAaAaA  where ax,ay and az are unit vectors along x, y and z-directions.
4. 4. Cylindrical Coordinates P (ρ, Φ, z) x= ρ cos Φ, y=ρ sin Φ, z=z z Φ z ρ x y P(ρ, Φ, z)    z   20 0 A vector A in Cylindrical coordinates can be written as ),,( zAAA  or zzaAaAaA   where aρ,aΦ and az are unit vectors along ρ, Φ and z-directions. zz x y yx   ,tan, 122 
5. 5. The relationships between (ax,ay, az) and (aρ,aΦ, az)are zz y x aa aaa aaa        cossin sincos zz yx yx aa aaa aaa        cossin sincos or zzyxyx aAaAAaAAA    )cossin()sincos( Then the relationships between (Ax,Ay, Az) and (Aρ, AΦ, Az)are
6. 6. zz yx yx AA AAA AAA        cossin sincos                                z y x z A A A A A A 100 0cossin 0sincos     In matrix form we can write
7. 7. Spherical Coordinates P (r, θ, Φ) x=r sin θ cos Φ, y=r sin θ sin Φ, Z=r cos θ   20 0 0    r A vector A in Spherical coordinates can be written as ),,(  AAAr or  aAaAaA rr  where ar, aθ, and aΦ are unit vectors along r, θ, and Φ-directions. θ Φ r z y x P(r, θ, Φ) x y z yx zyxr 1 22 1222 tan,tan,     
8. 8. The relationships between (ax,ay, az) and (ar,aθ,aΦ)are       aaa aaaa aaaa rz ry rx sincos cossincossinsin sincoscoscossin    yx zyx zyxr aaa aaaa aaaa      cossin sinsincoscoscos cossinsincossin    or Then the relationships between (Ax,Ay, Az) and (Ar, Aθ,and AΦ)are      aAA aAAA aAAAA yx zyx rzyx )cossin( )sinsincoscoscos( )cossinsincossin(   
9. 9.                                 z y xr A A A A A A 0cossin sinsincoscoscos cossinsincossin      In matrix form we can write      cossin sinsincoscoscos cossinsincossin yx zyx zyxr AAA AAAA AAAA   
10. 10. Cartesian Coordinates P(x, y, z) Spherical Coordinates P(r, θ, Φ) Cylindrical Coordinates P(ρ, Φ, z) x y z P(x,y,z) Φ z r x y z P(ρ, Φ, z) θ Φ r z y x P(r, θ, Φ)
12. 12. Cylindrical Coordinates ρρ ρ ρ ρρ ρ ρ ρρ ρ
14. 14. Spherical Coordinates
15. 15. Differential Length, Area and Volume Differential displacement   adrarddradl r sin Differential area   ardrdadrdraddrdS r  sinsin2 Differential Volume  ddrdrdV sin2  Spherical Coordinates
16. 16. Line, Surface and Volume Integrals Line Integral L dlA. Surface Integral Volume Integral  S dSA. dvp V v
17. 17. • Gradient of a scalar function is a vector quantity. • Divergence of a vector is a scalar quantity. • Curl of a vector is a vector quantity. • The Laplacian of a scalar A f Vector A. A A2  The Del Operator
18. 18. Del Operator Cartesian Coordinates zyx a z a y a x          Cylindrical Coordinates Spherical Coordinates za z aa            1   a r a r a r r          sin 11
19. 19. VECTOR CALCULUS GRADIENT OFA SCALAR DIVERGENCE OFAVECTOR DIVERGENCETHEOREM CURL OFAVECTOR STOKES’STHEOREM
20. 20. GRADIENT OF A SCALAR Suppose is the temperature at , and is the temperature at as shown.  zyxT ,,1  zyxP ,,1 2P dzzdyydxxT  ,,2
21. 21. The differential distances are 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)
22. 22. But, z y x ddz ddy ddx aL aL aL    So, previous equation can 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)
23. 23. The vector inside square brackets defines the change of temperature corresponding to a vector change in position . This vector is called Gradient of Scalar T. Ld dT For Cartesian coordinate: zyx z V y V x V V aaa          GRADIENT OF A SCALAR (Cont’d)
24. 24. For Circular cylindrical coordinate: z z VVV V aaa            1 For Spherical coordinate:   aaa          V r V rr V V r sin 11 GRADIENT OF A SCALAR (Cont’d)
25. 25. EXAMPLE Find gradient of these scalars: yxeV z cosh2sin   2cos2 zU   cossin10 2 rW  (a) (b) (c)
26. 26. SOLUTIONTO EXAMPLE (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             
27. 27. SOLUTIONTO EXAMPLE (Cont’d) (b) Use gradient for Circular cylindrical coordinate: z z zz z UUU U a aa aaa      2cos 2sin22cos2 1 2           
28. 28. (c) Use gradient for Spherical coordinate:       a aa aaa sinsin10 cos2sin10cossin10 sin 11 2            r r W r W rr W W SOLUTIONTO EXAMPLE (Cont’d)
29. 29. Illustration of the divergence of a vector field at point P: Positive Divergence Negative Divergence Zero Divergence DIVERGENCE OF AVECTOR
30. 30. 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
31. 31. What is ??  s dSA Vector field A at closed surface S DIVERGENCE OF AVECTOR (Cont’d)
32. 32. Where, dSdS bottomtoprightleftbackfronts            AA And, v is volume enclosed by surface S DIVERGENCE OF AVECTOR (Cont’d)
33. 33. For Cartesian coordinate: z A y A x A zyx          A For Circular cylindrical coordinate:   z AA A z               11 A DIVERGENCE OF AVECTOR (Cont’d)
34. 34. For Spherical coordinate:                  A r A r Ar rr r sin 1sin sin 11 2 2 A DIVERGENCE OF AVECTOR (Cont’d)
35. 35. Find divergence of these vectors: zx xzyzxP aa  2 zzzQ aaa   cossin 2    aaa coscossincos 1 2  r r W r (a) (b) (c) EXAMPLE
36. 36. 39 (a) Use divergence for Cartesian coordinate:       xxyz xz zy yzx x z P y P x P zyx                    2 02 P SOLUTIONTO EXAMPLE
37. 37. (b) Use divergence for Circular cylindrical coordinate:                   cossin2 cos 1 sin 1 11 22                    Q z z z z QQ Q z SOLUTIONTO EXAMPLE (Cont’d)
38. 38. (c) Use divergence for Spherical coordinate:                     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 SOLUTIONTO EXAMPLE (Cont’d)
39. 39. It states that the 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 DIVERGENCETHEOREM
40. 40. A vector field exists 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:  aD 3     S dsD   V DdV (a) (b) EXAMPLE
41. 41. (a) For two concentric cylinder, the left side: topbottomouterinner S d DDDDSD  Where,           10)( )( 2 0 5 0 1 4 2 0 5 0 1 3             z z inner dzd dzdD aa aa SOLUTIONTO EXAMPLE
42. 42.           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)(           z ztop z zbottom ddD ddD aa aa SOLUTIONTO EXAMPLE (Cont’d)
43. 43. Therefore   150 0016010   SD S d SOLUTIONTO EXAMPLE (Cont’d)
44. 44. (b) For the right side of Divergence Theorem, evaluate divergence of D   23 4 1       D So,        150 4 5 0 2 0 2 1 4 5 0 2 0 2 1 2                                    z r z dzdddVD SOLUTIONTO EXAMPLE (Cont’d)
45. 45. CURL OF AVECTOR 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.
46. 46. maxlim 0 a A AA n s s s dl Curl                Where, CURL OF AVECTOR (Cont’d) dldl dacdbcabs           AA
47. 47. 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. CURL OF AVECTOR (Cont’d)
48. 48. For Cartesian coordinate: 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                                 CURL OF AVECTOR (Cont’d)
49. 49. z z AAA z              aaa A 1   z zz AA z AA z AA a aaA                                             1 1 For Circular cylindrical coordinate: CURL OF AVECTOR (Cont’d)
50. 50. For Spherical coordinate:       ArrAA rr r r sin sin 1 2        aaa A               a aaA                                    r r r A r rA r r rAA r AA r )(1 sin 11sin sin 1 CURL OF AVECTOR (Cont’d)
51. 51. zx xzyzxP aa  2 zzzQ aaa   cossin 2    aaa coscossincos 1 2  r r W r (a) (b) (c) Find curl of these vectors: EXAMPLE
52. 52. (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                                   SOLUTIONTO EXAMPLE
53. 53. (b) Use curl for Circular cylindrical coordinate           z z z zz zz z z y Q x QQ z Q z QQ aa a aa aaaQ                cos3sin 1 cos3 1 00sin 11 3 2 2                                              SOLUTIONTO EXAMPLE (Cont’d)
54. 54. (c) Use curl for Spherical coordinate:                                  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 SOLUTIONTO EXAMPLE (Cont’d)
55. 55.     a aa a aa              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 SOLUTIONTO EXAMPLE (Cont’d)
56. 56. STOKE’STHEOREM The circulation 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
57. 57. STOKE’STHEOREM (Cont’d)
58. 58. By using Stoke’s Theorem, evaluate for   dlA   aaA sincos   EXAMPLE
59. 59. Stoke’s Theorem,    SL dSdl AA where, andzddd aS  Evaluate right side to get left side,   zaA   sin1 1  SOLUTIONTO EXAMPLE (Cont’d)
60. 60.     941.4 sin1 1 0 0 60 30 5 2       aA z S dddS    SOLUTIONTO EXAMPLE (Cont’d)
61. 61. Verify Stoke’s theorem for the vector field for given figure by evaluating:  aaB sincos   (a) over the semicircular contour.   LB d (b) over the surface of semicircular contour.    SB d EXAMPLE
62. 62. (a) To find  LB d   321 LLL dddd LBLBLBLB Where,        dd dzddd z sincos sincos   aaaaaLB SOLUTIONTO EXAMPLE
63. 63. So 20 2 1 sincos 2 0 2 0 0 0 0,0 2 01                                   LB zz L ddd   4cos20 sincos 0 0,2 0 0 2 22                                LB zz L ddd SOLUTIONTO EXAMPLE (Cont’d)
64. 64. 20 2 1 sincos 0 2 2 00,0 0 23                                r zz L ddd     LB Therefore the closed integral, 8242  LB d SOLUTIONTO EXAMPLE (Cont’d)
65. 65. (b) To find    SB d                     z z z zz a aaa a aa aaB                                                            1 1sin sinsin 1 00 cossin 1 0cossin0 1 sincos SOLUTIONTO EXAMPLE (Cont’d)
66. 66. Therefore   8 2 1 cos 1sin 1 1sin 0 2 0 2 0 2 0 0 2 0                                                             aaSB dd ddd zz SOLUTIONTO EXAMPLE (Cont’d)
67. 67. LAPLACIAN OF A 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
68. 68. For Cartesian coordinate: 2 2 2 2 2 2 2 z V y V x V V          For Circular cylindrical coordinate: 2 22 2 2 11 z VVV V                    LAPLACIAN OF A SCALAR (Cont’d)
69. 69. LAPLACIAN OF A SCALAR (Cont’d) For Spherical coordinate: 2 2 22 2 2 2 2 sin 1 sin sin 11                             V r V rr V r rr V
70. 70. EXAMPLE Find Laplacian of these scalars: yxeV z cosh2sin   2cos2 zU   cossin10 2 rW  (a) (b) (c) You should try this!!
71. 71. SOLUTIONTO EXAMPLE yxeV z cosh2sin22   02  U    2cos21 cos102  r W (a) (b) (c)
72. 72. ThankYou !!!