EELE 3331 Electromagnetic I Chapter 3. Vector Calculus. Islamic University of Gaza Electrical Engineering Department Dr.

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1 EELE 3331 Electromagnetic I Chapter 3 Vector Calculus Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik

2 Differential Length, Area, and Volume This chapter deals with integration and differentiation of vectors Applications: Next Chapter. Differential Length, Area, and Volume A. Cartesian Coordinate Systems: 1. Differential displacement: Differential displacement from point S(x,y,z) to point B(x+dx,y+dy,z+dz) is: dl=dx a x +dy a y +dz a z 2

3 Differential Length, Area, and Volume A. Cartesian Coordinate Systems: 2. Differential normal surface area (a) ds=dy dz a x (b) ds=dx dz a y (c) ds=dx dy a z 3

4 Differential Length, Area, and Volume A. Cartesian Coordinate Systems: 3. Differential volume dv=dx dy dz Notes: dl, ds Vectors dv Scalar 4

5 Differential Length, Area, and Volume B. Cylindrical Coordinate Systems: 1. Differential displacement: dl=dρ a ρ + ρ dφ a φ +dz a z 5

6 Differential Length, Area, and Volume B. Cylindrical Coordinate Systems: 2. Differential normal surface area Note: ds can be derived from dl (a) ds=ρ dφ dz a ρ (b) ds= dρ dz a φ (c) ds= ρ dρ dφ a z dv=ρ dρ dφ dz 3. Differential volume 6

7 Differential Length, Area, and Volume C. Spherical Coordinate Systems: 1. Differential displacement: dl = dr a r + r dθ a θ + r sinθ dφ a φ 7

8 Differential Length, Area, and Volume C. Spherical Coordinate Systems: 2. Differential normal surface area Note: ds can be derived from dl (a) ds=r 2 sinθ dθ dφ a r (b) ds=r sinθ dr dφ a θ (c) ds= r dr dθ a φ dv=r 2 sinθ dr dθ dφ 3. Differential volume 8

9 Example 3.1 Consider the object shown. Calculate : (a) The length BC (b) The length CD (c) The surface area ABCD (d) The surface area ABO (e) The surface area AOFD (f) The volume ABCDFO 9

10 Object has Cylindrical Symmetry Cartesian to Cylindrical: 0 A(5,0,0) A(5,0,0) B(0,5,0) B(5, /2,0) C(0,5,10) C(5, /2,10) 0 D(5,0,10) D(5,0,10) (a) along BC, 10 0 dl BC dl dz Example solution dz 10 Cylindrical Coordinates (b) Along CD, dl d CD d /2 /

11 Example solution (c) for ABCD, ds d dz, =5 Area ABCD = ds 10 /2 z00 d dz (5)( / 2)(10) =25 (d) for ABO, ds= d d, z=0 area ABO= 5 /2 00 d d 2 5 ( / 2)

12 (e) for AOFD, ds d dz, = area AOFD= d dz 50 0 z0 Example solution (f) For volume ABCDFO, dv d dz d /2 10 v dv d dz d 0 0 z0 12

13 Line, Surface, and Volume Integrals (Line=Curve=Contour) Integral: The Line integral A dl is integral of the tangential component of vector A along L. L b A dl A cos dl L a Line integral of A around L. If the path of integration is closed, such as abca, Circulation of A along L. L Adl 13

14 Line, Surface, and Volume Integrals Surface Integral: Given vector A continuous in a region containing the surface S The surface integral or the flux of A through S is: A ds S = A cos ds, ds= ds a n S Flux across ds is: d= A cos ds A ds Total Flux d A ds S 14

15 Line, Surface, and Volume Integrals Surface Integral: For a closed surface (defining a volume) : S A ds The net outward flux of A from S Notes: A closed path defines an open surface. A closed surface defines a volume. 15

16 Line, Surface, and Volume Integrals Volume Integral: We define: v v dv as the volume integral of the scalar ρ v over the volume v. 16

17 Example 3.2 Given that F=x 2 a x xz a y y 2 a z. Calculate the circulation of F around the (closed) path shown in the Figure. 17

18 The circulation of F around L is: F dl= F dl L For segment 1, y=0, z=0 F= x a xz a y a x a x y z x dl dx a (+ve direction) F x 0 2 dl x dx Example solution Segment 2, x 0, z 0, dl dya, F dl 0 F dl 0 y 18 2

19 2 2 Segment 3: y 1, F= x ax xz a y y az, dl dx ax dz az F 2 dl x dx dz 1 3 x z Segment 4: x 1, F= a z a y a dl dy a dz a y z 2 x y z 2 F l, but on 4,, d z dy y dz z y dz dy Example solution F dl ( y y ) dy, F dl= 19 L

20 The del operator, written as, is the vector differential operator. In Cartesian coordinates: Useful in defining: Del Operator a a a x y z x y z (1) The gradient of a scalar V, written as V (2) The divergence of a vector A, written as A (3) The curl of a vector A, written as A (4) The Laplacian of a scalar V, 2 V 20

21 Del Operator Del operator in Cylindrical Coordinates: (,,z) x y x y, =tan x cos, y sin Cartesian a a a x y z x y z use chain rule differentiation: sin cos x cos sin y 21

22 but a cos a sin a, x Del Operator a sin a cos a, a a y z z in Cylindrical: sin = cos cos a sin a Cartesian a a a x y z x y z cos + sin sin a cos a a z 1 In Cylindrical = a a az z 1 1 In Spherical, = ar a a r r r sin z 22

23 Del Operator Cartesian ax ay az x y z Cylindrical 1 = a a az z Spherical, 1 1 = ar a a r r r sin 23

24 Gradient of a Scalar The gradient of a scalar V is a vector that represents both the magnitude and the direction of the maximum space rate of increase of V V V V grad V = V = a a a x y z x y z V maximum rate of change in V V points in the direction of the max. rate of change in V Example: Room with temperature given by a scalar field T(x,y,z). At each point in the room, the gradient of T at that point will show the direction the temperature rises more quickly. The magnitude of the gradient will determine how fast the temperature rises in that direction. 24

25 Gradient of a Scalar The gradient of the function f(x,y) = (cos 2 x + cos 2 y) 2 depicted as a projected vector field on the bottom plane 25

26 V V V In Cartesian: V = a a a x y z For Cylindrical coordinates, 1 Recall = a a az z V 1 V V V = a a a z For Spherical coordinates, Recall = a r Gradient of a Scalar r a x y z 1 1 a r rsin V 1 V 1 V V = ar a a r r r sin 26 z

27 Notes: (a) (b) (c) (d) Gradient of a Scalar V U V U VU VU UV V UV VU 2 U U n n1 V nv V where U and V are scalars and n is an integer The gradient at any point is perpendicular to the constant V surface that passes through that point. 27

28 f(x,y) = (cos 2 x + cos 2 y) 2 f(x,y) =constant contour f ( x, y) x y 2 2 f ( x, y) C constant Contour f ( x, y, z) x y z f ( x, y, z) C level surface (ex: surface of sphere) 28

29 Example 3.3 Find the gradient of the following scalar fields: z 2 (a) V e sin 2 x cosh y (b) U z cos2 2 (c) W 10 r sin cos V V V (a) V = a a a x y z x y z -z -z -z 2e cos 2 x cosh y a x e sin 2 x sinh y a y e sin 2 x cosh y a z U 1 U U (b) U= a a a z 2 2 z cos 2 a 2 z sin 2 a cos 2 a z W 1W 1 W (c) W = ar a a r r r sin sin cos cos a 10sin sin a 2 10sin cos a r 20 z 29

30 2 2 2x 0.5x 10 x 2 point (2,2,1) f x y z x y z (,, ) 2 10 Example 3.5 Find the angle at which line x y 2 z intersects the ellipsoid x y 2z 10 To find the point of intersection: x x 2( x / 2) 10 surface of ellipsoid f ( x, y, z) constant f 2 x a 2 y a 4 z a x y z At (2,2,1), f 4 a x 4 a y 4 a z 30

31 Example continued Hence, a unit vector normal to the ellipsoid surface at P(2,2,1) is: a n f f cos a a a a a 0 3 r r 4 a a a 0 48 (1/ 3) 1,1,1 2, 2, x y z n n x y z 31