Fundamental Electromagnetics (Chapter 4: Vector Calculus)

Transcription

1 Fundamental Electromagnetics (Chapter 4: Vector Calculus) Prof. Kwang-Chun Ho Tel: Fax:

2 Key Point Study differential elements in length, area, and volume useful in vector calculus. Consider concept of line, surface, and volume integrals including a vector. What is gradient ( :del) operator? Gradient of a scalar in Cartesian, cylindrical, and spherical systems. Mathematical and physical concepts of divergence of a vector and divergence theorem. Mathematical and physical concepts of curl of a vector and Stokes s theorem. Laplacian of a scalar and classification of vector fields. Dept. Electronics and Information Engineering

3 Differential Length, Area, and Volume Cartesian coordinates Differential displacement: if moving from P to Q, it is given by if moving from S to B, it is dl dxa dya dza x y z dl dya y a x a y a z a x a z a y dl ( Differential normal area ) ( Differential elements in Cartesian system ) Dept. Electronics and Information Engineering 3

4 Differential Length, Area, and Volume Differential normal area: The differential surface element ds may be defined as ds dsa n where ds is the area of surface element, and a n is a unit vector normal to the surface ds Convention for the positive direction of ds : Be directed away from the volume if it is a part of the surface describing a volume As examples, if considering surface ABCD, ds dydza x if considering surface PQRS, ds dydza x dv dxdydz Differential volume(scalar): Cylindrical coordinates Differential displacement: if moving from R to A, it is given by dl d a da dza z Dept. Electronics and Information Engineering 4

5 Differential Length, Area, and Volume Differential normal area: if considering surface ABPQ, ds ddza. Differential volume: dv d ddz a a z dl a ds ddza ds d dza ds dd a z a z a a ( Differential normal area ) ( Differential elements in cylindrical system ) Dept. Electronics and Information Engineering 5

6 Differential Length, Area, and Volume Example 4.1: h Determine the volume of object shown in figure. Since the differential volume in cylindrical system is dv d ddz, the object volume becomes a V a v= dv= rdr df dz pah = 4 p òòò ò ò ò h MatLab Script: % Perform 3D symbolic integration syms a h intr = int('r','r',0,a); intp = int(intr,'p',0,pi/); intz = int(intp,'z',0,h); Dept. Electronics and Information Engineering 6

7 Differential Length, Area, and Volume Spherical coordinates Differential displacement: if moving from A to B, it is given by dl dra rd a r sinda r ds rdrd r sin r dr dl A rd ds r sin drd B ds r sin dd r rsin d ( Differential elements in spherical system ) Dept. Electronics and Information Engineering 7

8 Differential Length, Area, and Volume Differential normal area: if considering surface ds r, ds r sin ddar dv r sin drdd Differential volume: a r a a ds r sin ddar ds r sin drda ds rdrd a ( Differential normal area ) Dept. Electronics and Information Engineering 8

9 Differential Length, Area, and Volume Example 4.: The figure may be described as 3r 5,60 90,45 60, where surface r 3 is the same as AEHD, surface 60 is AEFB, and surface 45 is ABCD Calculate (a) The distance DH (b) The distance FG (c) The surface area AEHD (d) The surface area ABCD (e) The volume of the object rd dr rsind Dept. Electronics and Information Engineering 9

10 Differential Length, Area, and Volume Solution: (a) (b) 60 p DH = ò r sin qdf = = f= p (c) AEHD = ò ò r sin qdfdq = = q= 60 f= 45 r= 3, q= 90 5p FG = ò rdq = = q= 60 r= 5 r= 3 MatLab Script: syms th inth = int('r^*sin(th)','th',pi/3,pi/); intp = int(inth,'p',pi/4,pi/3); intr = int(intp,'r',3,5); p (d) ABCD = ò ò rdqdr = = 3 r= 3 q= p (e) Volume = ò ò ò r sin qdqdfdr= = r= 3 f= 45 q= 60 Dept. Electronics and Information Engineering 10

11 Differential Length, Area, and Volume Example 4.3: Determine the volume and the surface of a solid sphere of radius R Solution: The volume is R p p 4 3 òòò òòò ò ò ò V = dv= r sin qdrdqdf= r dr sin qdq df= pr 3 The total surface area can be found by p p òò òò ò ò S = ds = r sin qdqdf = R sin qdq df= 4pR r= R 0 0 Dept. Electronics and Information Engineering 11

12 Vector Line Integrals Line integral involving a vector : Work: If you have taken a physics class, you have probably encountered the notion of work in mechanics M F F L If a constant force of F (in the direction of motion) is applied to move an object a distance L in a straight line, then the work exerted is Work [Joule] Force [Newton] Distance [Meter] =FL Dept. Electronics and Information Engineering 1

13 Vector Line Integrals Now, suppose that the there is an angle theta between directions, in which the constant force is applied and the direction of motion is constant M F L In this case, the work is given by Work = F L= FLcosq The above formulas for work assume that the object moves in a straight line and that the force and the angle between the direction of the force and the direction of motion are constant Dept. Electronics and Information Engineering 13

14 Vector Line Integrals Suppose that the object is moved along a curve C in the xy-plane, and the force is given by the vector field What is the work required to move the object? Whole work along this path can be treated as the addition of a sequence of Work due to infinitesimal displacement dl along C : dw = F cosqdl = F dl Thus, we have W = dw = F dl ò C If the path of integration is a closed path, the line integral becomes a closed contour integral: W = ò C ò C F dl F Dept. Electronics and Information Engineering 14

15 Vector Line Integrals F dl F cos Dept. Electronics and Information Engineering 15

16 Vector Line Integrals Example 4.4: L Given that F xaxxzay y az, calculate the circulation of F around the closed path Solution: The circulation around path is given by æ ö F dl = ò ç ò ò ò ò Fdl çè ø For segment 1, since y=0=z, F dl = x axdxax = x dx For segment, since x=0=z, Fdl =- y a dya = 0 z y Dept. Electronics and Information Engineering 16

17 Vector Line Integrals For segment 3, since x=z (dx=dz) and y=1, F dl= xa-xza - a dxa + dza = xdx-dz ( x y z) ( x z) For segment 4, since y=z, and x=1, F dl = a -za - y a dya + dza =-zdy-ydz ( x y z) ( y z) Thus, putting all these together, we have x dx + x - dx + -y - y dy =- 6 ò ò ( 1) ò ( ) Note: is always taken as along positive, and dl the direction is taken care of by the limits of integration. Dept. Electronics and Information Engineering 17

18 Vector Line Integrals Example 4.5: Calculate the circulation of around the edge L of the wedge defined by Solution: For line 1: 0,0 60, z 0 A cosa zsinaz For line : For line 3: Summing them up gives Line 3 Line Line 1 Dept. Electronics and Information Engineering 18

19 Vector Surface Integrals Surface integral: Suppose that the velocity of a fluid in 3D-space is described by the vector field F Let S be a surface in 3D-space Then, the volume of fluid crossing S per unit time is called as the total flux across S F Fa n a n a F n ds Dept. Electronics and Information Engineering 19

20 Vector Surface Integrals What is the formula for the flux? Consider a open surface S, dividing into vector elements of area ads n In one unit of time, a blob of fluid of length F will pass through the surface ds The blob may be in the shape of a parallelepiped The volume of this fluid is then Volume = F ads= FcosqdS Consequently, the total flux through the surface S is Y= Fa ds = FdS ò S n n This is an example for surface integral of a vector field! ò S Dept. Electronics and Information Engineering 0

21 Vector Surface Integrals For a closed surface (defining a volume), the surface integral (net outward flux of from S ) becomes Y=ò What is open or closed surface? A closed path define an open surface whereas a closed surface defines a volume S F F ds Positive direction: the outward direction from the volume a n a n (Closed surface) a n Positive direction: Apply right-hand rule to the perimeter of the open surface a n a n (Open surface) Dept. Electronics and Information Engineering 1

22 Vector Surface Integrals Example 4.6: z Determine the flux of G 10e a a z over the closed surface of the cylinder 1, 0 z 1 Solution: The total flux is Y= ò G ds =Y t +Y b+ys For t, z 1 and ds ddaz 1 p - Y = G ds = 10e rdrdf t ò t ( p) ò ò r= 0f= r - = 10e = 10pe 0 Dept. Electronics and Information Engineering

23 Vector Surface Integrals For, z 0 and ds dd b az 1 p 1-0 r Y b = ò G ds =- ò ò10e rdrdf=- 10( p) =-10p b r= 0f= 0 For, 1 and ds dzda s Y s = ò G ds = ò ò 10e r dzdf= 10 1 p = 10p 1-e - s 1 p -z 1 -z e - z= 0f= 0 Thus, the total flux through the top, bottom, and sides of the cylinder is t b s e e 0 () ( ) Dept. Electronics and Information Engineering 3

24 DEL Operator Del operator, written, is the vector differential operator In Cartesian coordinates, it is a a a x y z x y z In cylindrical coordinates, it is 1 a a a z z In spherical coordinates, it is 1 1 ar a a r r rsin The proof will be discussed later! Dept. Electronics and Information Engineering 4 D Alembertian operator:

25 Gradient of a Scalar Physical Meaning Conduction heat transfer The direction of heat transfer will be opposite to the temperature gradient since the net energy transfer will be from high temperature to low one. This direction of maximum heat transfer will be perpendicular to the equal-temperature surfaces surrounding a source of heat. Source Dept. Electronics and Information Engineering 5

26 Gradient of a Scalar Rain on the mountain Water running down the mountain will follow the streamlines. The streamlines are vectors. The steeper the gradient, the larger the vectors. Contours far apart Contours close together steeper Gradient Contours (Scalar Fields) Vector Fields flatter The gradient of a scalar field is a vector field! Side View Dept. Electronics and Information Engineering 6

27 Gradient of a Scalar Definition: Physical definition: The gradient of a scalar field is a vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change. Mathematical definition: Can be obtained by evaluated the difference dv in the field between points P and on 1 P the contours V dl G (Depict the Gradient on contours) dl Dept. Electronics and Information Engineering 7

28 Gradient of a Scalar From differential calculus, the difference is dl V V V dv ( x, y, z) = dx + dy + dz x y z dn adn æ V V V ö a a a = + + dxa + dya + dza x y z ç è ø For convenience, let V V V G ax ay az x y z Then, dv = G dl dv = G( dl)cosq G cos dl ( ) x y z x y z n V1 V 1 V Notice that when dl dv dl is a maximumwhen 0 is in the direction of G a G n dv dv V V V G dl max dn x y z Dept. Electronics and Information Engineering 8

29 Gradient of a Scalar By definition, a n is the direction of gradient of a scalar field, which describes a unit vector normal to constant V scalar surface G is the magnitude of gradient Thus, dv grad V an ang V dn = Gradient V is a vector whose magnitude is equal to the maximum rate of change of the physical quantity per unit distance and whose direction is along the direction of maximum change. A good example is a mountain: The contour map of the terrain is the height coordinates of the given point V defined by the The gradient of at a point is a vector which presents the direction of the greatest rate of change The magnitude indicates how steep the slope is V Dept. Electronics and Information Engineering 9

30 Gradient of a Scalar dn 1 V1 V 1 V dn dn 1 (Slope 1 > Slope ) Thus, Gradient 1 > Gradient ( Gradients illustrated by vectors on contours ) dn V1 V 1 V Dept. Electronics and Information Engineering 30

31 Gradient of a Scalar Example 4.7: Calculate the angle between unit vectors normal to the surfaces xyz 3 and xln( z) y 4 at the point of intersection (-1,,1). Solution: Use a property of gradient: V at any point is perpendicular to the constant V surface that passes through that point Let f xyz3 and g xln( z) y 4 Since vectors normal to the surfaces are f xyax x ay az, x g ln( z) ax yay az z MatLab Script: syms x y z jacobian(x^*y+z-3,[x y z]) jacobian(x*log(z)-y^+4,[x y z]) Dept. Electronics and Information Engineering 31

32 Gradient of a Scalar the unit vectors at (-1,,1) become a nf f ( 4,1,1) g (0, 4, 1), ang f 18 g 17 Then, the angle between two unit vectors is given by -5 cos q= anf ang =, q= Dept. Electronics and Information Engineering 3

33 Gradient of a Scalar Example 4.8: Given xy yzxz : Find gradient at point (1,,3) Find the directional derivative of at the same point in the direction toward point (3,4,4) Solution: Since at point (1,,3) Using the unit vector a l of direction toward point (3,4,4) at point (1,,3), it becomes where the distance vector is The rate of change of in the direction of a l Dept. Electronics and Information Engineering 33

34 Visual EMT using MatLab Depict the gradient of the following scalar field x y V x, y xe The gradient is V V x y V a a e 1x a xya x y x y x y Run scalargrad.m Dept. Electronics and Information Engineering 34

35 Visual EMT using MatLab clear; % Clear variables v = -:0.:; [x,y] = meshgrid(v); % Create a grid of points z = x.* exp(-x.^ - y.^); [px,py] = gradient(z,.,.); % Compute gradient contour(x,y,z,10); % Draw contours hold on; quiver(x,y,px,py); % Plot vectors hold off; xlabel('x'); ylabel('y'); title('gradient of the scalar field: \nabla V'); [FX,FY] = GRADIENT(F,HX,HY), when F is -D, uses the spacing specified by HX and HY. HX and HY can either be scalars to specify the spacing between coordinates or vectors to specify the coordinates of the points. QUIVER(X,Y,U,V) plots velocity vectors as arrows with components (u,v) at the points (x,y). Dept. Electronics and Information Engineering 35

36 Divergence of a Vector What does the divergence mean physically? Measure how much the vector field diverges (or emanates) from a point P in a given region of space Measure the change rate of density in a given region of space F P div FP ( ) 0 ( P is a source) div FP ( ) 0 ( P is a sink) div FP ( ) 0 Dept. Electronics and Information Engineering 36

37 Divergence of a Vector As example, look at water flowing When the flow in equals the flow out as in a water pipe, then the divergence is zero There is no the change of density in a region of space lying entirely within the water! Next, consider the velocity of the air in a tire, which has just been punctured by a nail The density within a region of space changes because the air is expanding (indicating source) as the pressure drops Thus, the divergence is not zero (greater than zero)! By measuring the net flux of air passing through a surface surrounding the region of space, we can say how the density of the interior has changed Similarly, a negative divergence indicates a sink Dept. Electronics and Information Engineering 37

38 Divergence of a Vector Dept. Electronics and Information Engineering 38

39 Divergence of a Vector Definition: Physical definition: F Divergence of at a given point P is the outward of flux from a small closed surface per unit volume as the volume shrinks to zero Mathematical definition: div F v ò lim D v 0 F ds F F F = F = + + D v x y z S x y z where is the volume element enclosed by the closed surface S in which P is located Dept. Electronics and Information Engineering 39

40 Divergence of a Vector Transforming it into the other coordinates gives for cylindrical coordinates, and for spherical coordinates Example 4.9: 1 1 Ff Fz F = ( rfr ) + + r r r f z 1 1 é F ù f F = ( r Fr ) + ( Fq sin q) + r r rsin q ê q f ú ë û Suppose F xax yay zaz is the velocity field for an expanding air Determine the divergence of the velocity field Dept. Electronics and Information Engineering 40

41 Divergence of a Vector Solution: Since F = 3, the air is expanding at the rate of 3 cubic units per unit of volume Gauss Divergence theorem: Consider a source with volume V Total outward flux of a vector field F through the closed surface S is the same as the volume integral of the divergence of F F ds = F dv ò S ò v ( ) Gauss: German mathematician, lived Closed surface S Dept. Electronics and Information Engineering 41

42 Divergence of a Vector The contributions from the internal surfaces of adjacent elements cancel each other Because the flux of the adjacent elements are opposite directions Thus, the net contribution of volume integral of divergence is that of the external surface bounding the volume Example 4.10: z If Gr () 10e a a z, determine the flux of G out of the entire surface of the cylinder 1, 0 z 1 Solution: If is the flux of G through the given surface, then Dept. Electronics and Information Engineering 4

43 Divergence Theorem then Using the divergence theorem, it is Y= GdS = G dv= ò because G Y= ò G ds =Y +Y +Y ò v ( ) Gf G = ( rgr ) + + r r r f z 1 -z -z = ( 10e r ) + ( 10e ) = 0 r r z There is no the outflow of flux! z t b s Dept. Electronics and Information Engineering 43

44 Divergence Theorem Example 4.11: (a) Determine the flux of D cos a zsina over the closed surface of the cylinder 0 z 1, 4 (b) Verify the divergence theorem for this case Solution: (a) The flux is Since there is no z-component of D, and Dept. Electronics and Information Engineering 44

45 Divergence Theorem (b) Considering by the divergence theorem, we have Dept. Electronics and Information Engineering 45

46 Divergence Theorem Sketches of vector fields: (a): One very poor sketch (b) and (c): Two fair sketches (d): Usual form of flux-line sketch Arrows show the direction of the field, and spacing of the lines is inversely proportional to the strength of the field Dept. Electronics and Information Engineering 46

47 Visual EMT using MatLab Depict the divergence of the following vector field x P 1x ax xya y e The divergence is P x P y P= + = 4x( x + y -) e x y y ( x y ) - + Run vectordiv.m! Dept. Electronics and Information Engineering 47

48 Visual EMT using MatLab clear; x = -:.1:; y = -1:.1:1; [xx,yy] = meshgrid(x,y); px = (1-*xx.^).*exp(-xx.^-yy.^); py = -*xx.*yy.*exp(-xx.^-yy.^); div = divergence(x,y,px,py); % Compute divergence surf(x,y,div); % Plot 3D colored surface xlabel('x'); ylabel('y'); title('divergence of -D vector field: \nabla \cdot {\bf P}'); DIV = DIVERGENCE(X,Y,U,V) computes the divergence of a -D vector field U,V. The arrays X,Y define the coordinates for U,V. Dept. Electronics and Information Engineering 48

49 Curl of a Vector What does the curl mean physically? There are two kinds of sources: Flow source: causes a net outflow flux of a vector Vortex source: causes a net circulation of a vector An example is the phenomenon of water whirling down a sink drain! The velocity of water is a vector field The scalar circulation of a vector field around a closed path is defined as the line integral of the vector over the path a n Circulation of F around contour ò L Fdl S dl Dept. Electronics and Information Engineering 49

50 Curl of a Vector Measure of the scalar circulation of the vector : Use curl meter: Drop into vector field and imagine the torque generated on the meter ( curl meter ) Vortex source : Whirlpool, eddy, and flush toilet Spin fast in the middle Spin slower with translation It is a device to probe the field for studying the curl of the field. It responds to the circulation of the field. Dept. Electronics and Information Engineering 50

51 Curl of a Vector Constant velocity flow : Constant field has no curl, therefore the circulation is zero No curl! Variable velocity parallel flow : Curl! Physical definition: A vector whose magnitude is the maximum net circulation of per unit area as the area tends to zero and whose direction is the normal direction of the area when the area is oriented to make the net circulation maximum F Dept. Electronics and Information Engineering 51

52 Curl of a Vector Mathematical definition: curl æ ax ay az F dl ö ò F a ç lim = F = D S x y z L n DS0 ç çè ø max F F F x y z for Cartesian coordinate For cylindrical and spherical coordinates, it becomes F a a a 1 z F F F z z 1 F r sin r a ra rsina r F rf rsin F r Dept. Electronics and Information Engineering 5

53 Curl of a Vector Example 4.1: The water flows along a velocity V V sin in a 0 az a sluice box Determine the curl of the velocity vector Solution: ax ay az V x y z 0 0 Vz Vz V0 x ay cos a x a a (The curl varies with x position across sluice box) y z y a x z t 0 t t 0 x a x Dept. Electronics and Information Engineering 53

54 Stokes s Theorem Definition: The circulation of a vector field F around a closed path L is equal to the surface integral of F over the open surface S bounded by L Fdl = F ds ò L Proof: ò S ( ) Assume that the surface S is subdivided into a large number of cells If the j-th cell has surface area and is bounded by path C j Stokes: Irish mathematician, lived S j L C j Dept. Electronics and Information Engineering 54

55 Stokes s Theorem we have Adding the contributions of all the differential areas to the flux, it becomes lim å Example 4.13: æ ö ( F) D S = lim F dl å ò ç è j ø F ds = F dl j DSj0 j j DSj0 j C x G dl Given that G x ya ya, find ò (a) where L is shown in the figure L ò ( ) (b) where S is the area bounded by L S ò S ( ) D j = j ò C ( ) G ds F S Fdl y L ò j Dept. Electronics and Information Engineering 55

56 Stokes s Theorem Solution: (a) Consider æ ö G dl ò = ç ò + ò + ò Gdl L çè 1 3 ø (1) () For line (1), since y=x, dy=dx, and dl ds dl dxax dyay ò G dl = ò ( x - x) dx=- z (3) 4 L 0 For line (), since y= x+, dy= dx, anddl dxax dya 3 17 ò G dl = ò (- x + x - x+ ) dx= 1 L 1 For line (3), since y=0, and dl dxa x 0 G dl = x ydx = 0 ò L ò y= 0 y Dept. Electronics and Information Engineering 56

57 Stokes s Theorem Thus, putting all these together, we obtain ò G dl = = L (b) Since G x a, ds dxdy a, the integral becomes S ( ) ( ) x - x+ x z 1 x - x+ G ds = - - x dxdy = x dydx + x dydx ò òò òò ò ò z = ò x y dx + ò x y dx = + x (- x + ) dx = 4 ò 6 Dept. Electronics and Information Engineering 57

58 Stokes s Theorem Example 4.14: Use Stokes s theorem to confirm your result in Example 4.5 Solution: Since A cosa zsinaz Line 3 Line Line 1 using Stokes s theorem we have Dept. Electronics and Information Engineering 58

59 Stokes s Theorem A The line integral of around L is equal to the surface integral of A Thus, Stokes s theorem is verified! Some useful properties: Divergence of the curl of any vector field is zero: F = 0 Curl of the gradient of any scalar field vanishes: V 0 Dept. Electronics and Information Engineering 59

60 Visual EMT using MatLab Depict the curl of the following vector field The curl is x y P e a sin( xy) a x y P y e cos( xy) a z x y Run vectorcurl.m! Dept. Electronics and Information Engineering 60

61 Visual EMT using MatLab clear; x = -1:0.:1; y = -1:0.:1; [xx,yy] = meshgrid(x,y); px = exp(-(xx.^+yy.^)); py = sin(xx.*yy); curl_z = curl(x,y,px,py); % Compute curl [m,n] = size(curl_z); % Return the number of rows and columns curl_x = zeros(m,n); % Create an m-by-n matrix of zeros curl_y = zeros(m,n); quiver3(xx,yy,zeros(m,n),curl_x,curl_y,curl_z,'r'); pause; surf(x,y,curl_z); xlabel('x');ylabel('y'); zlabel(' \nabla \times {\bf P} '); title('magnitude of Curl of -D vector field: \nabla \times {\bf P} '); CURLZ = CURL(X,Y,U,V) computes the curl z component of a D vector field U,V. The arrays X,Y define the coordinates for U,V QUIVER3(X,Y,Z,U,V,W) plots the vectors as arrows with components (u,v,w) at the points (x,y,z). Dept. Electronics and Information Engineering 61

62 Laplacian of a Scalar Definition: Divergence of the gradient of any scalar field, i.e. V = V Thus, in Cartesian coordinates, Laplacian V is V V V x y z V In cylindrical coordinates, In spherical coordinates, Note: Laplacian of a scalar field is another scalar field. z 1 V 1 V V V r r r r sin r sin V sin V V r V Dept. Electronics and Information Engineering 6

63 Laplacian of a Scalar If Laplacian vanishes in any region, a scalar field V is said to be harmonic (it is of the form of sine or cosine) in a given region. That is, V 0 (Laplace s equation) Now, let s define the Laplacian of a vector A A = A - A where in Cartesian system the Laplacian becomes Example 4.15: Determine the Laplacian of the scalar field Solution: ( ) A Aa Aa Aa x x y y z z If U is contour, the Laplacian presents the change rate of slope! U x yxyz y x y z U x y xyz Dept. Electronics and Information Engineering 63

64 Laplacian of a Scalar Example 4.16: For a scalar field V, show that V 0 ; that is, the curl of the gradient of any scalar field vanishes. Solution: a a a x y z V x y z V / x V / y V / z V V V V V V ax ay az z y y z x z z x x y y x 0 Dept. Electronics and Information Engineering 64

65 Classification of Vector Fields All vector fields can be classified as A = ka x, A= kr, A = k r, A = k r + cr, A = 0, A = 3 k, A = 0, A = 3, c A = 0 A A = 0 A = k A= k A = 0 A A 0 A vector field is said to be solenoidal (or divergenceless) if A vector field is said to be irrotational (or potential) if An irrotational field is also known as a conservative(path independent) field such as gravitational field and electrostatic field Dept. Electronics and Information Engineering 65

66 Classification of Vector Fields Example 4.17: Show that B y zcos xz ax xay xcos xzaz is conservative, without computing any integrals Solution: If B is conservative, then B 0 must be satisfied. ax ay az B x y z y zcos xz x xcos xz 0a cos xzxzsin xzcos xzxzsin xz a (1 1) a 0 x y z Dept. Electronics and Information Engineering 66

67 Homework Assignments Problem 4.1: Given that, calculate r ds over the region s x xy s S y x,0 x1 Problem 4.: If H x y ax x zy ay 5yzaz evaluate H dl along the contour of figure L Problem 4.3: Find the gradient of the following scalar fields: (a) U 4xz 3yz z (b) T 5e sin at, /3,0 (c) H r cos cos ò ò (Problem 3.) Dept. Electronics and Information Engineering 67

68 Homework Assignments Problem 4.4: Find the divergence and curl of the following vectors (a) (b) Problem 4.5: A e a sin xya cos xza B z cosa zsin az xy x y z Given that F sina a, find F dl where L is the L contours in figures (Problem 3.5) ò Dept. Electronics and Information Engineering 68

69 Homework Assignments Problem 4.6: cos z (a) D ds Let D za a. Evaluate (b) ò Problem 4.7: S ò V D dv over the region defined by 5, 1 z 1,0 If the vector field is irrotational, determine, and 3 T xy z a 3x za 3xz ya x y z Dept. Electronics and Information Engineering 69