EELE 3331 – Electromagnetic I Chapter 3 Vector Calculus Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik 2012 1 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 +dy a +dz a x y z 2 Differential Length, Area, and Volume A. Cartesian Coordinate Systems: 2. Differential normal surface area (a) dS=dy dz ax (b) dS=dx dz ay (c) dS=dx dy az 3 Differential Length, Area, and Volume A. Cartesian Coordinate Systems: 3. Differential volume dv=dx dy dz Notes: dl, dS → Vectors dv → Scalar 4 Differential Length, Area, and Volume B. Cylindrical Coordinate Systems: 1. Differential displacement: dl=dρ aρ+ ρ dφ aφ +dz az 5 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 aρ (b) dS= dρ dz aφ (c) dS= ρ dρ dφ az 3. Differential volume dv =ρ dρ dφ dz 6 Differential Length, Area, and Volume C. Spherical Coordinate Systems: 1. Differential displacement: dl = dr ar + r dθ aθ + r sinθ dφ aφ 7 Differential Length, Area, and Volume C. Spherical Coordinate Systems: 2. Differential normal surface area Note: dS can be derived from dl 2 (a) dS=r sinθ dθ dφ ar (b) dS=r sinθ dr dφ aθ (c) dS= r dr dθ aφ 2 3. Differential volume dv =r sinθ dr dθ dφ 8 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 Example 3.1 - solution Object has Cylindrical Symmetry Cylindrical Coordinates Cartesian to Cylindrical: A(5,0,0) A(5,00 ,0) B(0,5,0) B(5, /2,0) C(0,5,10) C(5, /2,10) D(5,0,10) D(5,00 ,10) (a) along BC, dl dz 10 BC dl dz 10 0 /2 (b) Along CD, dl d CD d 5 /2 2.5 0 10 0 Example 3.1 - solution (c) for ABCD, dS d dz, =5 Area ABCD = dS 10 /2 d dz (5)( / 2)(10) z00 =25 (d) for ABO, dS= d d , z=0 5 /2 area ABO= dd 00 5 2 ( / 2) 6.25 2 0 11 Example 3.1 - solution (e) for AOFD , dS d dz, =0 5 10 area AOFD=d dz 50 00z (f) For volume ABCDFO, dv d dz d 5 /2 10 v dv d dz d 0 0z 0 62.5 12 Line, Surface, and Volume Integrals (Line=Curve=Contour) Integral: The Line integral A d l is integral of the tangential component L of vector A along L. b • Ad l A cos dl La Line integral of A around L. • If the path of integration is closed, such as abca, Al d L Circulation of A along L. 13 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: AS d S = A cos dS, dS= dS an S Flux across dS is: d= A cos dSA d S Total Flux dd A S S 14 Line, Surface, and Volume Integrals Surface Integral: For a closed surface (defining a volume) : AS d → The net outward flux of A from S S Notes: A closed path defines an open surface. A closed surface defines a volume. 15 Line, Surface, and Volume Integrals Volume Integral: We define: v dv v as the volume integral of the scalar ρv over the volume v. 16 Example 3.2 2 2 Given that F=x ax – xz ay – y az. Calculate the circulation of F around the (closed) path shown in the Figure. 17 Example 3.2 - solution The circulation of F around L is: Fdd l= F l L 1 2 3 4 For segment 1, y=0, z=0 2 2 2 F=x ax xz a y y a z x a x dl dx ax (+ve direction) 0 1 Fld x2 dx 11 3 Segment 2, x0, z 0, dl dya , F dl 0 F d l 0 y 2 18 Example 3.2 - solution 22 Segment 3: y1, F= x ax xz a y y a z, dl dx a x dz a z Fld x2 dx dz 3 3 1 x 1 12 z 1 0 30 3 3 2 Segment 4: x1, F= ax z a y y a z dl dy ayz dz a Fd l z dy y2 dz, but on 4, z y, dz dy 4 512 5 1 Fd l ( y y2 ) dy , F d l= 0 4 6 L 3 3 619 6 Del Operator The del operator, written as , is the vector differential operator. In Cartesian coordinates: ax a y a z x y z Useful in defining: (1) The gradient of a scalar VV, written as (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 VV, 2 20 Del Operator Del operator in Cylindrical Coordinates: ( , ,z) 2 2 -1 y Cartesian xy, =tan x a a a xycos , sin xx y y z z use chain rule differentiation: sin cos x cos sin y 21 Del Operator but ax cos asin a , aysin a cos a, a z a z Cartesian in Cylindrical: a a a sin xx y y z z = cos cos aa sin cos + sin sin a cos a az z 1 In Cylindrical =aa a z z 11 In Spherical, =ar a a r r r sin 22 Del Operator Cartesian a a a xx y y z z 1 Cylindrical =a a a z z 11 Spherical, =a a a r r r r sin 23 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 VVV grad V= V = a a a xx y y z z VV maximum rate of change in VV points in the direction of the max. rate of change in 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 Gradient of a Scalar The gradient of the function f(x,y) = −(cos2x + cos2y)2 depicted as a projected vector field on the bottom plane 25 Gradient of a Scalar VVV In Cartesian: V= a a a xx y y z z For Cylindrical coordinates, 1 Recall =a a a z z VVV1 V= a a a z z For Spherical coordinates, 11 Recall =aa a r r rr sin VVV11 V= ar a a r r r sin 26 Gradient of a Scalar Notes: (a) VUVU (b) VU V U U V VUVVU (c) UU2 (d) Vnn nV1 V where UV and 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 f(x,y) = −(cos2x + cos2y)2 f(x,y) =constant →contour 22 f(,) x y x y f( x , y ) C constant Contour ------------------------ f(,,) x y z x2 y 2 z 2 f(,,) x y z C level surface (ex: surface of sphere) 28 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 (c) Wr10 sin2 cos VVV (a) V= a a a xx y y z z ---z z z 2e cos2 x cosh y ax e sin 2 x sinh y a y e sin 2 x cosh y a z UUU1 (b) U= a a a z z 2 2zz cos2 a 2 sin 2 a cos2 a z WWW11 (c) W= a a a rr r rsin 2 29 10sin cos ar 20sin cos cos a 10sin sin a Example 3.5 Find the angle at which line x y2 z intersects the ellipsoid x2 y 2 2 z 2 10 To find the point of intersection: x2 x 2 2( x / 2) 2 10 2x22 0.5 x 10 x 2 point (2,2,1) f( x , y , z ) x2 y 2 2 z 2 10 surface of ellipsoid f( x , y , z ) constant f 2 x ax 2 y a y 4 z a z At (2,2,1), f 4 a 4 a 4 a x y z 30 Example 3.5 - continued Hence, a unit vector normal to the ellipsoid surface at P(2,2,1) is: f 4 ax a y a z a n f 48 ax a y a z 3 ar cos n arn (1/ 3) 1,1,1 2,2,1 5 9 3 3 15.790 90 74.20 31 .