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 ax+dy ay+dz az

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

a z 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 a + ρ dφ a +dz a ρ φ z

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) z00 =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 • Ad 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 , d S= dS a  n S

Flux across dS is: d = A cos dS A 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:

 dv  v 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:  Fdd 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 Fld  x2 dx   11 3  Segment 2, x 0, z  0, d l  dya , F  dl 0  F d l  0 y  2 18 Example 3.2 - solution 22 Segment 3: y 1, F= x ax  xz a y  y a z , d l  dx a x  dz a z Fld  x2 dx  dz 3

3 1 x 1 12  z  1   0 30 3 3 2 Segment 4: x 1, F= ax  z a y  y a z dl dy ayz dz a Fd l   z dy  y2 dz , but on 4, z  y , dz  dy 4 512 5 1 Fd 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 xycos  ,  sin  xx  y y  z z

 use chain rule differentiation:  sin  cos x      cos  sin y    

21 Del Operator but ax  cos a sin a , Cartesian aysin a  cos a , a z  a z     ax  a y  a z  in Cylindrical: x  y  z sin = 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  nV1  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 2 2 2 f(x,y) = −(cos x + cos y) 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) Wr 10 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 2zz 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 y 2 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