g 1 VECTOR ANALYSIS 3,Xa0=a6 agxao=a 6x8=6 cons =con: const. (a) (b) Flgure 1.6.Unit vectors for(a)Cartesian.(b)cylindrical.and (c)spherical coordinates. is important to recognize that the Cartesian coordinate system is the only one for which all hree unit vector are constant.All other coordinate systems possess at least one unit vector whose direction is not constant.When it becomes necessary unit vector ap depends on and this must be accounted for.The component of a in the direction of component of in the direction of ay is ap'ay =sin and the component of ap in the direction of a:is apa0. Thus,ap=ar cosφ+ay sin中，and the resolution we are seeking is apd中=(ax cos中+ay sin中)dφ. It is often necessary to express the components of a vector that are expressed in one coordinate system into those for another coordinate system.The required coordinate component transformations are listed below and on the inside front cover: Ax=Apcos中-Aosin中 =A,sin0cosb+Ae cos0cos中-Aosin中， Ay=Ap sin中+Aocos中 =A,sin8sin中+Ae cos0sin中+Adcos中， A:A,cos 0-Ae sin 6 Ap=Ax cos+Ay sin Ar sin 0 +Aa cos 0, (1.15) Ab=-Ax sin中+Ay cos中， A,=Axsin0cos中+A,sin0sin中+A.cos0 =Apsin 0 A:cos 0, Ag=Ar cos8cos中+Ay cos0sin中-A:sin0 =Ap cos 6-A:sin 0
1.3 COORDINATE SYSTEMS 7 These component transformations are relatively easy to derive with the aid of Figures 1.4 and 1.6 and Equation(1.5).Consider,for example,the transformation A=A,cos +Ay sin In general A=axAx +ayAy +a:A:.The component of the vector aA,in the directionsAA,cos the component of the vector ay Ay in the direction of apAp is ay Ayap=Ay sin and the component of the vector aA:in the direction of aA is zero.Therefore,the component of A in the direction ofap(that is,A)is Ap=axAx·ap+ayAy·ap+a,A:·ap An=A:cos中+Ay sin中. In the same way,we have Ap=aA,·ap+agAg·ap+aAap, Ap Ar sin 0 Ae cos 6. ☐Example1.l Suppose that a vector field is given by A(x,y,z)=2ax +x2ay +xya. What is the field at the point(1,1,2)?Substituting,we get A(1,1,2)=2ax+ay+az At the given point,0 tan-!(x2 +y2)1/2/z =tan-1 2/2 =35.3 and= tan-1 y/x tan-1 1 =45,so using Equations (1.15),we get Ap=Ax cosφ+Ay sin中=2.121， A0=-Ax sin中+A,c0s中=-0.707 A:=1, A,=Apsin 0 A:cos 0 2.041, Ae=ApCos 0-A:sin 0 1.155. Thus,A can also be expressed as A=2.121ap-0.707a++az Or A=2.041a,+1.155ag-0.707ab:
8 1 VECTOR ANALYSIS dl a,dr +aar de +aor sin do 小=axdk+aydy+adz 小=a dp +aopdo+azdz ds=a,r2 sin 0 do de ds=ax dy dz,ay dz dx, ds ap do dz,a dp dz. aer sin dr do. a:dx dy a:p dp do aar dr do dy sin d dy dx dy dz dv =p dp do dz dy=r2 sin 0 dr do do (a) (b) Figure 1.7.Differential elements of vector length,vector area,and scalar volume for(a)Cartesian (b)cylindrical,and (c)spherical coordinates. Differential elements of vector length,vector area,and scalar volume can be found using Figure 1.7.Note that each component of vector area is normal to its coordinate surface.In the case of cylindrical and spherical coordinates,the differential area and volume are obtained from the first-order approximation that the differential volumes are rectangular boxes. ■Example1.2 The component of differential area normal to a spherical surface is a,ds= r2 sin 0 do do.Thus,the surface area of a sphere is 5= 6=0J=0 2sin0d0db=2rr2。sin0d0=4r2(m Its volume(for a radius r=a)is v-(4/3)πa3(m3 Note that in this example the order of integration is unimportant.This is not alway the case
1.4 CIRCULATION AND FLUX 1.4 CIRCULATION AND FLUX Consider the expended in moving an object along a specified path C such as that shown in Figure 1.8.The total work can be appro mated as a finite sum of incremental amounts of work.The nth amount of incremental work is [see Equation(1.8】 △Wn=Fn△I cos=Fn·△ln, (1.16) where is the angle between F and Al as shown in Figure 1.8.The total work is approximately N N w≈∑AW=∑F·Al (1.17) n=1 =1 If we let N approach infinity in such a way that Al approaches zero for any n,the limit is an exact integral expression for W,called a line integral in vector analysis, w =F.d. (1.18) A force field F such that the line integral from P to P along the path Ci is equal to that along the path C2 for any Pi and P must then also be such that the line integral ar ound c any closed path is zero.The line integral around a closed path is called the circularion of F.A field F,whose circulation is zero is called a conservative field since energy is conserved in moving an object around the closed path 4 Path Q Flgure 1.8. ng objectalong a prescribed path
10 1 VECTOR ANALYSIS ΦF·dl=0 conservative field F (1.19) The differential vector length dl is listed in Figure 1.7 and on the inside front cover for our three standard coordinate systems. Example 1.3 Suppose we desire to find the work done in moving an object from (0,0,1)to (2.4.1)along the parabolic path described by y1 in the nonuniform field F=2yar 2xay za:(see Figure 1.9).Since Cartesian coordinates are appropriate,we get F·dl=(2yax+2xay+zaz)·(axdr+avdy+a,dz) 2y dx +2x dy +z dz and In the first integral we use the equation of the pathy=x2,while in the second we use dy 2x dx from the equation of the path.(Note that this changes the limits from those on y to those on x,i.e.,0 to 2.)The third integral is zero.Thus,we have (0,0,1) y=2x,2=1 (2,4,10 y"x2,2=1 Flgure 1.9.Geometry for the line integrals in Example 1.3