EE468G NOTES (2)Spatial differentiation and integration of scalar and vector functions Outline: Gradient, divergence, and curl operators Volume, surface, and line integral Theorems about integrals Gradient: Operated on a scalar function Result is a vector The magnitude of the gradient is the max rate of change The direction is the direction of max rate of change In Cartesian system: gradient of scalar function ( , , )f x y zis defined as zfayfaxfazyxfzyxfzyxˆˆˆ),,(),,(grad Contour of a scalar function Gradient: A vector Magnitude is the rate of change In the direction of max rate of change.Solution: (1) By definition         yzazxaxyazyzyxayyzyxaxyzyxafzyxzyx10ˆ53ˆ6ˆ53ˆ53ˆ53ˆ22222222 (2) The max rate of change at P(1,2,0) is given by     36.1215302100513216222220,2,1f (3) First we need to determine the direction of P1-P2: yxzyxaaaaalˆˆ)22(ˆ)12(ˆ)01(ˆ  2ˆˆ||ˆyxlaalla  The rate of change in the direction of P1-P2 is:    253262/ˆˆˆ10ˆ53ˆ6ˆ2222zxxyaaayzazxaxyaFyxzyxl 6.10203212ˆ)0,2,1(laf Example: given  22, , 3 5f x y z x y yz, (1) Calculate f. (2) Calculate the max rate of change of fat point P(1,2,0). (3) Calculate the rate of change of falong the direction from point P1(0,1,2) to P2(1,2,2) evaluated at P(1,2,0).The gradient operation in cylindrical system: zfafafafzˆˆˆ The gradient operation in sphereical system: sinˆˆˆrfarfarfafr Solution: In Cartesian system: yzyxazfayfaxfafˆˆˆˆ In cylindrical system:  , , sinfz    cosˆsinˆ)sin(ˆ)sin(ˆaaaaf  In spherical system: sinsin),,( rrf  )(cosˆ)sin(cosˆ)sin(sinˆ sin)sinsin(ˆ)sinsin(ˆ)sinsin(ˆaaarrarrarrafrr Example: given  ,,f x y z y, Calculate fin Cartesian, cylindrical, and spherical systems:Divergence: Operated on a vector function The result is a scalar function. Definition in Cartesian system is given by yxzBBBBx y z      Solution: (1) By definition:     232681212 12 24xy xyzxzBx y zz xy xyz          (2) When evaluated at P(1,4,2), the above result is 212 2 12 1 4 24 1 4 2 312.B            Example: given, zyxaxyzaxyaxzBˆ8ˆ6ˆ1232 (1) Calculate B. (2) Evaluate B at P(1,4,2).Curl operator: Operated on vector functions The result is a vector Definition in Cartesian system: zyxzyxAAAzyxaaaA ˆˆˆ Solution:    00ˆ4ˆ3ˆ034ˆˆˆ33zyxzyxaxayayzzxzyxaaaA Details: x-component: yyzzy303 y-component:  33404xzxzx z-component: 00343yzzxyx Example: GivenyxayzazxAˆ3ˆ43, calculate A Solution: By definition:Volume integral: ( , , )VF x y z dV The integration should be performed in a specific coordinate system, because the detail of dVdepends on coordinate system. For example, dV dxdydzif the integration is performed in the Cartesian system. Example: 2ˆ( , , ) 3zF x y z x ya, integrate this function in the volume region specified by: 1 2, 0 1, 2 2x y z       . Solution: The volume region is specified in the Cartesian system,  zzzzVaadzydydxxadxdydzayxdVFˆ1842139ˆ3 ˆ3 )ˆ3(22102122210212   Surface integral: ( , , )SF x y z dS The direction of dSis the outward normal direction of the surface element. Example: given yxaxaxFˆ2ˆ32, calculate the surface integral ( , , )SF x y z dSon the surface of a cube specified by 1 1, 1 1, 1 1x y z         Solution: The cube has six faces (top, bottom, left, right, front, and rear). Hence the integral is the sum of six integrals. We consider then separately. On the top face (1, 1 1, 1 1z x y      ):ˆzdS a dxdy Since ˆ0zFa, this integration is zero. On the bottom face: the integration is zero (same reason as for top) On the front face: 1, 1 1, 1 1x y z      , ˆxdS a dydz 12)03( ˆ)ˆ2ˆ3(111111112front    dydzxdydzaaxaxSdFxyx On the rear face: the integration is also 12. One the right face: 1, 1 1, 1 1y x z      , ˆydS a dxdz 3/8)2( ˆ)ˆ2ˆ3(1111211112right    dxdzxdxdzaaxaxSdFyyxOn the left face, : 1, 1 1, 1 1y x z       , ˆydS a dxdz, the integral will be 8/3 In the same way, the integrals on the top and bottoms will be 0 as the function has no z components. Summarize: the result is (12)+(12)+(-8/3)+8/3)+(0)+(0)=24Line integral: ( , , )CF x y z dl Integration result may depend on path dlis in the direction of path: dzadyadxaldzyxˆˆˆ Steps to evaluate line integral: (1) Identify path-condition, (2) Identify a variable and its start-ending values, (3) calculate dl A B C D 1 1 2 2 y x Solution: (1) The integral result is the sum of the integrals on AB, BC, CD, and DA. On AB: dxalddyyxˆ ,0 ,1 , hence 55ˆ)ˆ3ˆ5(2121dxdxaaxyayldFxyxAB On BC, dyaldxyˆ,2   96ˆˆ23ˆ52121 ydydyaayayldFyBCyx On CD, ˆ2,xy dl a dx Example: given yxaxyayFˆ3ˆ5 , calculate the line integral on two paths: (1) A-B-C-D-A, (2) A-B-D-A.1010ˆ)ˆ3ˆ5(1212dxdxaaxyayldFxyxCD On DA, ˆ1,yx dl a dy