MIT OpenCourseWare http://ocw.mit.edu Haus, Hermann A., and James R. Melcher. Electromagnetic Fields and Energy. Englewood Cliffs, NJ: Prentice-Hall, 1989. ISBN: 9780132490207. Please use the following citation format: Haus, Hermann A., and James R. Melcher, Electromagnetic Fields and Energy. (Massachusetts Institute of Technology: MIT OpenCourseWare). http://ocw.mit.edu (accessed [Date]). License: Creative Commons Attribution-NonCommercial-Share Alike. Also available from Prentice-Hall: Englewood Cliffs, NJ, 1989. ISBN: 9780132490207. Note: Please use the actual date you accessed this material in your citation. For more information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms2 APPENDIX 2.1 LINE AND SURFACE INTEGRALS Consider a path connecting points (a) and (b) as shown in Fig. A.2.1. Assume that a vector field A(r) exists in the space in which the path is situated. Then the line integral of A(r) is defined by � (b) A ds (1) (a) · To interpret (1), think of the path between (a) and (b) as sub divided into differential vector segments ds. At every vector segment, the vector A(r) is evaluated and the dot product is formed. The line integral is then defined as the sum of these dot products in the limit as ds approaches zero. A line integral over a path that closes on itself is denoted by the symbol � A ds. · Fig. A.2.1 Configuration for integration of vector field A along line having differential length ds between points (a) and (b). 12 Appendix Chapter 2 Fig. A.2.2 Integration line having shape of quarter segment of a circle with radius R and differential element ds. To perform a line integration, the integral must first be reduced to a form that can be evaluated using the rules of integral calculus. This is done with the aid of a coordinate system. The following example illustrates this process. Example 2.1.1. Line Integral Given the two-dimensional vector field A = xix + axy iy (2) find the line integral along a quarter circle of radius R as shown in Fig. A.2.2. Using a Cartesian coordinate system, the differential line segment ds has the components dx and dy. ds = ixdx + iydy (3) Now x and y are not independent but are constrained by the fact that the integration path follows a circle defined by the equation x 2 + y 2 = R2 (4) Differentiation of (4) gives 2xdx + 2ydy = 0 (5) and therefore x dy = dx (6)− y Thus, the dot product A ds can b e written as a function of the variable x alone.· A ds = xdx + a xydy = (x − ax 2)dx (7)· When the path is described in the sense shown in Fig. A.2.4, x decreases from R to zero. Therefore, 02 ax 30 aR3 R2� A ds = � (x − ax 2)dx = � x �����= (8)· R 2 − 3 R 3 − 2 If the path is not expressible in terms of an analytic function, the evaluation of the line integral becomes difficult. If everything else fails, numerical methods can be employed.� � 3 Sec. 2.2 Appendix Surface Integrals. Given a vector field A(r) in a region of space containing a specified (open or closed) surface S, an important form of the surface integral of A over S is A da (9) S · The vector da has a magnitude that represents the differential area of a surface element and a direction that is normal to that area. To interpret (9), think of the surface S as subdivided into these differential area elements da. At each area element, the differential scalar A da is evaluated and the surface integral is defined · as the sum of these dot products over S in the limit as da approaches zero. The surface integral �S A da is also called the “flux” of the vector A through the surface · S. To evaluate a surface integral, a coordinate system is introduced in which the integration can be performed according to the methods of integral calculus. Then the surface integral is transformed into a double integral in two independent variables. This is best illustrated with the aid of a specific example. Example 2.1.2. Surface Integral Given the vector field A = ixx (10) find the surface integral � A da, where S is one eighth of a spherical surface of S · radius R in the first octant of a sphere (0 ≤ φ ≤ π/2, 0 ≤ θ ≤ π/2). Because the surface lies on a sphere, it is best to carry out the integration in spherical coordinates. To transform coordinates from Cartesian to spherical, recall from (A.1.3) that the x coordinate is related to r, θ, and φ by x = r sin θ cos φ (11) and from (A.1.6), the unit vector ix is ix = sin θ cos φ ir + cos θ cos φ iθ − sin φ iφ (12) Therefore, because the area element da is da = irR2 sin θdθdφ (13) the surface integral becomes � � π/2 � π/23 3 2A da = dθ dφR sin θ cos φ· S 0 0 πR3 � π/2 πR3 (14) = dθ sin3 θ = 4 60 A surface integral of a vector A over a closed surface is indicated by A da (15) S ·4 Appendix Chapter 2 Note also that we use a single integral sign for a surface integral, even though, in fact, two integrations are involved when the integral is actually evaluated in terms of a coordinate system. 2.2 PROOF THAT THE CURL OPERATION RESULTS IN A VECTOR The definition [curl A]n = lim 1� A ds (1) a 0 a · → assigns a scalar, [curl A]n, to each direction n at the point P under consideration. The limit must be independent of the shape of the contour C (as long as all its points approach the point P in the limit as the area a of the contour goes to zero). The identification of curl A as a vector also implies a proper dependence of this limit upon the orientation of the normal n of a. The purpose of this appendix is to show that these two requirements are indeed satisfied by (1). We shall prove the following facts: 1. At a particular point (x, y, z) lying in the plane specified by its normal vector n, the quantity on the
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