Summary of Class MW 02/ TR 03 8.02 Topics: Coordinate Systems; Gradients; Line and Surface Integrals Related Reading: Spring 2006 Math Review PresentationHale Bradt’s Spring 2001 8.02 Mathematics Supplement Topic Introduction Today we go over some of the more advanced mathematical concepts we will need in the course, so that you see the mathematics before being introduced to the physics. Maxwell’s equations as we will state them involve line and surface integrals over open and closed surfaces. A closed surface has an inside and an outside, e.g. a basketball, and there is no two dimensional contour that “bounds” the surface. In contrast, an open surface has no inside and outside, e.g. a flat infinitely thin plate, and there is a two dimensional contour that bounds the surface, e.g. the rim of the plate. There are four Maxwell’s equations: 0000(1) (2) 0(3) (4) inSSBEencCCQddddddIdt dtεµµε⋅= ⋅=ΦΦ⋅=− ⋅= +∫∫ ∫∫∫∫EA BAEs BsGGGGGGGGwwvv Equations (1) and (2) apply to closed surfaces. Equations (3) and (4) apply to open surfaces, and the contour C represents the line contour that bounds those open surfaces. There is not need to understand the details of the electromagnetic application right now; we simply want to cover the mathematics in this problem solving session. Line Integrals The line integral of a scalar function along a path C is defined as (,,)fxyz 10(, ,) lim ( , , )iNiii iCNisfxyzds f x y z s→∞=∆→=∆∑∫ where C has been subdivided into N segments, each with a length is∆. Line Integrals Involving Vector Functions For a vector function ˆˆˆxyzFFF=++FijkG the line integral along a path C is given by ()()ˆˆ ˆˆˆˆxyz x y zCC Cd F F F dx dy dz F dx F dy F dz⋅= + + ⋅ + + = + +∫∫ ∫Fs i j k i j kGG where ˆˆˆddxdydz=++sijkG is the differential line element along C. Summary of Class MW 02/ TR 03 W01D3 p. 1/2Summary of Class MW 02/ TR 03 8.02 Surface Integrals A function of two variables can be integrated over a surface S, and the result is a double integral: (, )Fxy (, ) (, )SSF x y dA F x y dxdy=∫∫ ∫∫ where is a (Cartesian) differential area element on S. In particular, when , we obtain the area of the surface S: dA dx dy=(, ) 1Fxy= SSAdA dxdy==∫∫ ∫∫ Surface Integrals Involving Vector Functions For a vector function (, ,)xyzFG, the integral over a surface S is is given by ˆnSS SddA⋅= ⋅ =∫∫ ∫∫ ∫∫FA Fn FdAGGG where and is a unit vector pointing in the normal direction of the surface. The dot product is the component of FˆddA=AGnˆnˆnF =⋅FnGGparallel to . The above quantity is called “flux.” For an electric field E , the electric flux through a surface is ˆnG ˆEnSSdA E dAΦ= ⋅ =∫∫ ∫∫EnG Important Equations The line integral of a vector function: ()()ˆˆ ˆˆˆˆxyz x y zCC Cd F F F dx dy dz F dx F dy F dz⋅= + + ⋅ + + = + +∫∫ ∫Fs i j k i j kGG The flux of a vector function: ˆEnSSdA E dAΦ= ⋅ =∫∫ ∫∫EnG Summary of Class MW 02/ TR 03 W01D3 p.
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