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17.5 Prel 1-6, Ex 1-2

17.5 Prel 1-6, Ex 1-2 - S E C T I O N 17.5 Surface...

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S E C T I O N 17.5 Surface Integrals of Vector Fields (ET Section 16.5) 1141 Using the Surface Integral over a Graph we have: Area ( S ) = S 1 dS = D 1 + g 2 x + g 2 y d A (1) In parametrizing the surface by φ ( x , y ) = ( x , y , g ( x , y )) , ( x , y ) = D , we have: T x = x = 1 , 0 , g x T y = y = 0 , 1 , g y Hence, n = T x × T y = i j k 1 0 g x 0 1 g y = − g x i g y j + k = − g x , g y , 1 n = g 2 x + g 2 y + 1 n k k There are two adjacent angles between the normal n and the vertical, and the cosines of these angles are opposite numbers. Therefore we take the absolute value of cos φ to obtain a positive value for Area ( S ) . Using the Formula for the cosine of the angle between two vectors we get: | cos φ | = | n · k | n k = | − g x , g y , 1 · 0 , 0 , 1 | 1 + g 2 x + g 2 y · 1 = 1 1 + g 2 x + g 2 y Substituting in (1) we get: Area ( S ) = D d A | cos φ | 17.5 Surface Integrals of Vector Fields (ET Section 16.5) Preliminary Questions 1. Let F be a vector field and ( u , v) a parametrization of a surface S , and set n = T u × T v . Which of the following is the normal component of F ? (a) F · n (b) F · e n SOLUTION The normal component of F is F · e n rather than F · n . 2. The vector surface integral S F · d S is equal to the scalar surface integral of the function (choose the correct answer): (a) F (b) F · n , where n is a normal vector (c) F · e n , where e n is the unit normal vector SOLUTION The vector surface integral S F · d S is defined as the scalar surface integral of the normal component of F on the oriented surface. That is, S F · d S = S ( F · e n ) dS as stated in (c). 3. S F · d S is zero if (choose the correct answer): (a) F is tangent to S at every point. (b) F is perpendicular to S at every point.
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