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17.5 Ex 18-21 - S E C T I O N 17.5 Surface Integrals of...

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S E C T I O N 17.5 Surface Integrals of Vector Fields (ET Section 16.5) 1155 Step 1. Compute the tangent and normal vectors. We have, T u = u = u u 3 v, u + v, v 2 = 3 u 2 , 1 , 0 T v = ∂v = ∂v u 3 v, u + v, v 2 = − 1 , 1 , 2 v T u × T v = i j k 3 u 2 1 0 1 1 2 v = ( 2 v) i 6 u 2 v j + 3 u 2 + 1 k = 2 v, 6 u 2 v, 3 u 2 + 1 Since the normal is pointing downward, the z -coordinate is negative, hence, n = − 2 v, 6 u 2 v, 3 u 2 1 Step 2. Evaluate the dot product F · n . We first express F in terms of the parameters: F ( ( u , v)) = y , z , 0 = u + v, v 2 , 0 We compute the dot product: F ( ( u , v)) · n ( u , v) = u + v, v 2 , 0 · − 2 v, 6 u 2 v, 3 u 2 1 = − 2 v( u + v) + 6 u 2 v · v 2 + 0 = − 2 v u 2 v 2 + 6 u 2 v 3 Step 3. Evaluate the surface integral. The surface integral is equal to the following double integral: S F · d S = D F ( ( u , v)) · n ( u , v) du d v = 3 0 2 0 2 u v 2 v 2 + 6 u 2 v 3 du d v = 3 0 u 2 v 2 v 2 u + 2 u 3 v 3 2 u = 0 d v = 3 0 16 v 3 4 v 2 4 v d v = 4 v 4 4 3 v 3 2 v 2 3 0 = 270 18. Let S be the oriented half-cylinder in Figure 14. In (a)–(f), determine whether S F · d S is positive, neg- ative, or zero. Explain your reasoning. (a) F = i (b) F = j (c) F = k (d) F = y i (e) F = − y j (f) F = x j y x n z FIGURE 14 SOLUTION
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1156 C H A P T E R 17 LINE AND SURFACE INTEGRALS (ET CHAPTER 16) y x n 3 1 z 1 S is parametrized by: ( θ , z ) = ( cos θ , sin θ , z ), 0 z 3 , π 2 θ π 2 Hence, T θ = θ = − sin θ , cos θ , 0 T z = z = 0 , 0 , 1 T θ × T z = i j k sin θ cos θ 0 0 0 1 = ( cos θ ) i + ( sin θ ) j = cos θ , sin θ , 0 The normal to S is pointing in the outward direction, hence the x -coordinate of n is positive. Since π 2 θ π 2 , we have cos θ 0, hence, n = cos θ ,
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