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chap16-et-student-solutions

chap16-et-student-solutions - LINE AND 17 SURFACE INTEGRALS...

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17 LINE AND SURFACE INTEGRALS 17.1 Vector Fields (ET Section 16.1) Preliminary Questions 1. Which of the following is a unit vector field in the plane? (a) F = y , x (b) F = y x 2 + y 2 , x x 2 + y 2 (c) F = y x 2 + y 2 , x x 2 + y 2 SOLUTION (a) The length of the vector y , x is y , x = y 2 + x 2 This value is not 1 for all points, hence it is not a unit vector field. (b) We have y x 2 + y 2 , x x 2 + y 2 = y x 2 + y 2 2 + x x 2 + y 2 2 = y 2 x 2 + y 2 + x 2 x 2 + y 2 = y 2 + x 2 x 2 + y 2 = 1 Hence the field is a unit vector field, for ( x , y ) = ( 0 , 0 ) . (c) We compute the length of the vector: y x 2 + y 2 , x x 2 + y 2 = y x 2 + y 2 2 + x x 2 + y 2 2 = y 2 + x 2 ( x 2 + y 2 ) 2 = 1 x 2 + y 2 Since the length is not identically 1, the field is not a unit vector field. 2. Sketch an example of a nonconstant vector field in the plane in which each vector is parallel to 1 , 1 . SOLUTION The non-constant vector x , x is parallel to the vector 1 , 1 . y x 3. Show that the vector field F = z , 0 , x is orthogonal to the position vector −→ O P at each point P . Give an example of another vector field with this property. SOLUTION The position vector at P = ( x , y , z ) is x , y , z . We must show that the following dot product is zero: x , y , z · − z , 0 , x = x · ( z ) + y · 0 + z · x = 0
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S E C T I O N 17.1 Vector Fields (ET Section 16.1) 563 Therefore, the vector field F = z , 0 , x is orthogonal to the position vector. Another vector field with this property is F = 0 , z , y , since 0 , z , y · x , y , z = 0 · x + ( z ) · y + y · z = 0 4. Give an example of a potential function for yz , xz , xy other than ϕ ( x , y , z ) = xyz . SOLUTION Since any two potential functions of a gradient vector field differ by a constant, a potential function for the given field other than φ ( x , y , z ) = xyz is, for instance, φ 1 ( x , y , z ) = xyz + 1. Exercises 1. Compute and sketch the vector assigned to the points P = ( 1 , 2 ) and Q = ( 1 , 1 ) by the vector field F = x 2 , x . SOLUTION The vector assigned to P = ( 1 , 2 ) is obtained by substituting x = 1 in F , that is, F ( 1 , 2 ) = 1 2 , 1 = 1 , 1 Similarly, F ( 1 , 1 ) = ( 1 ) 2 , 1 = 1 , 1 x 1 1 1 y F ( P ) = 1, 1 F ( Q ) = 1, 1 Compute and sketch the vector assigned to the points P = ( 1 , 2 ) and Q = ( 1 , 1 ) by the vector field F = y , x . 3. Compute and sketch the vector assigned to the points P = ( 0 , 1 , 1 ) and Q = ( 2 , 1 , 0 ) by the vector field F = xy , z 2 , x . SOLUTION To find the vector assigned to the point P = ( 0 , 1 , 1 ) , we substitute x = 0, y = 1, z = 1 in F = xy , z 2 , x . We get F ( P ) = 0 · 1 , 1 2 , 0 = 0 , 1 , 0 Similarly, F ( Q ) is obtained by substituting x = 2, y = 1, z = 0 in F . That is, F ( Q ) = 2 · 1 , 0 2 , 2 = 2 , 0 , 2 F ( P ) = 0, 1, 0 F ( Q ) = 2, 0, 2 y x z Compute the vector assigned to the points P = ( 1 , 1 , 0 ) and Q = ( 2 , 1 , 2 ) by the vector fields e r , e r r , and e r r 2 . In Exercises 5–13, sketch the following planar vector fields by drawing the vectors attached to points with integer coor- dinates in the rectangle 3 x , y 3 . Instead of drawing the vectors with their true lengths, scale them if necessary to avoid overlap.
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