Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch19 solutions

Calculus: Multivariable

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19.1 SOLUTIONS 1307 CHAPTER NINETEEN Solutions for Section 19.1 Exercises 1. (a) The fux is positive, since ~ F points in direction oF positive x -axis, the same direction as the normal vector. (b) The fux is negative, since below the xy -plane ~ F points toward negative x -axis, which is opposite the orientation oF the surFace. (c) The fux is zero. Since ~ F has only an x -component, there is no fow across the surFace. (d) The fux is zero. Since ~ F has only an x -component, there is no fow across the surFace. (e) The fux is zero. Since ~ F has only an x -component, there is no fow across the surFace. 2. The vector ±eld ~ F = F 1 ~ i + F 2 ~ j + F 3 ~ k = - z ~ i + x ~ k is a ±eld parallel to the xz -plane that suggests swirling around the origin From the positive x -axis to the positive z -axis. (a) The fux going through this surFace is negative, because ~ F · ~n = ( - z ~ i + x ~ k ) · ~ i = - z , z is positive here. (b) The fux going through this surFace is positive, because ~ F · ~n = - z , z is negative here. (c) The fux through this surFace is negative, because ~ F · ~n = ( - z ~ i + x ~ k ) · ( - ~ k ) = - x , x is positive. (d) The fux through this surFace is negative, because ~ F · ~n = - x , x is positive. (e) The fux through this surFace is zero, because it is in the xz -plane, which is parallel to the vector ±eld. 3. The vector ±eld ~ r is a ±eld that always points away From the origin. (a) The fux through this surFace is zero, because the plane is parallel to the ±eld. (b) The fux through this surFace is zero also, For the same reason. (c) The fux through this surFace is zero also, For the same reason. (d) The fux through this surFace is negative, because the ±eld in that quadrant is going up and away From the origin, and since the orientation is downward, the fux is negative. (e) The fux through this surFace is zero also. 4. ~v · ~ A = (2 ~ i + 3 ~ j + 5 ~ k ) · ~ k = 5 . 5. ~v · ~ A = (2 ~ i + 3 ~ j + 5 ~ k ) · 2 ~ i = 4 . 6. The rectangle lies in the plane z + 2 y = 2 . So a normal vector is 2 ~ j + ~ k and a unit normal vector is 1 5 (2 ~ j + ~ k ) . Since this points in the positive z -direction it is indeed an orientation For the rectangle. Since the area oF this rectangle is 5 we have ~ A = 2 ~ j + ~ k , ~v · ~ A = (2 ~ i + 3 ~ j + 5 ~ k ) · (2 ~ j + ~ k ) = 6 + 5 = 11 . 7. The rectangle lies in the plane z + 2 x = 2 . So 2 ~ i + ~ k is a normal vector and 1 5 (2 ~ i + ~ k ) is a unit normal vector. Since this points in both the positive x -direction and the positive z -direction, it is an orientation For this surFace. Since the area oF the rectangle is 5 , we have ~ A = 2 ~ i + ~ k and ~v · ~ A = (2 ~ i + 3 ~ j + 5 ~ k ) · (2 ~ i + ~ k ) = 4 + 5 = 9 .
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Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch19 solutions

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