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 flux is positive, since ~ F points in direction of positive x -axis, the same direction as the normal vector. (b) The flux is negative, since below the xy -plane ~ F points toward negative x -axis, which is opposite the orientation of the surface. (c) The flux is zero. Since ~ F has only an x -component, there is no flow across the surface. (d) The flux is zero. Since ~ F has only an x -component, there is no flow across the surface. (e) The flux is zero. Since ~ F has only an x -component, there is no flow across the surface. 2. The vector field ~ F = F 1 ~ i + F 2 ~ j + F 3 ~ k = - z ~ i + x ~ k is a field parallel to the xz -plane that suggests swirling around the origin from the positive x -axis to the positive z -axis. (a) The flux going through this surface is negative, because ~ F · ~n = ( - z ~ i + x ~ k ) · ~ i = - z , z is positive here. (b) The flux going through this surface is positive, because ~ F · ~n = - z , z is negative here. (c) The flux through this surface is negative, because ~ F · ~n = ( - z ~ i + x ~ k ) · ( - ~ k ) = - x , x is positive. (d) The flux through this surface is negative, because ~ F · ~n = - x , x is positive. (e) The flux through this surface is zero, because it is in the xz -plane, which is parallel to the vector field. 3. The vector field ~ r is a field that always points away from the origin. (a) The flux through this surface is zero, because the plane is parallel to the field. (b) The flux through this surface is zero also, for the same reason. (c) The flux through this surface is zero also, for the same reason. (d) The flux through this surface is negative, because the field in that quadrant is going up and away from the origin, and since the orientation is downward, the flux is negative. (e) The flux 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 . 8. The square has area 9, so its area vector is 9 ~ i . Since ~ F = - 5 ~ i on the square, Flux = - 5 ~ i · 9 ~ i = - 45 . 9. The square has area 16, so its area vector is 16 ~ j . Since ~ F = 5 ~ j on the square, Flux = 5 ~ j · 16 ~ j = 80 . 10. Since the square, S , is in the plane y = 0 and oriented in the negative y -direction, d ~ A = - ~ j dxdz and Z S ~ F · d ~ A = Z S (0 + 3) ~ j · ( - ~ j dxdz ) = - 3 Z S dxdz = - 3 · Area of square = - 3(2 2 ) = - 12 .

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1308 Chapter Nineteen /SOLUTIONS 11. The disk has area 25 π , so its area vector is 25 π ~ j . Thus Flux = (2 ~ i + 3 ~ j ) · 25 π ~ j = 75 π.
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