14519 on the face of the cube in the xy plane z 0 and

This preview shows page 49 - 52 out of 92 pages.

14.5.19: On the face of the cube in the xy -plane, z = 0, and so F ( x, y, z ) · n = x, 2 y, 0 · 0 , 0 , 1 = 0 , and hence the flux of F across that face is zero. Similarly, the flux across the faces in the other two coordinate planes is zero. On the top face we have z = 1, and hence F ( x, y, z ) · n = x, 2 y, 3 · 0 , 0 , 1 = 3 . Similarly, the flux across the face in the plane y = 1 is 2 and the flux across the face in the plane x = 1 is 1. Hence the total flux of F across S is 3 + 2 + 1 = 6. 14.5.20: The hemispherical surface z = 4 x 2 y 2 has unit normal vector n = 1 2 x, y, z 1876
and parametrization x = 2 sin φ cos θ, y = 2 sin φ sin θ, z = 2 cos φ, 0 φ 1 2 π, 0 θ 2 π. The usual computation of | r φ × r θ | (see the solution of Problem 6, 11, 12, 13, or 16) yields dS = 4 sin φ dA (provided that 0 φ π ), and thereby F ( x, y, z ) · n = 4 sin φ cos θ, 6 sin φ sin θ, 2 cos φ · sin φ cos θ, sin φ sin θ, cos φ = 2 cos 2 φ + 4 sin 2 φ cos 2 θ 6 sin 2 φ sin 2 θ, and hence the flux of F across the upper hemispherical surface H is H F · n dS = 2 π 0 π/ 2 0 (8 cos 2 φ sin φ + 16 sin 3 φ cos 2 θ 24 sin 3 φ sin 2 θ ) dφ dθ = 2 π 0 8 3 cos 3 φ 16 cos φ cos 2 θ + 16 3 cos 3 φ cos 2 θ + 24 cos φ sin 2 θ 8 cos 3 φ sin 2 θ π/ 2 0 = 2 π 0 8 3 + 32 3 cos 2 θ 16 sin 2 θ = 40 3 sin θ cos θ 2 π 0 = 0 . On the circular disk D that forms the base of the hemispherical solid, F · n = 2 x, 3 y, 0 · 0 , 0 , 1 = 0 , and therefore D F · n dS = 0 . Hence the total flux of F across the surface S is zero. 14.5.21: For the same reasons given in the solution of Problem 19, F · n = 0 on the three faces of the pyramid in the coordinate planes. On the fourth face T a unit normal vector is n = 1 26 3 , 4 , 1 , and because this face is the graph of z = h ( x, y ) = 12 3 x 4 y , we have dS = 1 + ( h x ) 2 + ( h y ) 2 dA = 26 dx dy. Therefore F · n dS = (3 x 4 y ) dy dx , and consequently T F · n dS = 4 0 (12 3 x ) / 4 0 (3 x 4 y ) dy dx = 4 0 3 xy 2 y 2 (12 3 x ) / 4 0 dx = 4 0 18 + 18 x 27 8 x 2 dx = 18 x + 9 x 2 9 8 x 3 4 0 = 0 . 1877
Thus the total flux of F across S is zero. If you now turn two pages ahead in your textbook, you will see how the divergence theorem enables you to obtain the same result in less than two seconds and without need of pencil, paper, or computer. 14.5.22: On the circular disk D that forms the base of the given parabolic solid, we easily see that F · n = 2 x, 2 y, 3 · 0 , 0 , 1 = 3, a constant, and therefore the flux of F across D is simply the product of 3 and the area of D : 12 π . An upward-pointing unit vector normal to the upper curved surface C described by z = h ( x, y ) = 4 x 2 y 2 is n = 1 1 + 4 x 2 + 4 y 2 2 x, 2 y, 1 , and dS = 1 + ( h x ) 2 + ( h y ) 2 dA = 1 + 4 x 2 + 4 y 2 dA. Therefore on C we have F · n dS = 4 x 2 + 4 y 2 + 3. Thus the flux of F across C is C F · n dS = D (4 x 2 + 4 y 2 + 3) dA = 2 π 0 2 0 (4 r 2 + 3) · r dr dθ = 2 π r 4 + 3 2 r 2 2 0 = 44 π, and therefore the total flux of F across S is 44 π 12 π = 32 π 100 . 530964914873383630804588 .

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture