(950, #23) Evaluate
where F = <x, 2 y, 3 z> and S is the S is the
cube with vertices 
Solution: Tabulating the normals, chosen so that
all are pointining outward from the surface, we have:
| Plane | Normal
| F * Normal |
| x = 1, y = -1..1, z = -1..1
| <1,0,0>
| x = 1 (x = 1)
|
| x = -1, y = -1..1, z = -1..1
| <-1,0,0>
| -x = 1 (x = -1)
|
| y = 1, x = -1..1, z = -1..1
| <0,1,0>
| 2y = 2 (y = 1)
|
| y = 1, x = -1..1, z = -1..1
| <0,-1,0>
| -2y=2 (y = -1)
|
| z = 1, x = -1..1, y = -1..1
| <0,0,1>
| 3z = 3 ( z = 1)
|
| z = 1, x = -1..1, y = -1..1
| <0,0,-1>
| -3z = 3 (z = -1)
|
Since the area of each face is 4, it follows that
