need help on divergence theorem. 8. Let S be the part of the surface z = 4 - x^2+ y^2 above the xy-plane. Let F =. (a) If no is the upward pointing unit normal to S explain why

Question

need help on divergence theorem. 8. Let S be the part of the surface z = 4 - x^2+ y^2 above the xy-plane. Let F =< xy, y, z >.
(a) If no is the upward pointing unit normal to S explain why

anonymous · Answer

opps i mean z=4-x^2-y^2. I am not sue how to tell if no is the upward pointing unit normal to S

anonymous · Answer

$$n _{o}=no$$

anonymous · Answer

The solid is bounded by the plane z = 0 and S. The orthogonal projection of S is the disk x^2 + y^2 <= 4.  The union of S and the disk forms a simple closed surface, and we work with unit normal vectors directed outward.  So any vector that is normal to S will have a positive z component.  At (0, 0, 4), for example, the unit normal vector n is <0, 0, 1>.

anonymous · Answer

so since any vector that is normal to S is positive it must be directed upward?

anonymous · Answer

The z components of all vectors normal to S are positive.

anonymous · Answer

ohhh okay i get what your saying now thanks!

anonymous · Answer

:^)

anonymous · Answer

thanks for the help!!!