Stokes theorem and downward orientation problem

[nlsherrill]

Homework Statement

From Calculus:Concepts and Contexts 4th Edition by James Stewart. Pg.965 #13

Verify that Stokes' Theorem is true for the given vector field F and surface S

F(x,y,z)= -yi+xj-2k

S is the cone z^2= x^2+y^2, o<=z<=4, oriented downwards

Homework Equations

The Attempt at a Solution

Now I have already solved this problem actually using Stokes Theorem and got -32*pi. My professor claims that it is just 32*pi, but I don't understand why. I thought if the problem stated it was oriented downward you had to multiply the normal vector by -1. Doing this gives me negative 32*pi. Also, if I try to solve the problem by just evaluating it as a line integral, I can't integrate it because the integrand ends up being 16*cos(t)^2 - 16*sin(t)^2, where it would clearly also be 32*pi if the negative was replaced with a positive. Could someone please clear things up for me?

 

[Office_Shredder]

I thought if the problem stated it was oriented downward you had to multiply the normal vector by -1.

There are two normal vectors at every point on a surface. With no further information there's no way to distinguish between them with Stoke's theorem. When you're told that a surface is oriented downwards, that means that you want to pick the normal vector who's z component is negative

S is the surface z²-x²-y²=0, so we can find normal vectors by using the gradient (-2x, -2y, 2z). At a point (x,y,z), this (after scaling) is a normal vector, and (2x, 2y, -2z) is also.How do we know which one to use? We're integrating for z between 0 and 4, and we want the z-component of our normal vector to be negative.
(-2x, -2y, 2z) has a positive z-component when z is positive, so that's not the right choice. So your normal vector should be (2x, 2y, -2z) (again, you have to scale it)

 

[nlsherrill]

What do you mean "scale it"?

Oh and I would like to mention when I compute it as a line integral I can integrate it and get -32*pi as well.