Stokes theorem and downward orientation problem

In summary, the conversation is discussing the verification of Stokes' Theorem for a given vector field and surface. The person has already solved the problem using the theorem and got a negative value, but their professor claims it should be positive. They are confused about whether to multiply the normal vector by -1 when the surface is oriented downwards. The conversation explains that there are two possible normal vectors for the given surface and we need to choose the one with a negative z-component for the integration to be correct. The person also mentions that they were able to get the same negative value by computing the line integral.
  • #1
nlsherrill
323
1

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?
 
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  • #2
nlsherrill said:
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)
 
  • #3
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.
 

1. What is Stokes theorem?

Stokes theorem is a mathematical formula that relates the surface integral of a vector field over a surface to the line integral of the same vector field around the boundary of the surface.

2. What is the significance of Stokes theorem?

Stokes theorem is significant because it allows for the conversion of a difficult surface integral into a more manageable line integral, making it a useful tool in solving various physics and engineering problems.

3. How is Stokes theorem related to the downward orientation problem?

The downward orientation problem refers to the question of how to orient a surface in order to correctly apply Stokes theorem. In order for Stokes theorem to be valid, the surface must be oriented in the direction of the outward normal vector. This ensures that the surface and its boundary have the same orientation, allowing for the application of the theorem.

4. What happens if the surface is not downward oriented?

If the surface is not downward oriented, the result obtained from applying Stokes theorem will be incorrect. This is because the theorem relies on the surface and its boundary having the same orientation, and a surface that is not downward oriented will have an inward normal vector instead of an outward one.

5. How can one ensure that the surface is downward oriented?

To ensure that the surface is downward oriented, one must check that the direction of the normal vector is pointing outward. If it is not, the orientation of the surface must be reversed. This can be achieved by either flipping the direction of the surface or by changing the direction of the normal vector to point outward.

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