# Homework Help: Stokes' theorem application

1. Nov 29, 2013

### mahler1

The problem statement, all variables and given/known data.
Let $F$ be the vector field defined by $F(x,y,z)=(-y,yz^2,x^2z)$ and $S \subset \mathbb R^3$ the surface defined as $S=\{x^2+y^2+z^2=4, z\geq 0\}$, oriented according to the exterior normal vector. Calculate:
$\iint_S (\nabla\times F).dS$.

The attempt at a solution.

I've calculated the curl, it's not an easy integral to calculate.

I can't apply Stokes' theorem because it is not a closed surface, but if I consider the surface $S^{*}=\{x^2+y^2+z^2=4, z\geq 0\} \cup \{ x^2+y^2\leq 4, z=0\}$, then this is a closed surface and $F$ is of class $C^1$, so Stokes'theorem says that:

$\iint_S^{*} (\nabla\times F).dS=\int_CF.ds$ where $C$ is the boundary of the surface $S^{*}$.

Now, my original integral is

$\iint_S (\nabla\times F).dS=\iint_S^{*} (\nabla\times F).dS-\iint_D (\nabla\times F).dS$, where $D=\{ x^2+y^2\leq 4, z=0\}$. But as $D$ is a closed surface, I can also apply Stokes' theorem, so

$\iint_D (\nabla\times F).dS=\int_{C'} F.ds$, where $C'$ is the boundary of $D$.

Now, my question is: isn't $C=C'$?, I mean, the curve boundary of $S^{*}$ is the same boundary than the one of $D$. If this is the case

$\iint_S (\nabla\times F).dS=\int_C F.ds-\int_{C'} F.ds=\int_C F.ds-\int_{C} F.ds=0$.

Could someone tell me if I am doing something wrong?

Last edited: Nov 29, 2013
2. Nov 29, 2013

### tiny-tim

hi mahler1!
i don't understand what you've done here, in particular what is the difference between ∫∫S and ∫∫*S

S and D have the same boundary, so stokes' says ∫∫S = ∫∫D

carry on from there

3. Nov 29, 2013

### LCKurtz

Say what???

4. Nov 29, 2013

### mahler1

Yes, sorry, I've corrected it and now it makes sense

5. Nov 29, 2013

### mahler1

$S$ is the upper semisphere of radius$2$ centered at the origin. I cannot apply Stokes' theorem on $S$ because $S$ is not a closed surface. Then, I define a new surface (I call it $S^{*}$ which is the union of two surfaces: the upper semisphere and the disk of radius $2$ centered at the origin and with $z=0$. This new surface is a closed surface so here I can apply Stokes' theorem. I hope that now is clear what $S^{*}$ is. Thanks for your suggestion, but I think that it isn't true that ∫∫S=∫∫D since $S$ is not closed.

6. Nov 29, 2013

### LCKurtz

No, it is an open surface with boundary, just what you would want for Stokes theorem.

But now the surface is closed, with no boundary, so Stokes theorem does not apply directly. Perhaps you are meaning to apply Stokes theorem to a second surface with the same boundary as the first?

7. Nov 29, 2013

### mahler1

This means I didn't get at all the hypothesis of Stokes, the surface doesn't need to be closed? If this is the case, I don't even need to define $S^{*}$ or $D$, I could just parametrize the boundary of $S$, which is the set $C=\partial S=\{(x,y,z) \in \mathbb R^3 : x^2+y^2=4, z=0\}$, and then, by Stokes' theorem:

$\iint_S (\nabla \times F)dS=\int_C F.ds$, would this be correct?

8. Nov 29, 2013

### tiny-tim

stokes' is supposed to apply to non-closed surfaces …

it relates an integral over the surface to an integral over its boundary (for a closed surface, the boundary is a null set, so the integral over the boundary is zero)

and now i'm off to bed :zzz:

9. Nov 29, 2013

### mahler1

tiny-tim:

Hmmm, what I've said before is completely wrong. I've read Stokes theorem in Tromba's textbook and evidently I mixed up all the concepts, I'll have to reread all the hypothesis of the theorem carefully, sorry about that and thanks for the previous suggestion.

10. Nov 29, 2013

### BruceW

yeah, to make use of stokes theorem, you need a surface that is not closed. just use the given surface $S=\{x^2+y^2+z^2=4, z\geq 0\}$. So from here, what is the boundary of this surface? and use that boundary for the line integral.

11. Nov 29, 2013

### LCKurtz

Perhaps you should look up the statement of Stokes theorem?

Now you're getting somewhere.

12. Nov 29, 2013

### mahler1

Thanks, I've read Stokes and Gauss theorem and I've mixed up the hypothesis of the two. I know how to solve the problem from here.

13. Nov 30, 2013

### tiny-tim

stokey-dokey!

14. Nov 30, 2013

### BruceW

I gauss that pun was inevitable...