# Using Stokes' theorem to find a value

1. Nov 13, 2014

### InclusiveMonk

1. The problem statement, all variables and given/known data
. In the diagram below, the line AB is at x = 1 and the line BD is at y = 1.

Use Stokes’ theorem to find the value of the integral ∫s2(∇xV)⋅dS where S2 where S2 is
the curved surface ABEF, given that AF and BE are straight lines, and the curve EF is in the y-z plane (i.e. the plane x = 0), explaining how you are able to do this.
[Hint: Consider the value of ∇×V on the vertical surfaces OAF, BED and ODEF. You should not need to actually calculate any integrals.]

2. Relevant equations
Stokes' theorem - ∫s2(∇xV)⋅dS = ∫CV⋅dr

Earlier on in the question I used Stoke's theorem to calculate the area and line integral for the path C and the area enclosed by C. The answer was 2.

I also calculated that (∇xV)=(3x2+3y2)k

3. The attempt at a solution

I really have no idea for this part of the question. These are the types of questions I struggle at with vector calculus. I have a feeling it have something to do with the fact that at x=1 the line is straight. Also he has suggested we consider the value for (∇xV) when x and y are unchanged and we have already looked at when z is unchanged. Other than that I'm really not sure. A point in the right direction would be of great help.

2. Nov 13, 2014

### LCKurtz

You haven't told us what C is nor given us the equation for V. So I will assume C is the boundary of the square in the xy plane and that you have calculated curl V correctly. Also, with respect to what I highlighted in red, you are not calculating the area. That is a unit square and its area is 1.

The point of this problem is that if you added those three vertical sides to the curved surface, then your new surface would have C as its boundary. That's the setting you need to use Stoke's theorem. But what are you adding to the surface integral when you throw in those three sides? That is what the hint is about. So what about $\nabla \times \vec V\cdot d\vec S$ on those three surfaces? The hint notes you shouldn't have to do much work to figure out what you get for those three surfaces.

3. Nov 13, 2014

### InclusiveMonk

Sorry it was in an edit that I didn't save.

The vector field is V(r)=-y3i+x3j+z3k

The C from the first question was the square path OABD.

Also w.r.t. what you highlighted in red, in my notes ∫S(∇xV)⋅dS is called the area integral. That's what I was referring to.

Last edited: Nov 13, 2014
4. Nov 13, 2014

### InclusiveMonk

Okay, ADEF is in the x plane, which makes the dS=dydzi. This would mean ∫C(∇xV)⋅dS=∫S(3x2+3y2)k⋅dydzi. I know that dotting two different unit vectors = 0, so does that mean ∫C(∇xV)⋅dS is equal to 0 for ADEF and by extension, BED and OAF are 0 too?

5. Nov 13, 2014

### LCKurtz

You should, as a matter of course, quote the message to which you are replying.

The more common usage is to call such an integral a surface integral.

You mean dotting to perpendicular unit vectors gives zero. And, yes, any integral whose integrand is zero evaluates to zero. And I wouldn't say "by extension" the others are zero. Just check them.

6. Nov 13, 2014

### InclusiveMonk

Yes, apologies. I have a habit of explaining how my brain remembers things rather than the mathematical reasoning. It occasionally causes me to lose marks. OAF and BED are along the y plane, thus dS will contain the unit vector j which would cause the integral to be zero.

Okay, so I have shown that for the vertical sides (∇xV)⋅dS=0. Does this mean that the surface integral through the curved surface ABEF equals the same as the surface integral though OABD, the unit square?

7. Nov 13, 2014

### LCKurtz

Yes, because both equal $\int_C \vec V \cdot d\vec r$. And you do understand why it is important that the three vertical surfaces give zero, right?

8. Nov 13, 2014

### InclusiveMonk

I think so. No flux of V is passing through the sides so it all must be through the ABEF surface (as I type that I'm beginning to think I don't understand)