Intuition with applying Stoke's theorem to a cube.

In summary: I think you are understanding it. The above theorem is what you want to use, where the surface ##S_2## is the flat square ##-1\le x \le 1,\, -1\le y \le 1,\, z=-1##, oriented upward. (Assuming the original cube was directed outward, which you never specified, and should have). So integrate curl F over that instead of using...The other faces are not equal. The wrong equation is being used.
  • #1
Bill Nye Tho
48
0

Homework Statement



F(x,y,z) = xyzi+xyj+x^2yzk

Surface is the top and four sides of cube with vertices at <+/-1,+/-1,+/-1>


Homework Equations



∫∫curlF * ds = ∫F*dr

The Attempt at a Solution



At Z = 1, I broke up the surface into 4 lines, parameterized them and combined everything but it seems like the book may have been trying to get a simpler idea across in the case of a cube. I can't quite figure that out.

Is there a simpler intuitive approach to finding the surface integral of a cube?
 
Physics news on Phys.org
  • #2
Hrm, seems like if I apply Green's Theorem it's faster.
 
  • #3
Bill Nye Tho said:

Homework Statement



F(x,y,z) = xyzi+xyj+x^2yzk

Surface is the top and four sides of cube with vertices at <+/-1,+/-1,+/-1>

Homework Equations



∫∫curlF * ds = ∫F*dr

The Attempt at a Solution



At Z = 1, I broke up the surface into 4 lines, parameterized them and combined everything but it seems like the book may have been trying to get a simpler idea across in the case of a cube. I can't quite figure that out.

Is there a simpler intuitive approach to finding the surface integral of a cube?

The four lines at z = 1 are not the boundary of that open cube. The boundary of that cube surface is the square in the plane z = -1. The interior of that square has the same boundary as your open cube. So the integral over the 5 faces of the cube can be calculated by doing the integral over the square in the z = -1 plane since they share the same boundary. Just one easy surface.
 
  • #4
Bill Nye Tho said:
Hrm, seems like if I apply Green's Theorem it's faster.

I'd second that opinion. And I'd also think about the divergence theorem. What's the flux over the whole cube? That should reduce the problem to just computing the flux over the missing face.
 
  • #5
LCKurtz said:
The four lines at z = 1 are not the boundary of that open cube. The boundary of that cube surface is the square in the plane z = -1. The interior of that square has the same boundary as your open cube. So the integral over the 5 faces of the cube can be calculated by doing the integral over the square in the z = -1 plane since they share the same boundary. Just one easy surface.

It says that S consists of the top and the four sides "(but not the bottom)"

Doesn't that mean, if we ignored the fact that all faces will be equal, that we can not apply Stoke's since there is no surface at the intersecting plane of z=-1?
 
  • #6
Dick said:
I'd second that opinion. And I'd also think about the divergence theorem. What's the flux over the whole cube? That should reduce the problem to just computing the flux over the missing face.

Thanks, seems like calculating the flux works out pretty well here too.
 
  • #7
Bill Nye Tho said:
It says that S consists of the top and the four sides "(but not the bottom)"

Doesn't that mean, if we ignored the fact that all faces will be equal, that we can not apply Stoke's since there is no surface at the intersecting plane of z=-1?

No. I don't know what you mean by the other faces being equal. Think of a hollow cube sitting on the plane z = -1. The edge resting on the plane is the boundary of the cube that you would use for Stokes theorem. The square that edge describes is the missing face sharing the same boundary. Both flux integrals would be equal to the circuit integral around that edge so they are equal. It is similar to Dick's idea.
 
  • #8
LCKurtz said:
No. I don't know what you mean by the other faces being equal.

According to Stokes' the ∫∫s1 curlF * ds = ∫∫s2 curlF * ds.

So I'm basically saying that technically we are given that there is no surface at z = -1 but we can assume that because it's a cube that it would be the same surface as any other side.

Right?

I'm getting a little too technical when it comes to the wording of the question but I'm just trying to really lock in vector analysis.
 
  • #9
Bill Nye Tho said:
According to Stokes' the ∫∫s1 curlF * ds = ∫∫s2 curlF * ds.

So I'm basically saying that technically we are given that there is no surface at z = -1 but we can assume that because it's a cube that it would be the same surface as any other side.

Right?

I'm getting a little too technical when it comes to the wording of the question but I'm just trying to really lock in vector analysis.

I think you are understanding it. The above theorem is what you want to use, where the surface ##S_2## is the flat square ##-1\le x \le 1,\, -1\le y \le 1,\, z=-1##, oriented upward. (Assuming the original cube was directed outward, which you never specified, and should have). So integrate curl F over that instead of using the other five sides. You can do that because ##S_2## and the other five sides share the same boundary.
 
Last edited:
  • #10
Thanks!
 

1. What is Stoke's theorem?

Stoke's theorem is a mathematical theorem that relates the surface integral of a vector field over a closed surface to the line integral of the same vector field around the boundary of that surface.

2. How is Stoke's theorem applied to a cube?

To apply Stoke's theorem to a cube, we first need to define a vector field over the surface of the cube. This can be done by assigning a direction and magnitude to each point on the surface. Then, we calculate the surface integral of the vector field over the cube's six faces and add them together. This is equal to the line integral of the vector field around the boundary of the cube, which is simply the sum of the lengths of the six edges of the cube.

3. What is the significance of applying Stoke's theorem to a cube?

Applying Stoke's theorem to a cube allows us to calculate the circulation of a vector field around the cube's boundary based on the behavior of the vector field on the cube's surface. This can help us understand the flow of a fluid or the movement of a particle in a certain region.

4. Are there any limitations to using Stoke's theorem on a cube?

One limitation of using Stoke's theorem on a cube is that it only applies to closed surfaces, meaning that the cube cannot have any holes or openings. It also assumes that the vector field is continuous and differentiable over the entire surface of the cube.

5. Can Stoke's theorem be applied to other shapes besides a cube?

Yes, Stoke's theorem can be applied to any closed surface, not just a cube. This includes spheres, cylinders, and more complex shapes. However, the calculation may become more complex as the shape becomes more irregular.

Similar threads

  • Calculus and Beyond Homework Help
Replies
26
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
608
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
21
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
818
Replies
3
Views
2K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top