Calculating Surface Integral with Stokes' Theorem on a Cube?

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The discussion focuses on using Stokes' theorem to calculate a surface integral for the vector function F = <xy, 2yz, 3zx> over a cube with an open back side. The curl of F is calculated as <-2y, -3z, -x>, and there is confusion about which side of the cube to find the normal vector for. It is clarified that to apply Stokes' theorem, the line integral should be calculated around the boundary of the open surface, specifically the four edges of the back side. Alternatively, if integrating over the surface, one would need to find the unit normal vectors for all five faces and compute each integral separately. The discussion concludes with a better understanding of the application of Stokes' theorem in this context.
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Homework Statement



I have to use stokes' theorem and calculate the surface integral, where the function F = <xy,2yz,3zx> and the surface is the cube bounded by the points (2,0,0), (0,2,0),(0,0,2),(0,2,2),(2,0,2),(2,2,0),(2,2,2). The back side of the cube is open.
[/B]

Homework Equations


The Attempt at a Solution



I found the curlF = <-2y,-3z,-x>.

I know next I have to find a unit normal vector. But exactly which side of the cube am I supposed to find the normal vector to? Or would I HAVE to take the line integrals for the bottom side of the cube?
 
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If I understand correctly, you want to calculate

Int{F.dA}

I think that divergence theorem could be useful

Int{F.dA} = Int{(div F)dV}

where "div F" is gradient operator (Del) dotted with F.

Stokes/curl theorem helps if you need

Int{F.dl}

then you use

Int{F.dl} = Int{(curl F).dA}

and now "curl F" is cross product of Del and F.

Since the back side is open I think that either you need to do each of five sides in turn or maybe do the whole cube with divergence theorem and then subtract the integral over the back side from it.

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reb659 said:

Homework Statement



I have to use stokes' theorem and calculate the surface integral, where the function F = <xy,2yz,3zx> and the surface is the cube bounded by the points (2,0,0), (0,2,0),(0,0,2),(0,2,2),(2,0,2),(2,2,0),(2,2,2). The back side of the cube is open.
[/B]

Homework Equations





The Attempt at a Solution



I found the curlF = <-2y,-3z,-x>.

I know next I have to find a unit normal vector. But exactly which side of the cube am I supposed to find the normal vector to? Or would I HAVE to take the line integrals for the bottom side of the cube?
You don't find the normal vector to any side. To use Stokes theorem you find the line integral around the boundary of the surface. Here that means around the four edges of the missing back, not bottom.

If you did want to do this by integrating over the surface, you would find the unit normal vector of all of the five faces and integrate each one separately.
 
Ahh I get it now. Thanks!
 
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