Need help finding flux across a surface

In summary, the problem is to find the upward flux across the surface z = x for the vector field F=(y,-x,0). The solution involves parametrizing the surface and finding the normal vector using the fundamental vector product. The final step is to find the limits of integration and evaluate the double integral of the vector field dot the normal vector.
  • #1
nprince
3
0

Homework Statement



Find the upward flux across the surface z = x of the vector field F=(y,-x,0).

Homework Equations



∫∫(F[itex]\bullet[/itex]n)ds

The Attempt at a Solution



I know what to do once I parametrize z = x but I am stuck on the parametrization. Would it just be: r(x,y) = (zi + j + xk) ?
 
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  • #2
or could I say f(x,y,z) = z - x

so the normal vector would be : [itex]\nabla[/itex]f / ||[itex]\nabla[/itex]f||

= -i + k / [itex]\sqrt{}[/itex]-12+12
 
  • #3
The first thing you are going to have to do is put bounds on your surface. You don't want to integrate across the entire infinite plane, do you?

As far as the parameterization of the surface is concerned, it is r(x, y)= xi+ yj+ xk. You are missing the "y" but I suspect that was a typo.

The best way to find n, or better, [itex]\vec{n} dS= d\vec{S}[/itex], is to find the "fundamental vector product" of the surface. That is the cross product [itex]\vec{r}_x\times\vec{r}_y[/itex].

[tex]d\vec{S}= (\vec{r}_x\times\vec{r}_y)dxdy[/tex]

I personally don't like the notation [itex]\vec{n}dS[/itex] because if you take that literally, as you have here, you have to divide by the length of the normal vector to find [itex]\vec{n}[/itex] and then multiply by it to find dS- and, of course, those will cancel!
 
  • #4
So then it's just the double integral of the vector field dot that cross product and I just need to find the limits of integration. Thanks!
 

1. What is flux and why is it important?

Flux is the amount of flow per unit area through a surface. It is important because it helps us understand the rate at which a certain quantity, such as heat or electricity, is passing through a surface.

2. How is flux measured?

Flux is measured in units of flow per unit area, such as watts per square meter or coulombs per square meter. It can be calculated by taking the dot product of the flow vector and the surface normal vector.

3. Can flux be negative?

Yes, flux can be positive or negative depending on the direction of the flow. A positive flux indicates flow in the same direction as the surface normal, while a negative flux indicates flow in the opposite direction.

4. What factors affect the flux across a surface?

The flux across a surface is affected by the magnitude and direction of the flow, the orientation and area of the surface, and the properties of the medium through which the flow is passing.

5. How can I find the flux across a surface?

To find the flux across a surface, you can use the formula flux = flow x surface area x cos(theta), where theta is the angle between the flow vector and the surface normal vector. Alternatively, you can use Gauss's law, which states that the flux through a closed surface is equal to the enclosed charge or current divided by the permittivity or permeability of the medium.

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