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We have a vector field $\mathbf{F} = 0.4 (-\mathbf{i}-\mathbf{j}+\mathbf{k})$ and want to calculate the flux through a window which consists of joining, in turn, the following points with straight line segments: $(1,1,0),(1,0,0),(0,0,0),(0,1,1),(1,1,1),(1,1,0)$ Which gives a weird looking wedge shape pictured below: enter image description here

Since this loop does not have an easily definable normal vector which we can use to find the flux using $\int_S \mathbf{F} \cdot \mathbf{dS} = \int_S \mathbf{F} \cdot \mathbf{\hat n}\,dS$, I was unsure how to proceed. I know we basically want to find the cross section from the vector fields point of view, so to speak, and when viewing the loop along the direction of the force, we see: enter image description here

And I could find this cross section from here, but am looking for a method that doesn't involve looking at the physical cross section and summing triangle areas. Is there a way to do this "normally" (without heavily relying on geometry)?

Poo2uhaha
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  • You can divide it into two, one that is in $z$ plane (horizontal) and the part which is vertical. – Math Lover Feb 17 '21 at 21:58
  • @MathLover you're sure this won't double count some regions from the vector field's perspective? A small elucidation would be very helpful! – Poo2uhaha Feb 17 '21 at 22:00
  • Actually I misread one of the coordinates. So it is not just one horizontal and one vertical part. – Math Lover Feb 17 '21 at 22:15
  • If you are familiar with Stokes Theorem then this post might have what you're looking for in the second formula $\big($after replacing $B$ with $F\big);$ – greg Feb 19 '21 at 17:04
  • @greg, I realised it is a trivial problem if you consider three vector fields, each corresponding to the components in the $\mathbf{i ,j, k}$ direction of our original field, perform integrals for each and then add your answers. – Poo2uhaha Feb 20 '21 at 14:29
  • Not sure what you're talking about with adding three vector fields. Please read the linked post. It describes how to calculate the vector area (call it $S$) of a surface bounded by a closed-loop path. Then the flux through the surface is $,\big|F\cdot S\big|.;$ The nice thing is $S$ doesn't change, so you can calculate the flux produced by any other vector field, e.g. $G$, as the dot product $,\big|G\cdot S\big|;$ – greg Feb 22 '21 at 22:20

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