Help find the flux through the surface

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To find the flux of the vector field A=(2x,-z^2,3xy) through the specified surface, it is essential to integrate over the defined surface rather than treating it as a closed surface. The discussion clarifies that the surface is a cylindrical wedge, and converting the vector field to cylindrical coordinates simplifies the integration process. Participants agree that using divergence is not necessary since the surface is not closed, and the integration should focus solely on the defined surface area. Ultimately, integrating in cylindrical coordinates will make handling the surface normal and limits of integration more straightforward. The consensus is that this approach will yield the most efficient solution for calculating the flux.
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Homework Statement



Given a vector field A=(2x,-z^2,3xy), find the flux of A through a surface defined by ρ<br /> =2, 0&lt;\phi&lt;\pi/2, 0&lt;z&lt;1

Homework Equations



∇\bulletA?


The Attempt at a Solution



Can I use divergence method here?
This is a closed surface correct? A cylindrical wedge?
Also do I need to convert the vector field to cylindrical form? Or the defined surface to rectangle form?

If I used divergence do I divide my answer by 4 since the wedge is a 1/4 of the cylinder?

Thanks
 
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The surface is not closed.
 
I agree. I read the problem as asking for the flux through the round surface of the wedge and not the other four faces.
 
Thanks for the replies.

This makes a lot more sense now. So knowing this I would integrate over the surfaces separately.

So it appears it would be easier to integrate in cylindrical form correct? So I would want to change the vector field from rectangular to cylindrical?
 
Rombus said:
So knowing this I would integrate over the surfaces separately.

You should only need to integrate over the one surface that is defined by the equalities & inequalities given.

So it appears it would be easier to integrate in cylindrical form correct? So I would want to change the vector field from rectangular to cylindrical?

Yes, that would probably be the easiest way to do it since the surface normal and differential area, and limits of integration will all be much simpler in cylindrical coordinates than in Cartesian coordinates.
 
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