Help find the flux through the surface

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Homework Help Overview

The problem involves finding the flux of a vector field A=(2x,-z^2,3xy) through a specified surface defined in cylindrical coordinates. The surface is described by the conditions ρ=2, 0<φ<π/2, and 0

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the appropriateness of using the divergence method and whether the surface is closed. Questions arise about converting the vector field to cylindrical coordinates and the implications of the surface being a cylindrical wedge.

Discussion Status

There is a recognition that the surface is not closed, and some participants clarify that the flux should be calculated over the round surface of the wedge rather than the other faces. Guidance is offered regarding the integration approach, suggesting that integrating in cylindrical form may simplify the process.

Contextual Notes

Participants note the need to integrate over the defined surface only and discuss the implications of the surface's geometry on the integration process.

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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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