Net Outward Flux of Vector Field Across Surface Solid

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Homework Help Overview

The problem involves calculating the net outward flux of a vector field defined as F(x,y,z)=3xy^2 i +xcosz j +z^3k across a solid surface bounded by a cylinder and two planes.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of a triple integral of the divergence of the vector field, with one participant suggesting the use of an online tool for evaluating integrals. There are questions about the correctness of the integrand and the conversion to polar coordinates.

Discussion Status

Some participants have provided feedback on the approach taken, questioning specific steps in the integration process and discussing the factors involved in converting to polar coordinates. There is an ongoing exploration of the method without a clear consensus on the correctness of the initial answer.

Contextual Notes

Participants are navigating the complexities of integrating vector fields and ensuring proper conversion between coordinate systems, with some assumptions about the setup being questioned.

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Homework Statement



calculate net outward flux of vector field F(x,y,z)=3xy^2 i +xcosz j +z^3k, across surface solid bounded by cylinder y^2 +z^2 =1
planes x=-1, x=2

Homework Equations





The Attempt at a Solution


my attempt was triple integral of divF=> 3*triple integral r^3
ended up getting an answer =3pi + (3pi/2)

is the answer i got correct?

thank you
 
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did not work out the integral, but your approach seemed correct. there's this handy site called wolfram alpha you know...it evaluates integrals and (i haven't checked) might even do flux.
 
myfunkymaths said:

Homework Statement



calculate net outward flux of vector field F(x,y,z)=3xy^2 i +xcosz j +z^3k, across surface solid bounded by cylinder y^2 +z^2 =1
planes x=-1, x=2

Homework Equations





The Attempt at a Solution


my attempt was triple integral of divF=> 3*triple integral r^3
ended up getting an answer =3pi + (3pi/2)

is the answer i got correct?

thank you

No, I don't think so. 3r^2 is ok for an integrand. How did you do the rest?
 
Dick said:
No, I don't think so. 3r^2 is ok for an integrand. How did you do the rest?
You forgot to multiply by r when converting to polar
 
ƒ(x) said:
You forgot to multiply by r when converting to polar

I didn't quite forget. I meant 3r^2 for div(F). Sure, it becomes 3r^3 with the extra factor of r from the area element. Thanks.
 
ok, thanks for the comments.
 

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