Net Outward Flux of Vector Field Across Surface Solid

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SUMMARY

The net outward flux of the vector field F(x,y,z) = 3xy² i + x cos(z) j + z³ k across the surface solid bounded by the cylinder y² + z² = 1 and the planes x = -1 and x = 2 was calculated using the divergence theorem. The divergence of F is 3r², leading to a triple integral that evaluates to 3π + (3π/2). The correct approach involves ensuring the integrand accounts for the Jacobian when converting to polar coordinates.

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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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