Net Outward Flux of Vector Field Across Surface Solid

  • Thread starter myfunkymaths
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In summary, the conversation discusses calculating the net outward flux of a vector field across a solid surface bounded by a cylinder and two planes. The approach taken was to use a triple integral of the divergence of the vector field, resulting in an answer of 3pi + (3pi/2). However, it was pointed out that the integrand should be 3r^2 instead of 3r^3. From there, it was suggested to use polar coordinates and multiply by r, resulting in a correct answer of 3pi.
  • #1

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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  • #2
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.
 
  • #3
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?
 
  • #4
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
 
  • #5
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.
 
  • #6
ok, thanks for the comments.
 

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