# Net Outward Flux of Vector Field Across Surface Solid

• myfunkymaths
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.

## 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

## 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

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

## 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.