Question

Find the flux of b = 2xyax + z2ya.y + 3x2y'1z over the surface defined by z = 1, 0 < x < 1) 0 < y < 2. 3

Answer 1

I'm assuming
`\vec b` is the vector field I've suggested in my comment,

`\vec b=(2xy,z^2y,3x^2y)`

Parameterize the given surface - call it
`\mathcal S` - by

`\vec r(u,v)=(u,v,1)`

with
`0\le u\le1` and
`0\le v\le2`. The flux is given by the surface integral

`\displaystyle\iint_(\mathcal S)\vec b\cdot\mathrm d\vec S=\iint_(\mathcal S)\vec b\cdot\vec n\,\mathrm dS`

where the surface element is

`\vec n\,\mathrm dS=(\vec r_u*\vec r_v)/(\|\vec r_u*\vec r_v\|)\|\vec r_u*\vec r_v\|\,\mathrm du\,\mathrm dv=(\vec r_u*\vec r_v)\,\mathrm du\,\mathrm dv`

(or use
`\vec r_v*\vec r_u`, depending on the orientation of the surface)

We have

`\vec r_v*\vec r_u=(0,0,1)`

`\vec b=(2uv,v,3u^2v)`

so the surface integral reduces to

`\displaystyle\iint_(\mathcal S)\vec b\cdot\mathrm d\vec S=\int_(u=0)^(u=1)\int_(v=0)^(v=2)3u^2v\,\mathrm dv\,\mathrm du=2`

(or possibly -2, again depending on the orientation of
`\mathcal S`)