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

$\vec n\,\mathrm dS=\dfrac{\vec r_u\times\vec r_v}{\|\vec r_u\times\vec r_v\|}\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv=(\vec r_u\times\vec r_v)\,\mathrm du\,\mathrm dv$

(or use $\vec r_v\times\vec r_u$, depending on the orientation of the surface)

We have

$\vec r_v\times\vec r_u=(0,0,1)$

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

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 $\mathcal S$)