Question

Evaluate the surface integral. s x ds, s is the part of the plane 6x + 3y + z = 6 that lies in the first octant.

Answer 1

The plane has intercepts at (1, 0, 0), (0, 2, 0), and (0, 0, 6), so we can parameterize the surface $\mathcal S$ by

$\mathbf r(u,v)=((1,0,0)(1-u)+(0,2,0)u)(1-v)+(0,0,6)v$
$\mathbf r(u,v)=((1-u)(1-v),2u(1-v),6v)$

where $0\le u\le1$ and $0\le v\le1$. Now

$\|\mathbf r_u\times\mathbf r_v\|=2\sqrt{46}(1-v)$

so the surface integral reduces to

$\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{46}\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-u)(1-v)^2\,\mathrm du\,\mathrm dv=\frac{\sqrt{46}}3$