Question

Evaluate the surface integral. s x ds, s is the triangular region with vertices (1, 0, 0), (0, −2, 0), and (0, 0, 10).

Answer 1

Parameterize the region $\mathcal S$ by the vector-valued function,

$\mathbf s(u,v)=((1-u)(1-v),-2u(1-v),10v)$

with $0\le u\le1$ and $0\le v\le1$. Then the surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=6\sqrt{14}(1-v)\,\mathrm du\,\mathrm dv$

So the surface integral is

$\displaystyle\iint_{\mathcal S}x\,\mathrm dS=6\sqrt{14}\int_{v=0}^{v=1}\int_{u=0}^{u=1}\underbrace{(1-u)(1-v)}_x(1-v)\,\mathrm du\,\mathrm dv=\sqrt{14}$