Question

Evaluate the surface integral. ∫∫s xy ds s is the triangular region with vertices (1, 0, 0), (0, 6, 0), (0, 0, 6)

Answer 1

Parameterize the surface $\mathcal S$ by

$\mathbf s(u,v)=(\langle1,0,0\rangle(1-u)+\langle0,6,0\rangle u)(1-v)+\langle0,0,6\rangle v$
$\mathbf s(u,v)=\langle(1-u)(1-v),6u(1-v),6v\rangle$

with $0\le u\le1$ and $0\le v\le1$, which has surface element

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=6\sqrt{38}(1-v)\,\mathrm du\,\mathrm dv$

Then the surface integral becomes

$\displaystyle\iint_{\mathcal S}xy\,\mathrm dS=\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-u)(1-v)(6u(1-v))(6\sqrt{38}(1-v))\,\mathrm dv\,\mathrm du=3\sqrt{\dfrac{19}2}$