Question

Evaluate the surface integral. s xy ds s is the triangular region with vertices (1, 0, 0), (0, 8, 0), (0, 0, 8)

Answer 1

Parameterize $S$ by

$\vec s(u,v)=(1-u)v\,\vec\imath+8uv\,\vec\jmath+8(1-v)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$.

Then the surface element is

$\mathrm dS=\|\vec s_u\times\vec s_v\|\,\mathrm du\,\mathrm dv=8\sqrt{66}\,v\,\mathrm du\,\mathrm dv$

and the surface integral is

$\displaystyle\iint_Sxy\,\mathrm dS=8\sqrt{66}\int_0^1\int_0^18(1-u)uv^3\,\mathrm du\,\mathrm dv=\boxed{8\sqrt{\dfrac{22}3}}$