Question

Evaluate the surface integral. s x ds, s is the triangular region with vertices (1, 0, 0), (0, −2, 0), and (0, 0, 12).

Answer 1

Parameterize the surface by

$\mathbf r(s,t)=(1-s)(0,0,12)+s\bigg((1-t)(1,0,0)+t(0,-2,0)\bigg)$
$\implies\mathbf r(s,t)=(x(s,t),y(s,t),z(s,t))=(s-st,-2st,12-12s)$

where $(s,t)\in(0,1)\times(0,1)$. We have

$\|\mathbf r_s\times\mathbf r_t\|=\|(1-t,-2t,-12)\times(-s,-2s,0)\|=2\sqrt{181}s$

The surface integral is then

$\displaystyle\iint_Sx\,\mathrm dS=2\sqrt{181}\int_{s=0}^{s=1}\int_{t=0}^{t=1}s(s-st)\,\mathrm dt\,\mathrm ds$
$=\displaystyle2\sqrt{181}\left(\int_{s=0}^{s=1}s^2\,\mathrm ds\right)\left(\int_{t=0}^{t=1}(1-t)\,\mathrm dt\right)$
$=\dfrac{\sqrt{181}}3$