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

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

Answer 2

The surface integral for the given triangular region would be:

$\frac{\sqrt{181}}{3}$

In order to determine the surface integral, we will find the perimeter of the surface:-

r(s, t) $= (1 - s)(0,0,12) + s((1-t)(1,0,0) + t(0, -2, 0))$

∵ $r(s, t) = (x(s, t), y(s, t), z(s, t)$

= $( s - st, -2st, 12 - 12s)$

Now, we have

|| $r_{s}$ × $r_{t}$|| = $||($$1-t, -2t, -12)$ × ($-s, -2s, 0)||$

= $2\sqrt{181}s$

After solving the integral, we get the surface integral:

=  $\frac{\sqrt{181}}{3}$

Thus, $\frac{\sqrt{181}}{3}$ is the correct answer.

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