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alina1380 [7]
3 years ago
6

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

Mathematics
1 answer:
aniked [119]3 years ago
7 0
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
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