Evaluate the surface integral. s x ds, s is the triangular region with vertices (1, 0, 0), (0, −2, 0), and (0, 0, 10).

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01-06-2018
Mathematics

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Evaluate the surface integral. s x ds, s is the triangular region with vertices (1, 0, 0), (0, −2, 0), and (0, 0, 10).

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LammettHash LammettHash · Answer 1 · 2018-06-01T18:45:23+02:00

Parameterize the region [tex]\mathcal S[/tex] by the vector-valued function,

[tex]\mathbf s(u,v)=((1-u)(1-v),-2u(1-v),10v)[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. Then the surface element is

[tex]\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=6\sqrt{14}(1-v)\,\mathrm du\,\mathrm dv[/tex]

So the surface integral is

[tex]\displaystyle\iint_{\mathcal S}x\,\mathrm dS=6\sqrt{14}\int_{v=0}^{v=1}\int_{u=0}^{u=1}\underbrace{(1-u)(1-v)}_x(1-v)\,\mathrm du\,\mathrm dv=\sqrt{14}[/tex]