Exercise · ZW-SURFACE-INTEGRAL-001 · ★☆☆ Basic
For \(\mathbf r(u,v)=\langle u,\,v,\,u+v\rangle\) with \(0\le u,v\le1\), the surface area is:
Choose one option.
Show answer
- C. \(\sqrt3\)
Explanation
\(\mathbf r_u\times\mathbf r_v=\langle-1,-1,1\rangle\) has constant length \(\sqrt3\) and \(D\) has area \(1\): \(\iint_D\sqrt3\,dA=\sqrt3\).
Review the concept: Surface integral
The surface integral of a function \(f\) over a surface \(S\), parametrized by \(\mathbf r\) over a region \(D\), adds up the values of \(f\) weighted by the area of each piece of \(S\).
\[\iint_S f\,dS=\iint_D f(\mathbf r(u,v))\,\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\]
More practice problems: surface integral
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