Exercise · ZW-SURFACE-INTEGRAL-002 · ★☆☆ Basic
For \(\mathbf r(u,v)=\langle u,\,v,\,u^2+v^2\rangle\), the area element \(dS\) is:
Choose one option.
Show answer
- A. \(\sqrt{1+4u^2+4v^2}\,du\,dv\)
Explanation
\(\mathbf r_u\times\mathbf r_v=\langle-2u,-2v,1\rangle\), whose length is \(\sqrt{4u^2+4v^2+1}\). Dropping the root or the factor is the typical mistake.
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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