Exercise · ZW-SURFACE-INTEGRAL-008 · ★★☆ Intermediate
The area of the part of the plane \(z=x+2y\) over the square \(0\le x,y\le1\) is:
Choose one option.
A \(1\) B \(\sqrt6\) C \(\sqrt3\) D \(\sqrt5\)
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Explanation For \(z=f(x,y)\), \(dS=\sqrt{1+f_x^2+f_y^2}\,dA=\sqrt{1+1+4}\,dA=\sqrt6\,dA\), and the square has area \(1\).
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 \(S\).
\[\iint_S f\,dS=\iint_D f(\mathbf r(u,v))\,\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\]
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