= Solution
{c}
Parametrize the <function graph> by $\mathbf R(x,y)=(x,y,f(x,y))$. Its two tangent <vectors> are $\mathbf R_x=(1,0,f_x)$ and $\mathbf R_y=(0,1,f_y)$. A small parameter <rectangle> maps, to first order, to a tangent <parallelogram> with <area> $|\mathbf R_x\times\mathbf R_y|\,dx\,dy$. Since
$$
\mathbf R_x\times\mathbf R_y=(-f_x,-f_y,1),
$$
the <surface area of a graph> is
$$
\boxed{A=\iint_S\sqrt{1+f_x^2+f_y^2}\,dx\,dy.}
$$
For example, $f\in C^1$ ensures a <regular surface> chart; more general domains can be handled by the corresponding convergent <improper integral>.
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