= Solution
{c}
A <parametrization> of the <surface of revolution> is
$$
\mathbf R(x,\phi)=(x,f(x)\cos\phi,f(x)\sin\phi),\qquad a<x<b,\quad0\leq\phi<2\pi.
$$
Its tangent <vectors> are orthogonal, with <norms> $\sqrt{1+f'(x)^2}$ and $|f(x)|$. Thus
$$
\boxed{A=2\pi\int_a^b|f(x)|\sqrt{1+f'(x)^2}\,dx.}
$$
The <absolute value> is necessary if the <function graph> passes below the axis. The formula gives the curved <surface area>; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.
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