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Let $f(x)>0$ be a continuous differentiable function on $[a,b]$, set up an integral for the surface area generated by revolving the graph of $f$ about the $x$-axis.

$\pi\int_a^b f^2(x)dx$

$2\pi\int_a^b |f(x)|dx$

$2\pi\int_a^b f(x)\sqrt{1+(f'(x))^2}dx$

$\pi(\int_a^b\sqrt{1+(f'(x))^2}dx)^2$

None of the above