Find the surface area of the solid generated by revolving the curve represented by $y= \sqrt{x},$ from $x=0$ to $x=1$, about the $x$-axis.

Solution We begin with the graph of $y=\sqrt{x} ,$ $0\leq x\leq 1,$ and revolve it about the $x$-axis, as shown in Figure 23.

Figure 23

We use formula (3) with $f( x) = \sqrt{x}$ and $f’ ( x) =\dfrac{1}{2 \sqrt{x}}.$ The surface area $S$ we seek is

The surface area of the solid of revolution is $\dfrac{\pi }{6}(5 \sqrt{5} -1)\approx 5.330$ square units.