Answer
The surface area is $4\pi {R^2}$
Work Step by Step
A sphere of radius $R$ centered at the origin is generated by rotating the circle parametrized by $c\left( t \right) = \left( {R\cos t,R\sin t} \right)$ about the $x$-axis for $0 \le t \le \pi $.
So, we have
$x\left( t \right) = R\cos t$, ${\ \ \ }$ $x'\left( t \right) = - R\sin t$,
$y\left( t \right) = R\sin t$, ${\ \ \ }$ $y'\left( t \right) = R\cos t$.
Using Eq. (4) of Theorem 3, the surface area of the sphere is
$S = 2\pi \mathop \smallint \limits_0^\pi \left( {R\sin t} \right)\sqrt {{{\left( { - R\sin t} \right)}^2} + {{\left( {R\cos t} \right)}^2}} {\rm{d}}t$
$S = 2\pi {R^2}\mathop \smallint \limits_0^\pi \sin t{\rm{d}}t = 2\pi {R^2}\left( { - \cos t} \right)|_0^\pi = 4\pi {R^2}$.
Note that this the familiar surface area of the sphere of radius $R$.
