Answer
$Area\left( S \right) \simeq 36.177$
Work Step by Step
Using Eq. (14) we compute the surface area of $z = 4 - {y^2}$ for $0 \le y \le 2$ rotated about the $z$-axis:
$Area\left( S \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$
$Area\left( S \right) = 2\pi \mathop \smallint \limits_0^2 y\sqrt {1 + {{\left( { - 2y} \right)}^2}} {\rm{d}}y = 2\pi \mathop \smallint \limits_0^2 y\sqrt {1 + 4{y^2}} {\rm{d}}y$
Write $t = 1 + 4{y^2}$. So, $dt = 8ydy$. The integral becomes
$Area\left( S \right) = \frac{\pi }{4}\mathop \smallint \limits_1^{17} \sqrt t {\rm{d}}t$
$Area\left( S \right) = \frac{\pi }{4}\left( {\frac{2}{3}{t^{3/2}}|_1^{17}} \right) = \frac{\pi }{6}\left( {17\sqrt {17} - 1} \right) \simeq 36.177$