Answer
$Area\left( S \right) = \sqrt 2 \pi {d^2}$
Work Step by Step
Referring to Figure 6, we can consider the cone ${x^2} + {y^2} = {z^2}$ as a surface of revolution as is described in Exercises 42, where in this case $z = g\left( y \right) = y$. Thus, we can parametrize the cone ${x^2} + {y^2} = {z^2}$ for $0 \le y \le d$ as
$G\left( {y,\theta } \right) = \left( {y\cos \theta ,y\sin \theta ,y} \right)$
Using Eq. (14) we compute the surface area of the upper half of the cone:
$Area\left( S \right) = 2\pi \mathop \smallint \limits_c^d y\sqrt {1 + g'{{\left( y \right)}^2}} {\rm{d}}y$
$Area\left( S \right) = 2\pi \mathop \smallint \limits_0^d y\sqrt {1 + {1^2}} {\rm{d}}y = 2\sqrt 2 \pi \mathop \smallint \limits_0^d y{\rm{d}}y$
$Area\left( S \right) = 2\sqrt 2 \pi \left( {\frac{1}{2}{y^2}|_0^d} \right) = \sqrt 2 \pi {d^2}$
