Surface Area

Let us take the region $R = x \in [0,4], z \in [0,2], y = f(x,z) = \sqrt{2xz}$ . The required surface area of $R$ can be computed according to:

$\large S = \int \int_{R} \sqrt{1 + f^{2}_{x} + f^{2}_{z}} dA = \int_{0}^{4} \int_{0}^{2} \sqrt{1 + (\sqrt{\frac{z}{2x}})^{2} + (\sqrt{\frac{x}{2z}})^{2}} dz dx = \int_{0}^{4} \int_{0}^{2} \sqrt{ \frac{2xz + x^{2} + z^{2}}{2xz}} dz dx = \int_{0}^{4} \int_{0}^{2} \frac{x + z}{\sqrt{2xz}} dz dx = \boxed{16}.$