Answer
$\mathop \smallint \limits_0^8 \mathop \smallint \limits_1^2 \frac{{x{\rm{d}}x{\rm{d}}y}}{{\sqrt {{x^2} + y} }} \simeq 5.04615$
Work Step by Step
Write $\mathop \smallint \limits_0^8 \mathop \smallint \limits_1^2 \frac{{x{\rm{d}}x{\rm{d}}y}}{{\sqrt {{x^2} + y} }} = \mathop \smallint \limits_{y = 0}^8 \left( {\mathop \smallint \limits_{x = 1}^2 \frac{{x{\rm{d}}x}}{{\sqrt {{x^2} + y} }}} \right){\rm{d}}y$.
Step 1. We hold $y$ constant and evaluate the inner integral with respect to $x$
The antiderivative of $\frac{x}{{\sqrt {{x^2} + y} }}$ with respect to $x$ is $\sqrt {{x^2} + y} $ because
$\frac{\partial }{{\partial x}}\left( {\sqrt {{x^2} + y} } \right) = \frac{x}{{\sqrt {{x^2} + y} }}$
So,
$S\left( y \right) = \mathop \smallint \limits_{x = 1}^2 \frac{{x{\rm{d}}x}}{{\sqrt {{x^2} + y} }} = \sqrt {{x^2} + y} |_1^2 = \sqrt {4 + y} - \sqrt {1 + y} $
Step 2. Integrate $S\left( y \right)$ with respect to $y$
Thus,
$\mathop \smallint \limits_0^8 \mathop \smallint \limits_1^2 \frac{{x{\rm{d}}x{\rm{d}}y}}{{\sqrt {{x^2} + y} }} = \mathop \smallint \limits_{y = 0}^8 \left( {\sqrt {4 + y} - \sqrt {1 + y} } \right){\rm{d}}y$
$ = \left( {\frac{2}{3}{{\left( {4 + y} \right)}^{3/2}} - \frac{2}{3}{{\left( {1 + y} \right)}^{3/2}}} \right)|_0^8$
$ = \left( {\frac{2}{3}\cdot24\sqrt 3 - \frac{2}{3}\cdot27} \right) - \left( {\frac{2}{3}\cdot8 - \frac{2}{3}\cdot1} \right)$
$ = 16\sqrt 3 - 18 - \frac{{16}}{3} + \frac{2}{3}$
$ = 16\sqrt 3 - \frac{{68}}{3} \simeq 5.04615$
