Answer
We show that the domain ${\cal D}$ is bounded by $x=0$, $y=0$, and ${y^2} = 4 - 4x$.
This is illustrated in the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}x{\rm{d}}y = \frac{{56}}{{45}}$
Work Step by Step
The triangle ${{\cal D}_0} = \left\{ {\left( {u,v} \right)|0 \le v \le u \le 1} \right\}$ is bounded left by $u=0$ and bounded right by $u=1$. Whereas, the lower and upper boundaries are $v=0$ and $v=u$, respectively. Thus, it can be described as a vertically simple region given by
${{\cal D}_0} = \left\{ {\left( {u,v} \right)|0 \le u \le 1,0 \le v \le u} \right\}$
Step 1. Find the image of the boundary $0 \le u \le 1$
a. The image of $u=0$ is $T\left( {0,v} \right) = \left( { - {v^2},0} \right)$
So, $x = - {v^2}$, $y=0$.
Since $x \le 0$, $y=0$, the image of $u=0$ is the negative $x$-axis in the $xy$-plane.
b. The image of $u=1$ is $T\left( {1,v} \right) = \left( {1 - {v^2},2v} \right)$. So, $x = 1 - {v^2}$, $y = 2v$.
Using $y = 2v$, we get $v = \frac{y}{2}$. Substituting $v = \frac{y}{2}$ in $x = 1 - {v^2}$ gives $x = 1 - \frac{{{y^2}}}{4}$.
Thus, the image of $u=1$ is the curve ${y^2} = 4 - 4x$.
We conclude that the image of $0 \le u \le 1$ in ${\cal D}$ are the boundaries $y=0$ and the curve ${y^2} = 4 - 4x$.
Step 2. Find the image of the boundary $0 \le v \le u$
a. The image of $v=0$ is $T\left( {u,0} \right) = \left( {{u^2},0} \right)$.
So, $x = {u^2}$, $y=0$.
Since $x \ge 0$, $y=0$, the image of $v=0$ is the positive $x$-axis in the $xy$-plane.
b. The image of $v=u$ is $T\left( {u,u} \right) = \left( {0,2{u^2}} \right)$.
So, $x=0$, $y = 2{u^2}$.
Since $x=0$ and $y \ge 0$, the image of $v=u$ is the positive $y$-axis.
Hence, the domain ${\cal D}$ is bounded by $x=0$, $y=0$, and ${y^2} = 4 - 4x$.
Step 3. Evaluate the Jacobian of $T\left( {u,v} \right) = \left( {{u^2} - {v^2},2uv} \right)$:
${\rm{Jac}}\left( T \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{2u}&{ - 2v}\\
{2v}&{2u}
\end{array}} \right| = 4{u^2} + 4{v^2}$
Write
$f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} $
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = \sqrt {{{\left( {{u^2} - {v^2}} \right)}^2} + 4{u^2}{v^2}} $
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = \sqrt {{u^4} + 2{u^2}{v^2} + {v^4}} = \sqrt {{{\left( {{u^2} + {v^2}} \right)}^2}} = {u^2} + {v^2}$
Using the Change of Variables Formula, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}x{\rm{d}}y = 4\mathop \smallint \limits_{u = 0}^1 \mathop \smallint \limits_{v = 0}^u {\left( {{u^2} + {v^2}} \right)^2}{\rm{d}}v{\rm{d}}u$
$ = 4\mathop \smallint \limits_{u = 0}^1 \mathop \smallint \limits_{v = 0}^u \left( {{u^4} + 2{u^2}{v^2} + {v^4}} \right){\rm{d}}v{\rm{d}}u$
$ = 4\mathop \smallint \limits_{u = 0}^1 \left( {\left( {{u^4}v + \frac{2}{3}{u^2}{v^3} + \frac{1}{5}{v^5}} \right)|_0^u} \right){\rm{d}}u$
$ = 4\mathop \smallint \limits_{u = 0}^1 \left( {{u^5} + \frac{2}{3}{u^5} + \frac{1}{5}{u^5}} \right){\rm{d}}u$
$ = 4\mathop \smallint \limits_{u = 0}^1 \left( {\frac{{28}}{{15}}{u^5}} \right){\rm{d}}u = \frac{{56}}{{45}}\left( {{u^6}|_0^1} \right) = \frac{{56}}{{45}}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \sqrt {{x^2} + {y^2}} {\rm{d}}x{\rm{d}}y = \frac{{56}}{{45}}$.
