Answer
Please see the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^{ - 1}}{\rm{d}}x{\rm{d}}y = 1$
Work Step by Step
We have the region ${\cal D}$, bounded by $y = {x^2}$, $y = \frac{1}{2}{x^2}$, and $y=x$.
Write the mapping $G\left( {u,v} \right) = \left( {uv,{u^2}} \right)$. Let ${{\cal D}_0}$ denote the domain in the $uv$-plane such that ${\cal D} = G\left( {{{\cal D}_0}} \right)$.
Step 1. Find the domain ${{\cal D}_0}$
a. Find the boundary corresponding to $y = {x^2}$ in the $uv$-plane
Using $x = uv$, $y = {u^2}$ we get
${u^2} = {u^2}{v^2}$, ${\ \ \ \ \ }$ $v=1$
So, the image of the horizontal line $v=1$ is the curve $y = {x^2}$.
b. Find the boundary corresponding to $y = \frac{1}{2}{x^2}$ in the $uv$-plane
Using $x = uv$, $y = {u^2}$ we get
${u^2} = \frac{1}{2}{u^2}{v^2}$, ${\ \ \ \ \ }$ $v = \sqrt 2 $
So, the image of the horizontal line $v = \sqrt 2 $ is the curve $y = \frac{1}{2}{x^2}$.
c. Find the boundary corresponding to $y=x$ in the $uv$-plane
Using $x = uv$, $y = {u^2}$ we get ${u^2} = uv$ or $u=v$.
Thus, ${{\cal D}_0}$ can be described as a horizontally simple region in the $uv$-plane given by
${{\cal D}_0} = \left\{ {\left( {u,v} \right)|1 \le v \le \sqrt 2 ,0 \le u \le v} \right\}$
Step 2. Evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
v&u\\
{2u}&0
\end{array}} \right| = - 2{u^2}$
Write
$f\left( {x,y} \right) = {y^{ - 1}}$
$f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = {u^{ - 2}}$
Using the Change of Variables Formula, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{{\cal D}_0}}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^{ - 1}}{\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{v = 1}^{\sqrt 2 } \mathop \smallint \limits_{u = 0}^v {u^{ - 2}}\left| { - 2{u^2}} \right|{\rm{d}}u{\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 1}^{\sqrt 2 } \mathop \smallint \limits_{u = 0}^v {\rm{d}}u{\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 1}^{\sqrt 2 } \left( {u|_0^v} \right){\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 1}^{\sqrt 2 } v{\rm{d}}v = \left( {{v^2}|_1^{\sqrt 2 }} \right) = 1$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^{ - 1}}{\rm{d}}x{\rm{d}}y = 1$.
