Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}x{\rm{d}}y = 82$
Work Step by Step
We have the domain ${\cal R} = \left[ {1,2} \right] \times \left[ {0,6} \right]$ in the $uv$-plane.
Using the map $G\left( {u,v} \right) = \left( {{u^2},u + v} \right)$, we evaluate the Jacobian of $G$:
${\rm{Jac}}\left( G \right) = \left| {\begin{array}{*{20}{c}}
{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\
{\frac{{\partial y}}{{\partial u}}}&{\frac{{\partial y}}{{\partial v}}}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{2u}&0\\
1&1
\end{array}} \right| = 2u$
Write $f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right) = y\left( {u,v} \right) = u + v$.
Using the General Change of Variables Formula, we evaluate:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {x,y} \right){\rm{d}}x{\rm{d}}y = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal R}^{} f\left( {x\left( {u,v} \right),y\left( {u,v} \right)} \right)\left| {Jac\left( G \right)} \right|{\rm{d}}u{\rm{d}}v$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}x{\rm{d}}y = 2\mathop \smallint \limits_{v = 0}^6 \mathop \smallint \limits_{u = 1}^2 u\left( {u + v} \right){\rm{d}}u{\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 0}^6 \mathop \smallint \limits_{u = 1}^2 \left( {{u^2} + uv} \right){\rm{d}}u{\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 0}^6 \left( {\left( {\frac{1}{3}{u^3} + \frac{1}{2}{u^2}v} \right)|_1^2} \right){\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 0}^6 \left( {\frac{8}{3} + 2v - \frac{1}{3} - \frac{v}{2}} \right){\rm{d}}v$
$ = 2\mathop \smallint \limits_{v = 0}^6 \left( {\frac{7}{3} + \frac{3}{2}v} \right){\rm{d}}v$
$ = 2\left( {\left( {\frac{7}{3}v + \frac{3}{4}{v^2}} \right)|_0^6} \right)$
$ = 2\left( {\frac{{42}}{3} + 27} \right) = 82$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}x{\rm{d}}y = 82$.
