Answer
$\dfrac{3}{4}$
Work Step by Step
Since, $|J| =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{2u}{v}&-\dfrac{v}{u^2}\\\dfrac{-u^2}{v^2}& \dfrac{1}{u}\end{vmatrix}=v^{-1}$
$\iint_R y^2 dA=\int_1^{2} \int_{1}^{2}(\dfrac{v}{u})^2(v^{-1}) du dv$
and $\iint_R y^2 dA=\int_1^2 v dv \int_{1}^{2} u^{-2} du$
Therefore,
$\iint_R y^2 dA=[\dfrac{v^2}{2}]_1^2[\dfrac{-1}{u}]_1^2=\dfrac{3}{4}$
