Answer
$\dfrac{3}{4}$
Work Step by Step
$Jacobian =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} 2(u/v)&(-v/u^2)\\-u^2/v^2& 1/u \end{vmatrix}=\dfrac{1}{v}$
Now, we have $\iint_R y^2 dA=\int_1^{2} \int_{1}^{2}(\dfrac{v}{u})^2(\dfrac{1}{v}) du dv$
$\implies \iint_R y^2 dA=\int_1^2 v dv \int_{1}^{2} \dfrac{1}{u^2} du$
This gives: $\iint_R y^2 dA=[v^2/2]_1^2[\dfrac{-1}{u}]_1^2$
$\implies \iint_R y^2 dA=\dfrac{3}{4}$
