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