Answer
$2 \ln 3$
Work Step by Step
Here, we have: $|J| =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{1}{v}&-\dfrac{u}{v^2}\\0& 1\end{vmatrix}=v^{-1}$
$\iint_R xy dA=\int_1^{3} \int_{u^{1/2}}^{(3u)^{1/2}} [uv^{-1}] dv du=\int_1^3u[\ln v]_{u^{1/2}}^{(3u)^{1/2}} du$
and $\iint_R xy dA=\int_1^3u[\ln (3)^{1/2}]du=[\ln 3^{1/2}][\dfrac{u^2}{2}]_1^3$
Hence,
$\iint_R xy dA=(\dfrac{1}{2}) (\ln 3) [\dfrac{9}{2}-\dfrac{1}{2}]=2 \ln (3)$
