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