Answer
$192$
Work Step by Step
$J(u,v) =\begin{vmatrix} \dfrac{\partial x}{\partial u}&\dfrac{\partial x}{\partial v}\\\dfrac{\partial y}{\partial u}&\dfrac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix} \dfrac{1}{4} \cdot \dfrac{1}{4}-\dfrac{1}{4} \cdot \dfrac{-3}{4}\end{vmatrix}=\dfrac{1}{4}$
Here, we have $\iint_R (4x+8y) dA=\int_0^8 \int_{-4}^{4} (4x+8y) dA$
or, $=\int_0^8 \int_{-4}^{4} [(4) (\dfrac{1}{4}) (u+v)]+[(8)( \dfrac{1}{4}) (v-3u)] du dv $
or, $=\int_0^8 \int_{-4}^{4} [4 \dfrac{1}{4}(u+v)]+[8 \dfrac{1}{4}(v-3u)] du dv$
or, $=\int_0^8 \int_{-4}^{4}(3v-5u) \cdot du dv$
or, $=(\dfrac{1}{4}) \int_0^8[3uv-2.5u^2]_{-4}^4 dv$
or, $=(\dfrac{1}{4})\int_0^8 24 v dv$
Thus, $(\dfrac{1}{4}) [12v^2]_0^8=192$
