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