Answer
$24$
Work Step by Step
$\vec{r} (x,y) =x i+y j+\sqrt {4-y^2} k$
$\implies \vec{r_x} \times \vec{r_y}=\dfrac{y j }{\sqrt {4-y^2}} + k$
$\implies |\vec{r_x} \times \vec{r_y}| =\sqrt {\dfrac{y^2 }{4-y^2}+1}=\dfrac{2}{\sqrt {4-y^2}}$
Now, $\iint_{S} 6 (x,y,z) \ d \theta=
\int_{1}^2 \int_{0}^3 6 (x,y,z) \ d \theta $
and $\int_{1}^2 \int_{0}^3 6 (x,y,z) \ d \theta =\int_{1}^2 \int_{-2}^{2} \sqrt {4-y^2} (\dfrac{2}{\sqrt {4-y^2}}) \ dy \ dx$
Now, we evaluate the integral
$\int_{1}^2 \int_{-2}^{2} \sqrt {4-y^2} (\dfrac{2}{\sqrt {4-y^2}}) \ dy \ dx=24$
