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