Answer
$3\sqrt 3$
Work Step by Step
Consider $\vec{r} (x,y) =x i+y j+(4-x-y) k$
So, $\vec{r_x} \times \vec{r_y}=i+j+k$
and $|\vec{r_x} \times \vec{r_y}| =\sqrt {1^2+1^2+1^2}=\sqrt 3$
Now, $\iint_{S} F (x,y,z) \ d \theta=\int_{0}^{1} \int_{0}^{1} (4-x-y) \times \sqrt 3 \ dy \ dx$
or, $=\sqrt 3 \int_{0}^{1} [4y-xy-\dfrac{y^2}{2} ]_0^1 \ dx$
or, $=\sqrt 3 \int_{0}^{1} 4(1-0)-x(1-0)-(\dfrac{1}{2}-0) \ dx$
or, $=\sqrt 3 \int_{0}^{1} (\dfrac{7}{2} -x)] \ dx$
By using a calculator, we evaluate the integral as:
$\sqrt 3 \int_{0}^{1} (\dfrac{7}{2} -x)] \ dx= 3\sqrt 3$
