Answer
$4$
Work Step by Step
We have $A(S)= \iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|$
and $\iint_{D} dA$ is the area of the region $D$
Now, $\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=v^2 i-2uv j +2u^2 k$
and $|\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|=\sqrt {( v)^2+(-2u v)^2+(2u^2)^2}=v^2+2u^2$
Therefore, $A(S)=\iint_{D} v^2+2u^2 dA=\int_0^{1} \int_{0}^{2} v^2+2u^2 dv du$
or, $= \pi \int_0^1 [\dfrac{v^3}{3}+2vu^2]_0^2 du$
or, $= \int_0^{1} \dfrac{8}{3}+4u^2 du$
This implies that,
$A(S)=[\dfrac{8u}{3}+\dfrac{4u^3}{3}]_0^1=4$
