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