Answer
$$4$$
Work Step by Step
$\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=v^2 i-2uv j +2u^2 k$
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, $A(S)=\iint_{D} |\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|\\=\iint_{D} v^2+2u^2 dA \\=\int_0^{1} \int_{0}^{2} v^2+2u^2 dv du\\= \pi \times \int_0^1 [\dfrac{v^3}{3}+2vu^2]_0^2 du \\= \int_0^{1} \dfrac{8}{3}+4u^2 du \\\\=[\dfrac{8u}{3}+\dfrac{4u^3}{3}]_0^1 \\=4$
