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