Answer
$\dfrac{\pi}{2} [\sqrt 2+\ln (1+\sqrt 2)]$
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, $\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}=\sin v i-\cos v j +u k$
and $|\dfrac{\partial r}{\partial u} \times \dfrac{\partial r}{\partial v}|=\sqrt {(\sin v)^2+(-\cos v)^2+u^2}=\sqrt {1+u^2}$
Therefore, $A(S)=\iint_{D} \sqrt {1+u^2} dA=\int_0^{1} \int_{0}^{\pi} \sqrt {1+u^2} dv du$
or, $= \pi \int_0^1 \sqrt {1+u^2} du$
or, $= \pi [\dfrac{u}{2}\sqrt {u^2+1}+\dfrac{1}{2} \ln (u+\sqrt {u^2+1})]_0^{1}$
This implies that,
$A(S)=\dfrac{\pi}{2} [\sqrt 2+\ln (1+\sqrt 2)]$
