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