Answer
$\dfrac{11 \pi}{12}$
Work Step by Step
Consider $\vec{r} (r,\theta) =r \cos \theta i+r \sin \theta j+(1-r^2) k$
So, $\vec{r_{r}} \times \vec{r_{\theta}}=-r \sin \theta+r \cos \theta+0$
and $|\vec{r_{r}} \times \vec{r_{\theta}}| =\sqrt {4r^4+r^2}=r \sqrt {4r^2 +1}$
Now, $S=\int_{0}^{ 2 \pi} \int_{0}^{1} (\cos^2 \theta )(4r^5 +r^3) \ dr \ d \theta $
or, $= \int_{0}^{2 \pi} [(2/3) r^6+\dfrac{1}{4} r^4 \cos^2 \theta ]_0^1 \ dx$
or, $=\int_{0}^{2 \pi} \dfrac{11}{12} \cos^2 \theta d \theta$
By using a calculator, we evaluate the integral as:
$\int_{0}^{ 2 \pi} \int_{0}^{1} (\cos^2 \theta )(4r^5 +r^3) \ dr \ d \theta=\dfrac{11 \pi}{12}$
