Answer
$\dfrac{11 \pi}{12}$
Work Step by Step
$\vec{r} (r,\theta) =r \cos \theta i+r \sin \theta j+(1-r^2) k$
$\implies \vec{r_{r}} \times \vec{r_{\theta}}=-r \sin \theta+r \cos \theta+0$
$\implies |\vec{r_{r}} \times \vec{r_{\theta}}| =\sqrt {4r^4+r^2}=r \sqrt {4r^2 +1}$
$I=\int_{0}^{ 2 \pi} \int_{0}^{1} (\cos^2 \theta )(4r^5 +r^3) \ dr \ d \theta \\= \int_{0}^{2 \pi} [(\dfrac{2r^6}{3})+\dfrac{r^4 \cos^2 \theta}{4} ]_0^1 \ dx \\=\int_{0}^{2 \pi} \dfrac{11}{12} \times \cos^2 \theta d \theta$
Now, we will use a calculator to evaluate the integral:
$I=\int_{0}^{ 2 \pi} \int_{0}^{1} (\cos^2 \theta )(4r^5 +r^3) \ dr \ d \theta=\dfrac{11 \pi}{12}$
