Need help evaluating a surface integral

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We are looking to find

$\iint_S \mathbf F \bullet d\mathbf S$ where $\mathbf F$ $=$ $\langle 0 , y , -z \rangle$ and our surface $\mathbf S$ is the paraboloid $y = x^2 + z^2$ for $0 \le y \le 1$ and the disk $x^2 + z^2 = 1$ , $y

First i parameterized our first surface, $\mathbf S_1$ , $y = x^2 + z^2$ for $0 \le y \le 1$

Using polar coordinates we get

$y=r^2$ , $x=rcos(\theta)$ , $z=rsin(\theta)$

Then our parameterization as a vector function is

$\mathbf a(r,\theta)= \langle rcos(\theta) , r^2 , rsin(\theta) \rangle$

Then we find the normal vector function by taking the cross product between $\mathbf a_r$ and $\mathbf a_\theta$

$\mathbf a_r \mathbf \times \mathbf a_\theta = \langle -2r^2cos(\theta) , r , -2r^2sin(\theta) \rangle$

and we have that

$\mathbf F(\mathbf a) = \langle 0 , r^2, -rsin(\theta) \rangle$

thus

$\mathbf F(\mathbf a) \bullet (\mathbf a_r \mathbf \times \mathbf a_\theta) = r^3 + 2r^3sin^2(\theta)$

Thus the flux over $\mathbf S_1$ is

$\iint \mathbf S_1 (r^3+r^3sin^2(\theta)) d\mathbf A$ for $0 \le r \le 1$ and $0 \le \theta \le 2(pi)$

which i evaluated to $\frac {5(pi)}{4}$

However the the flux of $\mathbf F$ over $\mathbf S_1$ is incorrect, so my question is where did i go wrong? is the parameterization correct?