= Solution
The <divergence theorem> includes every part of the closed boundary, including the two vertical cut faces. Its volume integrand simplifies because the terms linear in $x$ cancel:
$$
\nabla\cdot\mathbf F=3x^2+(3y^2+2x)-2x=3(x^2+y^2).
$$
In <cylindrical coordinates>, the remaining <integration region> is $0\leq r\leq1$, $0\leq z\leq1-r^2$, $\pi/2\leq\theta\leq2\pi$. With the <Jacobian determinant> $r$, the outward <flux> is
$$
\boxed{\int_{\partial V}\mathbf F\cdot d\mathbf S=\int_{\pi/2}^{2\pi}\int_0^1\int_0^{1-r^2}3r^3\,dz\,dr\,d\theta=\frac{9\pi}{2}\left(\frac14-\frac16\right)=\frac{3\pi}{8}.}
$$
The angular range is three quarters of a full revolution. The volume calculation already accounts for the base and cut faces, so no additional surface terms should be appended.
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