I have been given a vector field $F(x,y,z)= (e^x+xyz^2,e^y+x^2+z^2, e^z+xz-3y^2)$ and $V$ the solid included between $y=1-x^2-z^2$ and the planes $z=0$, $y=0$.
I have determined $G=\operatorname{curl}F= (-2 (3 y + z), (-1 + 2 x y) z, -x (-2 + z^2))$ and I have to compute the following integrals $$\int_\Sigma G(x,y,z)\cdot \mathrm d\sigma(x,y,z)\quad\text{ and }\quad \int_\sigma G(x,y,z)\cdot \mathrm d\sigma(x,y,z)$$ where $\Sigma$ is the orientated boundary of $V$ and $\sigma\subset \Sigma$ is the surface of equation $$y=1-x^2-z^2, \quad y\ge0,\, z\ge0$$ orientated with $\mathbf n=(n_1,n_2, n_3)$ such that $n_3\ge 0$
My thoughts
Divergence theorem came to mind, so, since $G=\operatorname{curl}F$, $$\nabla \cdot G =\nabla \cdot \nabla \times F=0$$ Then both integrals are equal to zero?
$\Sigma$ is the boundary of $V$, and $V$ is bounded by the paraboloid and the planes, $\nu$ and $\pi$.
$\sigma$ is just the chunk of paraboloid where $y\ge0, \, z\ge0$ so it doesn't have the two planes. There is no intersection with the planes.
So $\Sigma=\sigma +(\nu+\pi)$ and then $$\int_\sigma=-\left(\int_\nu+\int_\pi\right)$$ So I have $$\nu:\int_{-1}^1-2x\left(\int_{0}^{1-x^2} \mathrm d y\right) \mathrm dx\quad\text{ and }\pi:\int_{-1}^1\left(\int_0^{\sqrt[]{1-x^2}}-z \mathrm d z\right)\mathrm d x$$ The first integral is 0, the second one is $-2/3$
Your first flux integral is zero because, yes, the integral of $G=\nabla \times F$ through the surface of the solid is equal to the triple integral of $\nabla \cdot G = \nabla \cdot (\nabla \times F) = 0$ over the solid region.
As for the second integral, if I understand your description correctly, $\sigma$ is just part of the boundary of the solid. It's missing the chuck of the $yz$-plane: $y=0$ and $x^2+z^2 \leq 1$ with $z \geq 0$ which caps off the solid region. This means it is not the surface of a solid and so the divergence theorem does not apply.
However, you do have that the boundary of the solid $\Sigma = \sigma - \pi - \nu$ where $\pi$ is $y=0$ such that $x^2+z^2 \leq 1$, $z \geq 0$ oriented in the positive $y$-axis direction and $\nu $ is the part of the plane $z=0$ below the paraboloid. Thus since the integral through $\Sigma = \sigma - \pi -\nu$ is $0$, you have that the integrals through both $\sigma$ and $\pi+\nu$ are the same (you can "swap surfaces").
The integral through $\pi$ is much easier to compute since on $\pi$ ($y=0$) we have $G = (-2z, -z, 2x-xz^2)$. You then have the integral of $-z$ over the half disk $x^2+z^2 \leq 1$ and $z \geq 0$ to compute.
The integral through $\nu $ doesn't look to bad either.