= Solution
Subtract the integrands before integrating: $\mathbf G_2-\mathbf G_1=(0,y^3,0)$. With the upward <vector area element>,
$$
(\mathbf G_2-\mathbf G_1)\cdot d\mathbf S=(x^2y+2y^2z)dx\,dy=(2-2x^2-x^2y)dx\,dy.
$$
This extends continuously to the lower edge, so
$$
\boxed{|I_1-I_2|=\left|\int_0^1\int_0^2(2-2x^2-x^2y)dy\,dx\right|=2.}
$$
The individual <flux integrals> also converge: $\mathbf G_1\cdot d\mathbf S=x(3+2y)dx\,dy$ gives $I_1=5$, and hence $I_2=7$. Reversing the surface orientation changes both signs and leaves the requested magnitude unchanged.
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