Consider the sets $$V_1 = \{ (x,y,z) : x^2 + y^2 + (z-1)^2 \leq 4 \} \ \& \ V_2 = \{ (x,y,z): x^2 + y^2 + (z + 1)^2 \leq 4 \}$$ Let $V = V_1 \cup V_2$, and the vector field $F(x,y,z) = (x + xy, yz, xz)$. Calculate $$\int_{\partial V} \mathbf{F}\cdot d\mathbf{S}$$
I thought Stokes' Theorem could be helpful, so I found the rotational of $F$: $$\text{rot} F = (\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}) = (-y,-z,-x)$$ And the Stokes' Theorem states that $$\iint_S \text{rot}\mathbf{F}\cdot d\mathbf{S} = \int_{\partial S} \mathbf{F}\cdot d\mathbf{S}$$
But I don't know how to deal with the fact the surfaces intersect each other, as well as the Stokes' Theorem requieres a surface rather than a volume. I thought of maybe applying The Divergence Theorem could also work, but I don't know how to. This question comes from a Spaninsh University Second Year exam. Any help?
It is right that you should be using the divergence theorem here: the Stokes' Theorem you cite applies when you compute a line integral on the right hand side!
The divergence theorem tells you that $$ \iint_{\partial V} \mathbf{F}\cdot d\mathbf{S} =\iiint_{V} \nabla\cdot \mathbf{F}. $$ The divergence of $F$ is given by $$ (\nabla\cdot F)(x,y,z) = 1+ y + z + x. $$ If you now take a careful look at the region $V$, you see that the region $V$ is symmetric in $x,y$ and $z$: $$(x,y,z)\in V \iff (-x,-y,-z) \in V.$$
As the functions $x,y,z$ are odd, the only term contributing to the integral on the right hand side is the constant 1, which means that you are just computing the volume of the figure, which is the volume of two balls of radius 2, minus the volume of their intersection, since you counted this twice.
To compute the volume of the intersection, I suggest drawing a vertical cross-section for $x = 0$ to get a better view of the situation.