Every surface integreal zero then divergence of function is zero I need help with this problem:
Let $D \subseteq \mathbb{R}^3$ be an open and connected region and $\vec{F} \in C^1(D,\mathbb{R}^3)$ a vector-valued function on $D$. Show that if 
$$\int_S \vec{F}\cdot d\vec{S} = 0$$
for every spherical surface $S$ contained in $D$ then $\mbox{div} \vec{F} = 0$.
Now, using Stoke's that would be equivalent to show that 
$$\iiint_B \mbox{div}\vec{F} = 0$$
where $\partial{B} = S$ (boundary of sphere $B$ would be $S$), but I completely stuck with this problem. 
 A: We are given that $\vec F$ is continuously differentiable, that is, $\vec F \in C^1(D, \Bbb R^3)$.
Then if there exists $p \in D$ with
$\nabla \cdot \vec F(p) \ne 0, \tag 1$
we must have either
$\nabla \cdot \vec F(p) > 0 \tag 2$
or
$\nabla \cdot \vec F(p) < 0; \tag 3$
if (2) binds, it follows from the continuity of $\nabla \cdot \vec F$ that 
there exists some $\epsilon > 0$ such that
$\nabla \cdot \vec F(x) > 0 \tag 4$
for
$x \in B(p, \epsilon) = \{x \in D \mid \Vert x - p \Vert < \epsilon \}; \tag 5$
then
$\displaystyle \int_{B(p, \epsilon)} \nabla \cdot \vec F(x) \; dV > 0; \tag 6$
by the divergence theorem (6) implies
$\displaystyle \int_{\partial B(p, \epsilon)} \vec F \cdot d \vec S > 0, \tag 7$
which contradicts the given hypothesis.  Thus we may rule out (2); likewise, if (3) holds, we can by a similar argument conclude that
$\displaystyle \int_{\partial B(p, \epsilon)} \vec F \cdot d \vec S < 0, \tag 8$
which also contradicts the assumption that
$\displaystyle \int_{\partial B(p, \epsilon)} \vec F \cdot d \vec S = 0; \tag 9$
thus we must have
$\nabla \cdot \vec F(p) = 0 \tag{10}$
for all $p \in D$.
