Let $V\subset\mathbb{R}^3$ be compact and possess a piecewise smooth boundary $\partial V$. Let $\mathbf F$ be a continuously differentiable vector field defined on a neighborhood of $V$. Suppose that for every piecewise smooth oriented path $C\subset V$ that starts and ends at $\partial V$ the line integral $$ \int_C \mathbf F \cdot d\mathbf r = 0. $$ If $\nabla\cdot\mathbf F = 0$, can we conclude that $\mathbf F = \mathbf 0$?
In case the above is not true, is there a simple counter-example?
My Attempt: Without the vanishing divergence assumption, I came up with the following counter-example. Let $V$ be the unit ball centered at the origin. Then, $\mathbf F = x \mathbf i + y \mathbf j + z\mathbf k$ satisfies the assumptions and $\mathbf F \neq \mathbf 0$.
On the other hand, assuming both divergence and curl vanishes (and $V$ is simply connected), then there exists $\Phi:V\longrightarrow\mathbb{R}$ such that $\nabla \Phi = \mathbf F$. Thus, $\nabla^2 \Phi = 0$ and since $\Phi$ is constant in $\partial V$, then we can conclude that $\Phi$ is constant (Dirichlet problem). Thus, $\mathbf F = \mathbf 0$.
Assuming that only the divergence vanishes, I imagine there is a counter-example, but I couldn't find one.
Update: $\mathbf F$ must be $\mathbf 0$.
Let $\Gamma\subset V$ be a closed path. Let $C_n$ be a path that starts at some point in $\partial V$ goes through $\Gamma$ $n$ times and returns to some point in $\partial V$. We know that the line integral over $C_n$ is $0$ for every $n$. Thus, the line integral over $\Gamma$ must also vanish. Since $\Gamma$ is an arbitrary closed path, then $\mathbf F$ is conservative and we end up with the Dirichlet problem.
Is this correct?
Edit: @TedShifrin made me realize that the above is valid only if $V$ is connected. However, I imagine we can apply the above reasoning for each connected component.