proving inverse of gradient operator I'm trying to prove the inverse of the gradient operation ($\nabla V$):
$\vec{E} = \nabla V$
is the line integral operation:
$\int\vec{E}\bullet d\vec{\mathcal{l}} = V$

Here's my work:
$\vec{E} = \nabla V$
take the line integral of both sides:
$\int\vec{E}\bullet d\vec{\mathcal{l}} = \int \nabla V\bullet d\vec{\mathcal{l}}$
then for some reason i don't fully understand, right hand side becomes equal to V:
$\int\vec{E}\bullet d\vec{\mathcal{l}} = V$
I guess my question is this.  why does:
$\int \nabla V\bullet d\vec{\mathcal{l}} = V$ ?
 A: Your statement "the inverse of the gradient operator is the line integral operator" is not rigorously correct, as you are not saying which line you integrate over. What is true is the gradient theorem: if $\Gamma$ is any curve that starts at a point $\vec{p}$ and ends at a point $\vec{q}$, then
$$V(\vec{p}) + \int_\Gamma \nabla V \bullet \;\mathrm{d}\vec{\ell} = V(\vec{q}).\quad\quad\quad\quad\quad\quad\quad\quad(1)$$
If you let $\vec{p}$ be a constant, the gradient theorem is exactly the same as the statement in your question $\int \nabla V\bullet d\vec{\mathcal{l}} = V$, with $V(\vec{p})$ playing the role of the constant of integration.
The proof of the gradient theorem is given in the Wikipedia link: if the curve $\Gamma$ is expressed as a function $\vec{\gamma}(t)$ with $t\in[0,1]$ and with $\vec{\gamma}(0)=\vec{p}$ and $\vec{\gamma}(1)=\vec{q}$, then
$$\int_\Gamma \nabla V \bullet \;\mathrm{d}\vec{\ell}=\int_0^1\nabla V\left(\vec{\gamma}(t)\right)\bullet \frac{\mathrm{d}}{\mathrm{d}t}\vec{\gamma}(t) \;\mathrm{d}t.\quad\quad\quad\quad\quad\quad(2)$$
We also know from the chain rule of differentiation that
$$\nabla V\left(\vec{\gamma}(t)\right)\bullet \frac{\mathrm{d}}{\mathrm{d}t}\vec{\gamma}(t) = \frac{\mathrm{d}}{\mathrm{d}t}\left\{ V\left(\vec{\gamma}(t)\right)\right\}.\quad\quad\quad\quad\quad\quad(3)$$
Substituting (3) in (2), we have
\begin{align}
\int_\Gamma \nabla V \bullet \;\mathrm{d}\vec{\ell} & =\int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}\left\{ V\left(\vec{\gamma}(t)\right)\right\}\:\mathrm{d}t= V\left(\vec{\gamma}(1)\right)-V\left(\vec{\gamma}(0)\right) \\ & = V(\vec{q})- V(\vec{p})
\end{align}
which proves the gradient theorem (1). 
You might also want to read this answer, which explains a method to invert a gradient using this idea.
A: I think the answer may lie in a deeper understanding of just what $d\vec{\mathcal{l}}$ means.  The integral
$\displaystyle \int \nabla V \cdot  d\vec{\mathcal{l}} \tag 1$
is generally taken over a differentiable path $\gamma(t)$ joining two points $P$ and $Q$ in the domain $\Omega$ of $V$; that is, we assume that, for some closed interval
$I = [t_0, t_1] \subset \Bbb R, \tag 2$
we have a differentiable function
$\gamma: I \to \Omega, \tag 3$
such that
$\gamma(t_0) = P, \; \gamma(t_1) = Q; \tag 4$
along such $\gamma(t)$, 
$\dfrac{dV(\gamma(t))}{dt} = \nabla V(\gamma(t)) \cdot \dfrac{d\gamma(t)}{dt} = \nabla V(\gamma(t)) \cdot \gamma^\prime(t); \tag 5$
it follows that
$V(Q) - V(P) = \displaystyle \int_{t_0}^{t_1} \dfrac{dV(\gamma(s))}{ds} \; ds = \int_{t_0}^{t_1} \nabla V(\gamma(s)) \cdot \gamma^\prime(s) \; ds; \tag 6$
if we now introduce the shorthand
$d\vec{\mathcal{l}} = \gamma^\prime(t) \; dt, \tag 7$
then (6) reads
$V(Q) - V(P) = \displaystyle \int_{t_0}^{t_1} \nabla V(\gamma(s)) \cdot d\vec{\mathcal{l}}, \tag 8$
and we have seen that the notation (7) gives rise to (8).  Note that (8) gives the difference 'twixt $V$ at the two points $P$ and $Q$; we may only affirm
$V(Q) = \displaystyle \int_{t_0}^{t_1} \nabla V(\gamma(s)) \cdot d\vec{\mathcal{l}} \; ds \tag 9$
in the event that $V(P) = 0$.
