The inverse of triple integrands Firstly, I am an engineer by profession and not a student of mathematics. So, my mathematical abilities are severely handicapped. Having said that I could say that indefinite single integrands have inverses. For example, if
$$\int f(x)\mathrm{d}x=g(x)+C$$
Then, I could write
$$f(x)=\frac{\mathrm{d}g(x)}{\mathrm{d}x}$$
I understand that I could write this because 
$$\int f(x)\mathrm{d}x=\int_{C_0}^{x}f(y)\mathrm{d}y=g(x)+C$$
Here in x represents a bound to the integration of the function $f(y)$ and hence the output is a function of that bound {$g(x)$}.
But, just as a thought experiment, what about $$\iiint_{V}f(\vec{r})\mathrm{d}^3\vec{r}$$
Here in $V$ is a bound to the function $f(\vec{r})$, so I should be able to write the output as 
$$\iiint_{V}f(\vec{r})\mathrm{d}^3\vec{r}=\rho\left(V\right)$$
Is there an inverse function that when applied on $\rho(V)$ with the knowledge of $V$ gives me back $f(\vec{r})$. I definitely don't know of an inverse function as such. If there isn't, my question is why isn't there one? Is it because, according to your mathemitical senses $f(\vec{r})$ lacks the uniqueness of being the only function which when bound into $V$ would result in $\rho(V)$?
Like I said, I'm mathematically illiterate, just saying it again at the risk of sounding like a fool.
 A: The fundamental theorem of calculus in one dimension can be expressed saying that
$$
\int_a^bf(x)dx=F(a)-F(b)
$$
and
$$
\frac{d}{dx}F(x)=f(x)
$$
Intuitively this say that the integration estabilish  a relation between the values of a function at the ''boundary'' of an interval $[a,b]$ and the the derivative of this function (in the interval). A relation that can be written  in the form:
$$
\int_a^b dF(x)=F(a)-F(b)
$$
The extension of this result to higher dimensions   can be done defining what is the ''boundary'' of a ''multi-dimensional'' region and if there is a  good generalization of the notion of derivative for such a situation.
All this can be done usinge the machinery of the  exterior forms and derivatives on differentiable manifolds. 
The main result is the (generalized)  Stoke's theorem that says that
$$
\int _{M}d\omega=\int_{\partial M} \omega
$$
Here M is a compact smooth orientable  $n$-dimensional manifold with ''boundary'' $\partial M$, and $\omega$ is an $(n − 1)$-form on M, and $d\omega$ is the exterior derivative of $\omega$.
You can note the analogy with the one dimensional situation:
$F(a)-F(b)$ is the analogous of the  $\int_{\partial M} \omega$ that generalizes the idea of the difference of the values of a function at the ''boundary''  and the usual derivative of a function of a single variable is substituted by the exterior derivative.
A: If I understand you correctly, you're asking about how to find the underlying function $f$, given a function $\rho(V)$ satisfying
$$ \rho(V) = \iiint_V f(\vec r)\, d^3 \vec r $$
where $V$ is any region in space, and so "invert" the triple integral. I believe it is possible to do this and will explain how below. Note that in all that follows, I will assume that the function $f$ is continuous and defined over all of space.
Let's first establish that for a given $\rho$, the function $f$ is unique: that is, if $\rho(V) = \iiint_V f(\vec r)\, d^3 \vec r$, then there is no other continuous function $g$ whereby $\rho(V) = \iiint_V g(\vec r)\, d^3 \vec r$ for all possible regions $V$. To prove this, let's suppose $f$ is not unique: that there is indeed another function $g$ different from $f$ whose triple integrals agree with those of $f$ for all regions $V$. Now, since both $f$ and $g$ are continuous, it must be true that one function is greater than the other over some (possibly small) region of space that I'll call $W$. For sake of concreteness, let's say that $f$ is the greater function within the region $W$. In that case, it must mean that
$$ \iiint_W f(\vec r)\, d^3\vec r > \iiint_W g(\vec r)\, d^3\vec r $$
because $f$ always has larger values than $g$ over $W$, and $W$ is a valid region of space that we can input into the function $\rho$. This contradicts the assumption that the triple integrals of $f$ and $g$ are supposed to agree for all possible regions. Hence it must be the case that no such $g$ can exist, and so $f$ must be unique after all.
Now, as for computing $f$ given $\rho$, we can use the following technique. First of all, the function $\rho$ can accept any kind of region $V$ for input, so let's consider how $\rho$ behaves when we give it a special type of region for input: rectangular boxes. First, let's fix a point $(a,b,c)$ in space to be one corner of a box and let's allow the opposite corner to vary and call its coordinates $(x',y',z')$ or $\vec r'$ for short. What does the function $\rho(V)$ look like when we give it these special boxes for input? Well, for a given $\vec r'$ we choose, we get that
$$ \rho(V) = \int_c^{z'} \int_b^{y'} \int_a^{x'} f(x,y,z)\, dx\, dy\, dz $$
This is an iterated integral of three single integrals! That means, via the ordinary Fundamental Theorem of Calculus, we can undo each of them in order by taking derivatives:
\begin{align}
f(x',y',z') &= \frac{\partial}{\partial x'} \frac{\partial}{\partial y'} \frac{\partial}{\partial z'} \int_c^{z'} \int_b^{y'} \int_a^{x'} f(x,y,z)\, dz\, dy\, dx \\
&= \frac{\partial^3}{\partial x'\, \partial y'\, \partial z'}\: \rho \Big(\operatorname*{Box}_{a,b,c}(x',y',z') \Big)
\end{align}
