Differential of the volume of a rectangular parallelepiped. Suppose I have a rectangular parallelepiped with base $A$ and height $l$. In order to calculate the volume I could use a double integral and integrate over the intervals $[0,l]$ and $[0,A]$ the infinitesimal volume $dV = dAdl$ and I would get the total volume as below
$$\int dV = \int_{0}^{l}\int_{0}^{A}dA'dl'$$
However, since there is a closed formula for the volume ($V = lA$), I could also say that the total differential is equal to
$$dV = \frac{\partial V}{\partial l}dl + \frac{\partial V}{\partial A}dA$$ 
however, by integrating this $dV$ I get another result as below
$$\int dV = \int_{0}^{l}Adl' + \int_{0}^{A}ldA'= 2lA$$
What am I missing within these steps?
 A: I see you are confused with total differential and its integration. Total differential is the infinitesimal difference of a function in terms of the infinitesimal difference of its variables. 
Let us assume a function $z=xy$. The total differential $dz$ contains differences of $z$ on two directions. In the region $S=\{0\leq x\leq 1, 0\leq y\leq 1\}$, the integral of $dz$ is $1$, which is the total difference of $z$ in this region. But when you write it in integral
$$\int dz=\int xdy+ydx$$
It does not make sense to put integration region $S$ on both sides. We have to choose a path between $z=0$ and $z=1$, since both $x$ and $y$ are variables here. So similarly, you cannot put $0$ and $1$ as integration limits for both of the two terms on the right hand side. Even if you do that, in the term $xdy$, $x$ is still a variable, and in the term $ydx$, $y$ is still a variable. 
We can choose a path for example, $y=0, x=0\rightarrow 1$ then $x=1, y=0\rightarrow 1$. This will give the right answer for this integral. 
For your case, $V=lA$ is fixed. Normally we don't use total differential to represent it. If you do consider $l,A$ as variable, remember they are still variables in the integrals. So you have to specify a path to do it.
