Find the quantity of charge - given potential function 
A potential function is given by $V(r)=\frac{Ae^{-\lambda r}}{r}$ Find charge density and hence charge.

I first took the gradient of potential to get $\vec{E}(r)=\frac{Ae^{-\lambda r}}{r}[\lambda+\frac{1}{r}] \hat{r}$
Now using Gauss's law $\nabla \cdot \vec{E}=\frac{\rho}{\epsilon_0} $
$\implies \rho=-\epsilon_0 \frac{Ae^{-\lambda r}}{r}[\frac{1}{r^2}+\frac{\lambda}{r}+\lambda^2]$
Now to find the charge $Q=\int \rho 4\pi r^2 dr$ .
Did I do it correctly? What limits should I choose for the integration?
 A: At least in the first $r$-derivation, you missed some $\lambda$'s. Generally, always check your units if you derive solutation like that, as for exmple "$1+\frac{1}{r}$" just must be flawed. 
And you'll solve your problem more directly using the Poisson equation $\Delta V(r)=-\frac{\varrho(r)}{\varepsilon_0}$, and that with a $\Delta$ in explicitly spherical coordinates. 
To take a short track here, your $V(r)$ looks like a Yukawa potential, i.e. 
$$G(r)=\frac{\text{e}^{-\lambda r}}{4\pi\ r}$$
 solves 
$(\Delta-\lambda^2)G(r)=-\delta(\vec r)$ and hence 
$$\Delta V(r)=4\pi A\left(-\delta(\vec r)+\frac{\lambda^2}{4\pi}\frac{\text{e}^{-\lambda r}}{r}\right)\equiv -\frac{\varrho(r)}{\varepsilon_0}.$$
Integrate over all of space, if you want to find the total charge. If you have a density like $\propto r^{-n}$, then clearly there are charges everywhere. Remark: For integration, notice that you're dealing with a three dimensional delta-function here (but from the Coulomb potential, you know that it gives you a single charge anyway). For the exp-integration, up to some small numbers, you can figure out the result just by power counting in $\lambda$'s.
A: Gauss' law relates the integrated flux through a closed surface to the integrated charge inside it.
The symmetry of the situation lets you do the  flux integral trivially, so I assume that you mean the volume one. Well, you integrate over the full angular ranges and from the center to the bounding surface.
