Electric field due to nonconducting sphere

For calculating electric field outside a nonconducting sphere with a hollow spherical cavity. When I use the rule (Charge density= $dQ/dV$), I don't know exactly what is $dV$, is the volume here refers to the volume of the Gaussian surface ($V= 4/3 \pi r^3$) so that $dV$ will be = $\pi r^2 dr$, or the $V$ is the volume containing the charges only, so it will be =$V_0 – V_1 = 4/3 \pi r_0^3 - 4/3 \pi r_1^3$. Thus, since $r_0$ and $r_1$ are constants, therefore $dV$ will be = 0?

Note: $r_0$ is the radius for the whole sphere, $r_1$ is the radius for the cavity, and $r$ is for the Gaussian surface.

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This case case, the volume of the object is $V_0 - V_1$, so if $Q$ denotes its total charge, then the charge density $\rho$ is given by $$\rho = \frac{Q}{V_0-V_1}.$$
Thnx alot, but What if it is not constant and given as a fuction for example $(ρ= (ρ_0 * r_1)/r)$ – Abdulrahman Hessen Mar 15 '13 at 16:08
Well if the density is given to you, then there is no longer any need to calculate the density...in this case, in your notation, you would just write $\rho = dQ/dV$ where $dQ$ here is a small amount of charge contained in $dV$ a small volume of the object. – joshphysics Mar 15 '13 at 16:14
Unfortunately Still not clear for me. Well, here is the question as it is in the text book: An electric charge $Q$ is distributed through out a nonconducting sphere of radius $r_0$ and has a spherical cavity of radius & $r_1$ centered at the sphere's centre. Assume the charge $Q$ is distributed in the shell (i.e. between $r=r_1$ and $r=r_0$) and that the charge density varies as ρ= $(ρ_0 * r_1 / r)$ Find the electric field as a fuction of $r$ for $r>r_0$ – Abdulrahman Hessen Mar 15 '13 at 17:01
@AbdulrahmanHessen: Can you write down an expression for $dV$? You can infer that $Q_{\rm inc} = \int \rho dV$, so if you can write down good limits for the integral, over a single variable (hint: what should that variable be?), you're good to go. – Jerry Schirmer Aug 12 '13 at 23:36