Why capacitance is constant? I am new to the field of electrodynamics and I recently came across the definition of capacitance as
$$C = \frac{Q}{V}$$
It is said that capacitance is constant, which implies that total charge of the one conductor is proportional to the voltage difference between the two conductors. But we know that
$$Q = \int ρ(\vec r')dV$$
where the integral is evaluated in one conductor, and
$$V(\vec r) = \int \dfrac{ρ(\vec r')}{|\vec r - \vec r'|}dV$$
where the integral is evaluated along the two conductors.
How can we prove from these that $Q$ and $V$ are proportional?
 A: Notice how in both $Q$ and $V$ the charge density is acting in the same way in the integrals.$^*$ Scaling $\rho$ by some constant will scale $Q$ and $V$ by that same constant, so that $Q/V$ will not change at all.

$^*$However, beware of blindly throwing equations around.
In $C=Q/V$, $Q$ is the amount of net charge on one conductor (so that one has $+Q$ charge and the other has $-Q$ charge), and the $V$ is the potential difference between the conductors.
On the other hand, the equation you give for $Q=\iiint\rho(\mathbf r')\,\text d\tau'$ is the total charge in the system (which we usually assume is $0$ for a overall neutral system), and $V=\iiint\rho(\mathbf r')/|\mathbf r-\mathbf r'|\,\text d\tau'$ is the potential at position $\mathbf r$ assuming that $V\to0$ as $\mathbf r\to\infty$ (this is not the same thing as the potential difference between the two conductors).
Of course you can tweak / qualify these equations so that they are valid for the capacitor, but since this was not specified I thought I should address it.
A: Capacitance is determined by the physical characteristics of the conductors and space between them. For a parallel plate capacitor the capacitance is
$$C=\frac{εA}{d}$$
Where $ε$ = the permittivity of the medium (dielectric) between the capacitor plates, $A$=the plate area and $d$ = the distance between the plates. So as long as these physical parameters involving the conductors and space between them do not change, the capacitance is constant.
Then, for a fixed physical device (capacitor), $C$ in the equation
$$C=\frac{Q}{V}$$
is constant, meaning only charge and voltage vary, with the ratio of charge to voltage being constant.
It is important to realize $Q$ is the net charge on either plate, positive on one and negative on the other and equal in magnitude. That makes the net overall charge on the capacitor zero (overall electrically neutral). In this regard @BioPhysicist has already correctly addressed the integral equations for $Q$ and $V$.
Hope this helps.
A: The definition of capacity states that its body is equipotential.  So C = Q / U , and you used the wrong formula for potential. U is the potential difference between the metal parts of the capacitor.
