Sure, let's proceed!
The concept behind this question is integrating the given charge density P_v = P_0x/a over the full volume of the cube. Those limits run from 0 to a, as given.
As specified by the question, we assume that the charge density increases linearly with x. That means the charge density is zero at the one face of the cube (where x=0), and maximizes at the opposite face (x=a).
1. First, determine the element of volume in the cube. Seeing as we are in 3D space, it would be appropriate to use the dimensions of x, y, and z. Thus, the volume element (also known as the differential volume) dV is equal to dx*dy*dz. That's a small cube within the larger one.
2. Identify the charge density ρ (rho). As per the question, the charge density ρ is given to be P_0*x/a.
3. Now, to find total charge in the cube, we "sum over" (integrate) this density through the whole cube, because the charge dQ contained in a small volume dV = dx*dy*dz is given by dQ = ρ(x)dV. Keep in mind that the density changes with the x-direction, thus why the density function is a function of x.
4. Now to calculate the total charge Q, we integrate the product of the charge density and the volume element over the entire volume of the cube, which means integrating over each dimension x, y, and z from 0 to a:
Q = ∫ ρ(x) dV over the cube
5. Substituting in ρ(x) = P_0*x/a and the differential volume dV we arrive at:
Q = ∫ from 0 to a ∫ from 0 to a ∫ from 0 to a of ρ(x) dx dy dz
Simplifying this would give:
Q = ∫ from 0 to a ∫ from 0 to a ∫ from 0 to a (P_0*x/a) dx dy dz
6. Solving this three-dimensional integral will result in the total charge in the cube.
It's important to note that P_0, a are constants and can be removed from the integral. The three integrals are the same, so this can be rewritten as (please calculate the integral yourself).
7. After finishing the integrations Q should then be the total electric charge in the cube.
Please note that you should have knowledge of multiple integration in order to solve this assignment. This response assumed that you understand how to perform the actual integration, which is pretty straightforward.