Continuous Surface Charge Source - Integral Confusion

[technicolour1]

1. Homework Statement
Find the total charge on a circular disc of radius ρ = a if the charge density is given by
ρs = ρs0 (e^−ρ) sin2 φ C/m2 where ρs0 is a constant.
Are the two limits of integration from 0 -> a for ρ and 0->2∏ for φ? In the example given in the notes, ρ varies, instead of being a constant value. Does this mean that I cannot integrate ρ in this problem?


2. Homework Equations

Q = ∫ρdS

3. The Attempt at a Solution

Q = ∫∫ρs0 (e^−ρ) sin^2 (φ) dφdρ...

 

[SammyS]

1. Homework Statement
Find the total charge on a circular disc of radius ρ = a if the charge density is given by
ρs = ρs0 (e^−ρ) sin2 φ C/m2 where ρs0 is a constant.
Are the two limits of integration from 0 -> a for ρ and 0->2∏ for φ? In the example given in the notes, ρ varies, instead of being a constant value. Does this mean that I cannot integrate ρ in this problem?

2. Homework Equations

Q = ∫ρdS   (This is the surface density ρ.)

3. The Attempt at a Solution

Q = ∫∫ρs0 (e^−ρ) sin^2 (φ) dφdρ...

You have the Greek letter rho, ρ, used in two very different ways here.

ρ as a coordinate: This is the role usually player by the letter, r, in polar coordinates.

ρ as (surface) charge density: Here ρ seems to be accompanied by what looks like should be a subscript. ρ_s and ρ_s0.

Yes, the limits of integration are as you indicate.

The charge density, ρ_s, is a function of the coordinate, ρ. (somewhat confusing).

For polar coordinates, the element (differential) for area, dS, is given by: \displaystyle dS=\rho\,d\varphi\,d\rho\,.

Other than that, you seem to be on the right track.