Is this all you have to do for this problem

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To find the total charge on a nonconducting solid sphere with a volume charge density proportional to the distance from the center, the charge density is expressed as ρ = Ar for r ≤ R. The total charge Q can be calculated by integrating the charge density over the sphere's volume using the formula dV = 4πr²dr. A triple integral approach is suggested, integrating Ar*r²*sin(θ) with limits from 0 to R for r, 0 to π for θ, and 0 to 2π for φ, resulting in the expression AR^4π. The integral of 4πr² from 0 to R is also discussed to clarify the volume calculation.
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Homework Statement


A nonconducting solid sphere of radius R has a volume charge density that is proportional to the distance from the center. That is rho=Ar for r is less than/equal to , A is a constant. Find the total charge on the sphere.


Homework Equations




The Attempt at a Solution


rho=Ar=Q/V
Q=Ar(4/3)*pi*R^3

Is that it?
 
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No.

Integrate the charge density over the volume of the sphere.
 
would've that just be the integral of (Ar dV) which equals Ar*V?
 
Use dV = 4πr2dr .

Integrate (A)r , with respect to r, from r = 0 to r = R .
 
I went in for help on this one and my TA told me that I could do a triple integral for this one or with the integral you provided, but I chose to do the triple integral.

What I did was take the triple integral of Ar*r^2 * sin(theta) in this order: dr,dtheta,dphi
with limits of integration: 0 to R, 0 to pi, 0 to 2pi, respectively
For the answer I got AR^4 * pi

Can you explain how you get the dV = 4πr2dr
 
The surface area of a sphere of radius r, is 4πr2.

So the volume of a spherical shell of thickness dr is given by multiplying the thickness by the surface area.

dV = 4πr2 dr.

What's the integral of 4πr2 dr, from 0 to R ?
 
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