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If I am given a certain funciton which describes the volume charge density within the sphere with respect to the radius (ρ(r)),how can I calculate the total charge within the sphere.My idea was by calculating the total charge of many infinitesimally small and thin circles. Initially, I though that I could use $$\int (πy^2)*(ρ(r)) \, dr$$ Where $$\ x^2 + y^2 = r^2$$ Because this idea it didnt work i thought of using the areas of small spheres times the volume charge density $$\int (4πr^2)*(ρ(r)) \, dr$$ It worked by I am not sure if this is the proper way or if it was pure luck due to the numbers that i got the answer right.

Even if the 2nd way is right can someone give me some more information in order to understand it better, or is there a better solution which is more understandable.

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Your last equation is the right approach.

Assuming that \$\rho_r\$ is only a function of \$r\$ and not a function of \$x,y,z\$, your first thought about the thin shell is about right, except misapplied due to ignoring \$z\$. Instead, you just take the shortcut that you know the surface area at \$r\$ is \$4\pi\,r^2\$ and that it's thickness is obviously \$\text{d}r\$ (you are only certain that \$\rho_r\$ is exactly true for an infinitesimally thin shell, which of course is \$\text{d}r\$ thick.) So the volume of that thin shell is \$4\pi\,r^2\:\text{d}r\$.

Given that \$\rho_r\$ is in the units of \$\frac{\text{charge}}{\text{volume}}\$, multiplying it by the volume of the thin shell provides the charge in that thin shell. So it is correct to multiply the volume of the thin shell by your \$\rho_r\$ factor to get charge in that thin shell.

Integrating the sum of the charge in these thin shells results in the total charge of the sphere.

I might have just as well have written your last equation as \$\int 4\pi\,r^2\:\text{d}r\:\rho_r\$ or \$\int \rho_r\:4\pi\,r^2\:\text{d}r\$. It's all the same thing. But yes, I think you used the right equation at the end.

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  • \$\begingroup\$ Thank you for your answer! Can you please elaborate in my first trial in order to add the missing z dimension because for me it is more intuitively than the second way. I try to generilise it in any shape, obiously for a sphere it made sense to use the areas of smaller spheres but how about if I am given an equation of a not known shape? \$\endgroup\$ – pj33 Dec 29 '19 at 16:43
  • \$\begingroup\$ @pj33 I can't elaborate it because it's not structured well. And doing it well would require far, far more writing than I've done above. You usually want to formulate your solution in a way that conforms well with the problem and using x, y, and z, does NOT conform well. You could incorporate z, but use a different r, this time only laying on a plane, and do a double integration, one sweeping r around in a circle, the other pushing z from one end to the other. That would be easier and would use z. But breaking it out into all three, x, y, and z, would be a long addition I'd like to avoid. \$\endgroup\$ – jonk Dec 29 '19 at 18:58

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