# Description of charged sphere with Heaviside function in cylindrical coordinates

I need to describe density of charge of uniformly charged sphere (radius R, total charge Q, position of centre (0,0,0)) with Dirac delta function and Heaviside step function. The hard part is to describe this in cylindrical coordinates.

Well, this is easy in spherical coordinates, result is:

$$\rho(r, \theta, \phi) = \frac{3 \cdot Q}{4 \cdot PI \cdot R^3} \cdot H(R-r)$$

It is obvious that only thing that will change in cylindrical coordinates is Heaviside step function and because the condition is:

$$R^2 = r^2+ z^2$$

the result should be:

$$\rho(r, phi, z) = (3 \cdot Q)/(4 \cdot PI \cdot R^3) \cdot H(R^2 - r^2 - z^2)$$

Well, this is correct result but it is not the correct way to how to solve this problem. I have problem with how this Heaviside function affects bounds of integrals.

My whole progress in spherical coordinates is:

$$\rho (r, \theta, \phi) = A \cdot H(R-r)$$ $$Q = \int_0^{2\pi}\int_0^{\pi}\int_0^{\infty} \rho \cdot r^2 \sin \theta~dr d\theta d\phi = \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta d\theta \int_0^{\infty} A \cdot H(R-r) \cdot r^2~dr = A \int_0^{2\pi} d\phi \int_0^{\pi} \sin \theta d\theta \int_0^{R} r^2~dr = A \frac{4 \cdot \pi \cdot R^3}{3} \Rightarrow A = \frac{3Q}{4 \pi R^3}$$ $$\rho (r, \theta, \phi) = A \cdot H(R-r) = \frac{3Q}{4 \pi R^3} \cdot H(R-r)$$

In cylindrical as far as I know how to solve:

$$\rho (r, \phi, z) = A \cdot H(R^2-r^2-z^2)$$ $$Q = \int_0^{2\pi}\int_{-\infty}^{\infty}\int_0^{\infty} \rho \cdot r~dr dz d\phi = \int_0^{2\pi} d\phi \int_{-\infty}^{\infty}\int_0^{\infty} A \cdot H(R^2-r^2-z^2) \cdot r~dr dz = 2 \pi A \int_{-\infty}^{\infty}\int_0^{\infty} r \cdot H(R^2-r^2-z^2)~dr dz$$

How to continue?

## 1 Answer

you can either evalute the integral numerically or search for a good coordinate transformation to evaluate it "by hand". This transformation is the one to spherical coordinates, though.

Also you could try this: \begin{align} Q & = 2 \pi A \int_{-\infty}^{\infty}\int_0^{\infty} r \cdot H(R^2-r^2-z^2)~dr dz \\ & = 2 \pi A \int_{-R}^{R}\int_0^{\sqrt{R^2-z^2}} r~ dr dz \\ & = 2 \pi A \int_{-R}^{R} \left[\frac{1}{2}r^2\right]_0^{\sqrt{R^2-z^2}}~dz \\ & = \pi A \int_{-R}^{R} (R^2-z^2)~dz \\ & = \frac{4}{3} \pi R^3 A \end{align}

You can get the integral bounds like this: $$H((R^2 - z^2) - r^2) \tag1$$ implies $r \in [0, \sqrt{R^2-z^2}]$. Furthermore for $r=0$ one has $z^2 \le R^2$ in order to get $H\ne0$, such that $z \in [-R,R]$. For $r=\sqrt{R^2-z^2}$ one has $z=0$ in order to have $H \ne 0$, but this is already in the above interval.