Finding the density of the matter

[dingo_d]

Homework Statement

Find the density of the matter $$\rho(\vec{r})$$ in spherical coordinate system, with the mass M, which is homogeneously placed on two rings with the radii of a (with the center in the origin) put in the xy and yz plane

Homework Equations

$$\int \rho(\vec{r})dV=M$$
For a unit mass in the center that would be:
$$\int dx\int dy\int dz M\delta(x)\delta(y)\delta(z)=M\cdot 1\cdot 1\cdot 1$$
in cylindrical coordinates:
$$\rho(\vec{r})=\frac{M}{2\pi \rho}\delta(\rho)\delta(z)$$
$$\int\rho(\vec{r})dV=\int_{-\infty}^\infty dz\int_0^{2\pi}d\phi\int_0^\infty \rho d\rho \cdot\frac{A}{2\pi\rho}\cdot\delta(\rho)\delta(z)=M\Rightarrow M=A$$

The Attempt at a Solution

I don't know even how to start :( Help please...

 

[anathema]

To find the density of the matter in spherical coordinates, we can use the following equation:

\rho(\vec{r})=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi a^3}=\frac{3M}{4\pi a^3}

Where M is the total mass and a is the radius of the ring. This equation assumes that the mass is uniformly distributed on the rings.

To find the volume V, we can use the following equation:
V=\int \int \int \rho(\vec{r})dV

Since the rings are in the xy and yz planes, the volume can be divided into two parts: the volume above the xy plane and the volume above the yz plane.

For the volume above the xy plane, we can use cylindrical coordinates with the limits:
0\leq \rho \leq a
0\leq \phi \leq 2\pi
0\leq z \leq \sqrt{a^2-\rho^2}

For the volume above the yz plane, we can use cylindrical coordinates with the limits:
0\leq \rho \leq a
0\leq \phi \leq 2\pi
-\sqrt{a^2-\rho^2}\leq z \leq \sqrt{a^2-\rho^2}

Plugging these limits into the volume equation, we get:

V=\int_0^{2\pi}d\phi\int_0^a\rho d\rho\int_0^{\sqrt{a^2-\rho^2}}dz+\int_0^{2\pi}d\phi\int_0^a\rho d\rho\int_{-\sqrt{a^2-\rho^2}}^{\sqrt{a^2-\rho^2}}dz=\frac{4}{3}\pi a^3

Finally, plugging this into the equation for density, we get:

\rho(\vec{r})=\frac{3M}{4\pi a^3}=\frac{3}{4\pi a^3}\cdot \frac{M}{V}=\frac{3}{4\pi a^3}\cdot \frac{M}{\frac{4}{3}\pi a^3}=\frac