Finding mass with dirac delta function

[dingo_d]

Homework Statement

Distribution of matter is given in cylindrical coordinates:

\rho(\vec{r})=\frac{1}{\rho}\delta(\rho^2-10\rho+9)\delta\left(\frac{z^2-a^2}{z^2+a^2}\right)\delta(\cot(\phi))

where a>0 is a constant. Find the complete mass of the object.

Homework Equations

The mass of the object is given as:

M=\int_{\partial V}\rho(\vec{r})dV

So in order to calculate the integral I need to expand the deltas with this formula:

\delta(f(x))=\sum_{i=1}^n\frac{\delta(x-x_i)}{|f'(x_i)|}, where |f'(x_i)| is the derivative of the function evaluated at the zeroes of the function x_i.

Integral in cylindrical coordinate system is:

\int_0^\infty\rho d\rho\int_0^{2\pi} d\phi\int_{-\infty}^\infty dz

The Attempt at a Solution

I have expanded the first delta:

\delta(\rho^2-10\rho+9)=\frac{\delta(\rho-1)}{8}+\frac{\delta(\rho-9)}{8}.

I'm having problem with the second one. Do I 'brake' it with partial fractions?

Third one (which I'm not sure I got it right):

\delta(\cot(\phi))=\sum_{m=-\infty}^\infty\delta\left(\phi-\left(m+\frac{1}{2}\right)\pi\right)

So my problem is: how do I solve the second delta and I'm not quite sure I got the third one right. How should I integrate it?

 

[Maxim Zh]

I see no problem in the second delta. Just find the zeros and calculate the derivatives.
Your answer for the third delta seems to be right.

 

[dingo_d]

So the second one is:
a\left[\delta(z-a)+\delta(z+a)\right]?

But how do I calculate the integral? With \phi part? When I put it into Mathematica I get different answers:
For putting sum in front of the integral:
\frac{1}{2}\theta(3 - 2 m, \pi + 2 m \pi)
And for putting integral in front of the sum I have \frac{7}{\pi}, which is weird because shouldn't I get the same results no matter which goes first?

But that's on the side note, on my exam I won't have Mathematica :D, so how to calculate that?

 

[Maxim Zh]

The integral of delta function is:

<br /> \int_a^b f(x) \delta(x-c)\,dx = \begin{cases}<br /> f(c), &amp; a &lt; c &lt; b; \\<br /> f(c)/2, &amp; c = a \;\text{or}\; c = b; \\<br /> 0, &amp; \text{otherwise}.<br /> \end{cases}<br />

I think it's not a problem to calculate it without Mathematica.

In cylindrical coordinates

<br /> 0 \leq \phi &lt; 2\pi.<br />

You should keep only two terms of the infinite sum.

 

[dingo_d]

The integral of delta function is:

<br /> \int_a^b f(x) \delta(x-c)\,dx = \begin{cases}<br /> f(c), &amp; a &lt; c &lt; b; \\<br /> f(c)/2, &amp; c = a \;\text{or}\; c = b; \\<br /> 0, &amp; \text{otherwise}.<br /> \end{cases}<br />

I think it's not a problem to calculate it without Mathematica.

In cylindrical coordinates

<br /> 0 \leq \phi &lt; 2\pi.<br />

You should keep only two terms of the infinite sum.

Those terms are random or? I'm confused because I have only solved simple integrals with delta function in my class, never with these sums :(

 

[Maxim Zh]

No, they are not random.
When you integrate the sum

<br /> \sum_m \delta(\phi - \phi_m)<br />

only two terms fit the condition

\phi_m \in (0, 2\pi).

The other terms are not integrated and give zero.

 

[dingo_d]

I see, that's because I'm only need two zeros in that interval, so I'm only using those two! Thanks!