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Finding mass with dirac delta function

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Distribution of matter is given in cylindrical coordinates:


    where [tex]a>0[/tex] is a constant. Find the complete mass of the object.

    2. Relevant equations

    The mass of the object is given as:

    [tex]M=\int_{\partial V}\rho(\vec{r})dV[/tex]

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

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

    Integral in cylindrical coordinate system is:

    [tex]\int_0^\infty\rho d\rho\int_0^{2\pi} d\phi\int_{-\infty}^\infty dz[/tex]

    3. The attempt at a solution

    I have expanded the first delta:


    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):


    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?
  2. jcsd
  3. Jan 24, 2010 #2
    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.
  4. Jan 24, 2010 #3
    So the second one is:

    But how do I calculate the integral? With [tex]\phi[/tex] part? When I put it into Mathematica I get different answers:
    For putting sum in front of the integral:
    [tex]\frac{1}{2}\theta(3 - 2 m, \pi + 2 m \pi)[/tex]
    And for putting integral in front of the sum I have [tex]\frac{7}{\pi}[/tex], 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?
  5. Jan 25, 2010 #4
    The integral of delta function is:

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

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

    In cylindrical coordinates

    0 \leq \phi < 2\pi.

    You should keep only two terms of the infinite sum.
  6. Jan 25, 2010 #5
    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 :(
  7. Jan 25, 2010 #6
    No, they are not random.
    When you integrate the sum

    \sum_m \delta(\phi - \phi_m)

    only two terms fit the condition

    [tex]\phi_m \in (0, 2\pi).[/tex]

    The other terms are not integrated and give zero.
  8. Jan 25, 2010 #7
    I see, that's because I'm only need two zeros in that interval, so I'm only using those two! Thanks!!
    Last edited: Jan 25, 2010
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