1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume integral of a function over tetrahedron

  1. Jan 19, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the volume integral of the function $$f(x,y,z)=xyz^2$$
    over the tetrahedron with corners at $$(0,0,1) (1,0,0) (0,1,0) (0,0,1)$$
    2. Relevant equations
    I was able to solve it mathematically, but still cant figure out why the answer is so small.
    I only understand that if f(x,y,z) is the density, then the triple integral is the mass.
    What is the physical significance for calculating the volume integral of an arbitrary function over a geometrical shape?

    3. The attempt at a solution

    $$\int_{0}^{1} \int_{0}^{1-y} \int_{0}^{1-x-y} xyz^2 dz dx dy =\frac{1}{2520}$$

    Big thanks in advance !
     
  2. jcsd
  3. Jan 19, 2016 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    I think one of those points should be the origin.
    It just means you are summing the value of the function at every point within the specified volume.
    For now I assume that your calculation hides no mistake. The minuscule value of the integral might be caused by the behavior of your ##f(x,y,z)## within the specified tetrahedron. Look, the biggest value of either ##x##, ##y##, and ##z## within this tetrahedron is unity, therefore ##f(x,y,z)=xyz^2## cannot be bigger than unity (although I haven't calculated what the maximum value is, but certainly the maximum value cannot be bigger than 1). In fact, the values of coordinates are multiplied in ##f(x,y,z)## which makes this function sufficiently small if you remember that multiplication between two or more numbers gives a number which is smaller than the smallest number being multiplied.
     
  4. Jan 19, 2016 #3
    Yes, sorry, I made a mistake while typing.

    You are right! the answer is so small because I am multiplying a fraction to a fraction and again to a fraction squared.

    So there is no specific physical significance (I am asking this because this is a problem from my EM Theory assignment), only pure mathematical calculation? I should simply treat this integral as a mathematical problem, rather than trying to combine it with physical quantities?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Volume integral of a function over tetrahedron
Loading...