Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volume integration where the radius limit depends on the polar angle

  1. Sep 3, 2013 #1


    User Avatar

    I want to calculate a volume integral of a function f(r,theta). The limits for azimuthal and polar angle is of course 0-2*pi and 0-pi, respectively. But the limits for the radius is 0 to an expression depending on the polar angle. Can I simply first integrate the r-part, say from r^4 to 1/5*r^5, and then put in the expression with the polar angle and then integrate?

    See attached Word-file for explanations.

    Many thanks!

    Attached Files:

  2. jcsd
  3. Sep 3, 2013 #2
    What do you mean by "outer radius"?

    That's one ugly looking integral. What is this for?
  4. Sep 3, 2013 #3


    User Avatar


    Thanks for your answer.

    By outer radius I simply mean the radius of the droplet.

    All this is for calculating the vibrational energy of a vibrating droplet. I you want I can send you a more complete document.
  5. Sep 3, 2013 #4
    I'm not really sure if what you're doing makes sense.

    If you wanted to evaluate that integral, you need to sub in the expression for ##r##. The limits of the ##dr## integral are just ##[r,0]##.

    It would look something like this:

    $$\int ^{2\pi} _0 \int ^\pi _0 \int ^r _0 r^4 cos^2 (\theta)sin^3 (\theta) dr d\theta d\phi$$

    And you need to plug in that hideous expression for ##r##.

    I'll be honest with you, the reason you haven't gotten many responses is because this looks like total nonsense.

    A tip: Use the built in math script, nobody wants to download a file from a random user on an internet forum. It has a bit of a learning curve, but you'll get used to it.
  6. Sep 4, 2013 #5


    User Avatar

    Hi again,


    The question is if I can integrate r4 to 1/5r5 and then put in the expression for r (Eq. 2).

    Best Regards
  7. Sep 4, 2013 #6
    Certainly not, you need to sub the expression in before you can integrate anything.
  8. Sep 4, 2013 #7


    User Avatar

    But if I plug in the expression for r, how should I proceed with the r-integration. Should I see it as the function 1 and integrate to the primitive function r? This will differ by a factor of 1/5 as compared to to first integrate r4 to 1/5r5 and then put in the expression for r (Eq. 2).
  9. Sep 4, 2013 #8
    Plug the expression for ##r## in, and integrate as you normally would. For example

    $$\int ^{2 \pi}_0 \int ^r _0 r dr d \theta$$

    Where ##r = krcos(\theta)## would be rewritten as:

    $$\int ^{2 \pi}_0 \int ^r _0 (kr)cos(\theta)dr d\theta$$

    You just sub it in, and THEN do the integration.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook