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Volume integration where the radius limit depends on the polar angle

  1. Sep 3, 2013 #1


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    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


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    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


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    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


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    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.
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