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!

Triple Integral: Convert from Cartesian to Cylindrical Coordinates

  1. Nov 10, 2008 #1
    1. The problem statement, all variables and given/known data
    This is my last question about triple integrals in cylindrical coordinates.

    Evaluate the integral by changing to cylindrical coordinates:

    [tex]\int _{-3}^3\int _0^{\sqrt{9-x^2}}\int _0^{9-x^2-y^2}\sqrt{x^2+y^2}dzdydx[/tex]


    2. Relevant equations
    In cylindrical coordinates, [tex]x^2+y^2=r^2[/tex] and [tex]x=r\cos{\theta}[/tex].


    3. The attempt at a solution
    My converted integral looks like this:

    [tex]\int _0^{\pi }\int _0^{\sqrt{\frac{18}{1+\text{Cos}[\theta ]^2}}}\int _0^{9-r^2}r^2dzdrd\theta [/tex]

    This isn't quite right. Any ideas?
     
  2. jcsd
  3. Nov 10, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi daveyman! :smile:

    Hint: If |x| ≤ 3 and |y|2 ≤ 9 - x2, then … r what, and θ what? :wink:
     
  4. Nov 10, 2008 #3
    [tex]r=\sqrt{18-x^2}[/tex] which, in cylindrical coordinates is [tex]r=\sqrt{18-r^2\cos^2 \theta}[/tex].

    So would the new integral be

    [tex]\int _0^{\pi }\int _0^{\sqrt{18-r^2*\text{Cos}[\theta ]^2}}\int _0^{9-r^2}r^2dzdrd\theta [/tex]?
     
  5. Nov 10, 2008 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    No!

    Get some graph paper and shade in all the points for which |x| ≤ 3 and |y|² ≤ 9 - x². ​
     
  6. Nov 10, 2008 #5
    It is a circle at the origin with radius of three. So the integral will be

    [tex]\int _0^{\pi }\int _0^3\int _0^{9-r^2}r^2dzdrd\theta[/tex]

    which yields the correct answer.

    Thank you!
     
  7. Nov 10, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yup! :biggrin:

    The moral of this … always draw the region first …

    it's almost impossible to work out the limits without a diagram

    (though of course, you must then prove them without a diagram … but that's much easier once the diagram has told you the answer! :wink:)​
     
  8. Jul 11, 2011 #7
    Hey- I know this is quite an old thread... But just wondering how you know that the dθ part is between 0 and pi??
    Cheers
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?