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Triple integral with cylindrical coordinates

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Use cylindrical coordinates to evaluate the triple integral

    [tex]\int[/tex][tex]\int[/tex][tex]\int[/tex] [tex]\sqrt{x^2+y^2}[/tex] dV in region E

    where E is the solid bounded by the circular paraboloid z=9-(x^2+y^2) and the xy-plane.

    2. Relevant equations

    knowing that x = rcos[tex]\theta[/tex]
    y= rsin[tex]\theta[/tex]
    z=z
    for a coordinates in clindrical

    3. The attempt at a solution

    I'm not sure how to get the bounds of integration for this problem. I know it has something to do with the paraboloid given but I am not very good at drawing such a figure. I do understand that you need to change the x and y into xcos and ysin in order to integral the problem. any help would be greatly appreciated!
     
  2. jcsd
  3. Oct 28, 2009 #2

    lanedance

    User Avatar
    Homework Helper

    hey MaterWu77, start by transforming your integrand and bounding surface into cylindircal coordinates, things should simplify a fair bit
     
  4. Oct 28, 2009 #3

    Mark44

    Staff: Mentor

    Another relevant equation, and one you didn't show, is r2 = x2 + y2.

    In cylindrical coordinates, dV can be represented as dz r dr d[itex]\theta[/itex]. Take note of that factor of r.

    You region E is such that 0 <= z <= 9 - (x2 + y2), 0 <= r <= 3, and 0 <= [itex]\theta[/itex] <= [itex]2\pi[/itex].

    The resulting integral in cylindrical form looks like this. The question marks are placeholders that you need to fill in.

    [tex]\int_{\theta = ?}^{?} \int_{r = ?}^? \int_{z = 0}^? ?~dz~r~dr~d\theta[/tex]

    Because of the symmetry of your integrand and the region E, you can integrate the part in the first quadrant and multiply your result by 4.
     
  5. Oct 28, 2009 #4
    ok i understand how to get the bounds of integration. would the function that i am integrating be r^2 which would come from the [tex]\sqrt{x^2+y^2}[/tex] ?
     
  6. Oct 28, 2009 #5

    Mark44

    Staff: Mentor

    No, it would be sqrt(r^2), which is just r for your region. There is also the r in dz r dr d(theta). Wasn't sure if you were including that one as well.
     
  7. Oct 28, 2009 #6
    yes i was including the r from the dz r dr d(theta) to get the r^2
     
  8. Oct 28, 2009 #7

    Mark44

    Staff: Mentor

    So this is what you have:
    [tex]\int_{\theta = ?}^{?} \int_{r = ?}^? \int_{z = 0}^? r^2~dz~dr~d\theta[/tex]
    Do you have the limits of integration worked out?
     
  9. Oct 28, 2009 #8
    yes i do! and i have the rest of the problem worked out so thanks a lot for helping me out!
     
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