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!

Resolution of a difficult Integral

  1. Apr 27, 2014 #1
    1. The problem statement, all variables and given/known data
    I need to solve this integral

    [itex]\int[/itex][itex]\int[/itex][itex]\int[/itex]dxdzdy+[itex]\int[/itex][itex]\int[/itex][itex]\int[/itex]dxdzdy

    First limits are:

    -[itex]\sqrt{z^{2}-y^{2}}[/itex][itex]\leq[/itex]x[itex]\leq[/itex][itex]\sqrt{z^{2}-y^{2}}[/itex]
    -y[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{y}{2}[/itex]
    -[itex]\frac{2}{3}[/itex][itex]\leq[/itex]y[itex]\leq[/itex]0

    Second limits are:

    -[itex]\sqrt{z^{2}-y^{2}}[/itex][itex]\leq[/itex]x[itex]\leq[/itex][itex]\sqrt{z^{2}-y^{2}}[/itex]
    y[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{y}{2}[/itex]
    0[itex]\leq[/itex]y[itex]\leq[/itex]2

    3. The attempt at a solution

    23ua1vp.jpg

    I think it's getting too difficult to solve, is there an easier way to solve it?
     
    Last edited: Apr 27, 2014
  2. jcsd
  3. Apr 27, 2014 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Why not try using substitutions?
     
  4. Apr 27, 2014 #3
    What do you mean with substitutions? Changing variables?
     
  5. Apr 27, 2014 #4

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Trig substitutions are popular.
    The idea is to get rid of that square-root sign when you go to evaluate the dz part.
     
  6. Apr 27, 2014 #5

    lurflurf

    User Avatar
    Homework Helper

    cylindrical coordinates look helpful

    $$x\rightarrow r \, \cos(\theta)\\
    y\rightarrow r \, \sin(\theta)\\
    z\rightarrow z$$
     
  7. Apr 27, 2014 #6
    If I use cylindrical coordinates x=ρcos(Θ) ; y=ρsin(Θ) ; z=z

    I get the integral of ρ but how do I change the limits?
     
  8. Apr 27, 2014 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    i.e. if a<x<b, then the limits of ρ would be (a/cosθ)<ρ<(b/cosθ)
    ... and your next integral will be wrt θ

    A sketch of the cartesian limits will help you work out what the cylindrical limits should be.
     
  9. Apr 27, 2014 #8
    Excuse me, what is wrt?
     
  10. Apr 27, 2014 #9
    I get:
    [itex]\frac{-2}{3}[/itex][itex]\leq[/itex]ρsin(Θ)[itex]\leq[/itex]0

    -[itex]\frac{2}{3sin(Θ)}[/itex][itex]\leqρ[/itex][itex]\leq[/itex]0

    -ρsin(Θ)[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{ρsin(Θ)}{2}[/itex]

    Θ:[0,2π]

    And in the other:

    0[itex]\leq[/itex]ρsin(Θ)[itex]\leq[/itex]2

    0[itex]\leq[/itex]ρ[itex]\leq[/itex][itex]\frac{2}{sin(Θ)}[/itex]

    ρsin(Θ)[itex]\leq[/itex]z[itex]\leq[/itex]1+[itex]\frac{ρsin(Θ)}{2}[/itex]

    Is that correct?
     
  11. Apr 27, 2014 #10

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    wrt = 'with respect to'
     
  12. Apr 27, 2014 #11
    "with respect to"

    It's just a shorthand notation that we use. Similar to WLOG (without loss of generality) and iff (if and only if).
     
  13. Apr 27, 2014 #12

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    ... and everybody leaps in to help with the easy question ;)
    http://en.wiktionary.org/wiki/WRT
    http://www.internetslang.com/WRT-meaning-definition.asp

    - that would require me to do the problem, the idea is that you do it.
    I'll look in more detail a bit later - meantime:
    It's not always a blind substitution. Which order do you want to do the integration? Like this:
    $$\int \int \int \text{d}\theta \text{d}z \rho \text{d}\rho$$

    Note: if x^2 = z^2-y^2 then isn't y playing the role of a radius? Or is it z?
    Have you sketched out the region of integration?

    Are you unfamiliar with integrating in polar coordinates?
    http://www.math.dartmouth.edu/~m13w12/notes/class7.pdf

    Note: if this is an integral you constructed yourself rather than being handed to you in this form - you should reconsider the way you are dividing the volume up instead. In fact - that may be a good idea anyway.
     
    Last edited: Apr 27, 2014
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: Resolution of a difficult Integral
  1. Difficult Integral (Replies: 10)

  2. Difficult Integral (Replies: 31)

  3. Difficult integral (Replies: 7)

  4. Difficult integral. (Replies: 5)

  5. Difficult Integration (Replies: 22)

Loading...