1. Limited time only! Sign up for a free 30min personal 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!

Reversing order of integration of double integral qns.

  1. Mar 22, 2010 #1
    [sloved]reversing order of integration of double integral qns.

    1. The problem statement, all variables and given/known data
    pls refer to attached picture.

    2. Relevant equations




    3. The attempt at a solution

    intially upper and lower limits are , x^2 < y< x^3 and -1<x<1
    sketched y=x^2 and y= x^3. => sqrt(y) =x and cube root (y) = x
    divide the area into 3 section.
    new limits of dxdy
    sqrt(y) <x< cube root (y) with 0<y<1 ,
    and 0<x< sqrt(y) with 0<y< -1, ( for -ve x and +ve y portion of x^2 graph)
    and cube root (y)< x< 0 with -1<y<0 ( for -ve x and -ve y portion of x^3 graph)

    but the answer i have shows a different answer. guess i am wrong, but anyone can tell me which part?
    attached is a graph i tried to draw( pardon my IT skills=P)
     

    Attached Files:

    Last edited: Mar 22, 2010
  2. jcsd
  3. Mar 22, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Your picture is good, but you should emphasize that the region is also bounded by a vertical segment at x=(-1). Doesn't that mean some of your lower limits in the x integration need to be -1?
     
  4. Mar 22, 2010 #3
    hmm..to find limit of x integration we draw vertical line? so when i split the area to three sections, their upper limits is cube root y or sqrt y and the lower limit is 0?
    yea, but according to the answer i have, it uses limits of x integration as -1..and i dont understand why?
    could you kindly explain to me?
     
  5. Mar 22, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Take your diagram and draw a segment connecting (-1,1) and (-1,-1). That's the boundary curve that determines your lower limit when you are doing the x<0 parts of the integration.
     
  6. Mar 22, 2010 #5
    opps..i was wrong, we should draw horizontal lines instead. and yup..
    i get jus (for the -ve x portion)
    -1 < x < sqrt y with 0< y< 1
    and -1 < x < -x^3 with -1< y < 0
    using the these lines as a guide..
     
  7. Mar 22, 2010 #6
    opps i only can attach files when i post a thread? cant seem to be able to upload the edited graph here??
     
  8. Mar 22, 2010 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You'll need to express x limits in terms of y. And for both of them, the upper limit should be a negative number, like your graph shows.
     
  9. Mar 22, 2010 #8
    yup.i got them now.
    i get (jus for the -ve x portion)
    -1 < x < -sqrt y with 0< y< 1
    and -1 < x < -cube root y with -1< y < 0

    thanks alot for your guidance=D
     
  10. Mar 22, 2010 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Are you super sure about that -cube root(y)?? Remember if y is negative then cube root(y) is already negative.
     
  11. Mar 23, 2010 #10
    ahah! my wrong..it should be jus cube root of y...
    thx for pointing out!!
    =D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Reversing order of integration of double integral qns.
Loading...