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!

Change order of integration and hence evaluate

  1. Dec 29, 2011 #1

    sharks

    User Avatar
    Gold Member

    The problem statement, all variables and given/known data
    http://s1.ipicture.ru/uploads/20111229/vqL4SkL9.jpg

    The attempt at a solution
    So, i drew the graph of y against x although i don't know initially if a>b or a<b.
    http://s1.ipicture.ru/uploads/20111229/S3y3oSFS.jpg
    Looking at the limits, i realized that a must be less than b. So, the shaded areas are the regions which i need to find.

    Description:
    For y fixed, y varies from y=0 to y=a
    and x varies from x=0 to x = (a/b)√(b^2 - y^2)

    To change the order of integration, d.w.r.t.y first and then d.w.r.t.x. The region is split into 2 parts.

    The region A is the red section on the graph and region B is the blue section on the graph.

    Region A:
    For x fixed, x varies from x=0 to x=(a/b)√(b^2 - a^2)
    and y varies from y=0 to y=a

    Region B:
    For x fixed, x varies from x = (a/b)√(b^2 - a^2) to x=a
    and y varies from y=0 to y=√[b^2 - (b^2.x^2)/a^2]

    At this point, i checked but not sure if my work above is correct. But i'm moving on.

    Thus, after changing the order, the double integral is equal to the sum of the double integral for region A and B.

    I did the calculations in my copybook and the answer has become so complicated. The answer in my notes points to something quite simple.
     
    Last edited: Dec 29, 2011
  2. jcsd
  3. Dec 29, 2011 #2
    It looks like you are integrating over an ellipse, you could do a change of variables and integrate over a circle. And then factor in the Jacobian.
     
  4. Dec 29, 2011 #3

    sharks

    User Avatar
    Gold Member

    That's what i did above. There's no need for Jacobian... My notes don't show anything related to Jacobian at this stage. I don't think you read my partial answer.
     
  5. Dec 29, 2011 #4
    on region A x goes from 0 to what ever that straight vertical lines is, not the equation for that curve. On region B x goes from that vertical line to x=a if your going to do y first.
    And when I meant change of variables, right now we have an ellipse
    [itex] \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 [/itex]
    we could substitute in R=x/a and Q=y/b and then we would get an equation for a circle and then integrate this in the R-Q plane. Not sure if it will help in this case though
     
    Last edited: Dec 29, 2011
  6. Dec 29, 2011 #5

    sharks

    User Avatar
    Gold Member

    The limits you posted above match the ones i used. The vertical line is obtained by substituting y=a into the equation of the ellipse, as i need the x-coordinate for the point of intersection so as to change the order.
    The resulting equation is a bit complex, so i didn't write it on the graph. Here it is:

    Replace y = a into [tex]x = \frac{a}{b}\sqrt (b^2-y^2)[/tex]
    Then, [tex]x = \frac{a}{b}\sqrt (b^2-a^2)[/tex]
    So, i have to find [tex]\int_0^a \! \int_0^\frac{a\sqrt (b^2-y^2)}{b}xy\,dy\,dx[/tex]

    For region A:
    x fixed
    The limits are:
    [tex]0≤x≤\frac{a}{b}\sqrt (b^2-a^2)[/tex]
    [tex]0≤y≤a[/tex]

    For region B:
    x fixed
    The limits are:
    [tex]\frac{a}{b}\sqrt (b^2-a^2)≤x≤a[/tex]
    [tex]0≤y≤\sqrt (b^2-\frac{b^2x^2}{a^2})[/tex]

    When evaluating the double integral for region A:
    [tex]\frac{a^4 (b^2-a^2)}{4b^2}[/tex]

    My problem is with evaluating the double integral for region B.
    I've tried many times and verified my calculations but i can't get the final answer:
    http://s1.ipicture.ru/uploads/20111229/rb3LRWcs.jpg
    And this tells me that i somehow inexplicably have the graph wrong, and i think in that case a>b or a=b? Can someone please help?

    Edit: OK, i see it now. a=b. When y=a, x=0. It is very confusing though and i don't think i would have been able to evaluate this problem without deducing the value of a relative to b, without analysing the answer. Trick question!
     
    Last edited: Dec 29, 2011
  7. Dec 29, 2011 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, they are not. In your limits of integration, x varies from 0 to a. For each x, y varies from 0 to the y value on the ellipse. The integration is over the entire ellipse in the first quadrant. y varies from 0 to b, and for each y, x varies from 0 to [itex](b/a)a\sqrt{a^2- x^2}[/itex]

    [quoter]Description:
    For y fixed, y varies from y=0 to y=a
    and x varies from x=0 to x = (a/b)√(b^2 - y^2)

    To change the order of integration, d.w.r.t.y first and then d.w.r.t.x. The region is split into 2 parts.

    The region A is the red section on the graph and region B is the blue section on the graph.

    Region A:
    For x fixed, x varies from x=0 to x=(a/b)√(b^2 - a^2)
    and y varies from y=0 to y=a

    Region B:
    For x fixed, x varies from x = (a/b)√(b^2 - a^2) to x=a
    and y varies from y=0 to y=√[b^2 - (b^2.x^2)/a^2]

    At this point, i checked but not sure if my work above is correct. But i'm moving on.

    Thus, after changing the order, the double integral is equal to the sum of the double integral for region A and B.

    I did the calculations in my copybook and the answer has become so complicated. The answer in my notes points to something quite simple.[/QUOTE]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Change order of integration and hence evaluate
Loading...