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 of Variables

  1. Oct 9, 2005 #1
    Which change of variable should I use to find:

    [tex]\int\int_D (x^4-y^4) dxdy[/tex]

    Where D is in the first quadrant with"

    [tex]1 \leq x^2-y^2 \leq 3, 2\leq xy \leq 3[/tex]
     
  2. jcsd
  3. Oct 9, 2005 #2

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Try [tex] u = x^2 - y^2 [/tex] and [tex] v = xy [/tex] since this makes the integration region simple.
     
  4. Oct 9, 2005 #3
    It does, but finding the Jacobian and solving for x and y are almost impossible. Or so I tried.
     
  5. Oct 9, 2005 #4

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Yeah, I just gave it a try and it looks that way doesn't it. Not impossible, but certainly a mess on the face of it.

    Edit: Ok, a better choice is [tex] u = x^2 + y^2 [/tex] and [tex] v = x^2 - y^2 [/tex].
     
    Last edited: Oct 9, 2005
  6. Oct 9, 2005 #5

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    It is quite simple:
    In order to change variables, you have been taught to calculate the quantities [itex]\frac{\partial{x}}{\partial{u}},\frac{\partial{x}}{\partial{v}},\frac{\partial{y}}{\partial{u}},\frac{\partial{y}}{\partial{v}}[/itex] then form a matrix, and then calculate the determinant of this matrix, right?

    But to go from the (x,y) description to the (u,v) description should be the INVERSE of going from the (u,v) description to the (x,y) description!.
    Agreed?

    Thus, the matrix you seek should be the inverse of the matrix you get by calculating the quantities [itex]\frac{\partial{u}}{\partial{x}},\frac{\partial{u}}{\partial{y}},\frac{\partial{v}}{\partial{x}},\frac{\partial{x}}{\partial{y}}[/itex]
    In particular, its determinant is the reciprocal of the other matrix' determinant.

    does that help you?
     
    Last edited: Oct 9, 2005
  7. Oct 9, 2005 #6
    Yes, the Jacobian I have. But how do I get the expression x^4-y^4 from the previous substitution?
     
  8. Oct 9, 2005 #7

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Are you talking about [tex] u = x^2 + y^2 [/tex] and [tex] v = x^2 - y^2 [/tex]? If you are then try multiplying u and v.
     
  9. Oct 9, 2005 #8
    No I meant u = x^2-y^2 and v = xy.
     
  10. Oct 9, 2005 #9

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    Well, if you have calculated correctly, your integrand should now be:
    [tex]\frac{x^{4}-y^{4}}{2(x^{2}+y^{2})}=\frac{(x^{2}-y^{2})(x^{2}+y^{2})}{2(x^{2}+y^{2})}=\frac{x^{2}-y^{2}}{2}=\frac{v}{2}[/tex]
     
  11. Oct 9, 2005 #10

    Physics Monkey

    User Avatar
    Science Advisor
    Homework Helper

    Nice, both ways work pretty well.
     
    Last edited: Oct 9, 2005
  12. Oct 9, 2005 #11
    Ah, of course! Man I should slap myself silly for not seeing that.
     
  13. Oct 9, 2005 #12

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    I'm not in a habit of slapping others, so I can't help you out on that account.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Change of Variables
  1. Change of Variables (Replies: 0)

  2. Change of Variables (Replies: 4)

  3. Change in variables (Replies: 7)

  4. Change of variable (Replies: 5)

  5. Change of Variables (Replies: 1)

Loading...