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!

Mass in a plane region

  1. Dec 24, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the mass of the plane region R in the first quadrant of the (x,y)-plane bounded by the hyperbolas

    [itex] xy=1 \,\,\,\,\,\,\,\,\,\, xy=2\,\,\,\,\,\,\,\,\,\, x^2-y^2=3\,\,\,\,\,\,\,\,\,\, x^2-y^2=5 [/itex]

    Assume the density at the point (x,y) is [itex] \rho=x^2+y^2 [/itex]

    2. Relevant equations

    [tex] m=\int \int_R \rho(x,y)dxdy [/tex]

    3. The attempt at a solution

    I am stuck at finding a suitable change of variables to transform this into a "nice" region so I don't have to perform 3 seperate integrals. Even if I took the long way (3 integrals) the point of intersection is not easy to find analytically. What is a clever change of variables that I can use?

    I have tried the following:

    [itex] u=xy \,\,\,\,\,\,\,\,\,\, v=x^2-y^2 [/itex]

    then I can't find a nice expression for [itex] \rho(u,v) [/itex]

    I also tried

    [itex] x=u/v \,\,\,\,\,\,\,\,\,\, y=v [/itex]

    but then solving for v is ugly.

    I even tried

    [itex] u=x^2 \,\,\,\,\,\,\,\,\,\, v=y^2 [/itex]

    which gave another ugly region.

    Please help, thank you.
     
    Last edited: Dec 24, 2011
  2. jcsd
  3. Dec 24, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Your first set of transformations is the one you want. Now consider ##4 u^2+v^2##.
     
  4. Dec 24, 2011 #3
    Im still lost. I want [itex]\rho=x^2+y^2[/itex].
    [itex]4u^2=x^2+y^2[/itex]
    [itex]v^2=x^4-2x^2y^2+y^4[/itex]

    I'm still getting nowhere.
     
  5. Dec 25, 2011 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    How did you get 4u2=x2+y2 from u=xy?
     
  6. Dec 25, 2011 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Isn't [itex]4u^2=4(xy)^2=4x^2y^2\,?[/itex]

    Now, add that to [itex]x^4-2x^2y^2+y^4\,?[/itex]

    Factor that !
     
  7. Dec 25, 2011 #6
    Sorry, I made a typo since I was copy pasting

    [itex] 4u^2=4x^2 y^2 [/itex]
     
  8. Dec 25, 2011 #7
    Ok, round 2, here it goes.

    [itex] 4u^2=4x^2 y^2 [/itex]
    [itex] v^2=x^4-2x^2y^2+y^4 [/itex]
    [itex] 4u^2+v^2=x^4+2x^2 y^2+y^4=(x^2+y^2)^2=\rho^2 [/itex]

    Thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Mass in a plane region
  1. Area of plane region (Replies: 6)

  2. Complex Plane Regions (Replies: 4)

Loading...