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Double integral new coordinate system calculation

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data
    This is a 2 part question. I'm fine with the first part but the 2nd part I'm struggling with.

    The first part asks us to calculate the double integral,
    [itex]\int\int[/itex]Dx2dA
    for, D = {(x,y)|0≤ x ≤1, x≤ y ≤1}
    For this part I got an answer of 1/4.

    For the 2nd part we introduce a new coordinate system for D,
    x = 1-st, 0≤ s ≤1
    y = s, 0≤ t ≤1

    3. The attempt at a solution
    [itex]\int\int[/itex]Dx2dA
    = [itex]\int_0^1\int_0^1[/itex](1-st)2dt ds
    = [itex]\int_0^1\int_0^1[/itex](1-2st+s2t2)dt ds
    = [itex]\int_0^1[/itex]t-st2+[itex]\frac{1}{3}[/itex]s2t3 ds, from t=0 to t=1
    = [itex]\int_0^1[/itex]1-s+[itex]\frac{1}{3}[/itex]s2 ds
    = s-[itex]\frac{1}{2}[/itex]s2+[itex]\frac{1}{9}[/itex]s3, from s=0 to s=1
    = 1-[itex]\frac{1}{2}[/itex]+[itex]\frac{1}{9}[/itex]
    = [itex]\frac{11}{18}[/itex]

    I feel like,
    dA [itex]\neq[/itex] dt ds
    I'm not sure what it equals though. I thought I could use polar coordinates but I don't have a constant radius.
     
  2. jcsd
  3. Apr 2, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Okay, so you got this wrong. In more detail this is
    [tex]\int_{x= 0}^1\int_{y= x}^1 x^2dydx= \int_0^1 (1- x)x^2 dx[/tex]
    What do you get for that?

    You can't just replace "dydx" with "dtds". Since you are now measuring distance differently, the area will be measured differently. Use the Jacobian:
    [tex]\left|\begin{array}{cc}\frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \\ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t}\end{array}\right|[/tex][tex]= \left|\begin{array}{cc}-s & -t \\ 0 & 1 \end{array}\right|= -s[/tex]

    So you have to replace "dxdt" with -s dsdt as well as replacing [itex]x^2[/itex] with [itex](1- st)^2[/itex].

    Yes, that is the difficulty. Use the Jacobian, as I said before.

     
  4. Apr 2, 2013 #3
    Sorry, the lower bound on y for the first part should have been 1-x but I should be able to work through the rest of the reply and get an answer. Thank you for the detailed reply.
     
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