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Integration using change of vars

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Evaluate double integral R of sin((x+y)/2)*cos((x+y)/2), where R is the triangle with vertices (0,0), (2,0), and (1,1) using change of variables u = (x+y)/2 and v = (x-y)/2


    2. Relevant equations

    Are my integrands correct? I'm getting the wrong answer...the solution is 1 - (sin 2)/2 but I'm getting 1 + (sin2)/2 - 2cos(1)sin(1)

    3. The attempt at a solution
    (1) Solve for x and y using the change of variables and you get y = u-v and x = u+v. Convert the (x,y) coordinates into (u,v) coordinates and for (u,v) coordinates you get new vertices (0,0), (1,1), and (1,0).

    (2) SOlve for the Jacobain factor --> Factor = 2.

    (3) Set up the integral: so according to the new triangle in (u,v) coordinates we see that v ranges from 0 to 1 and u ranges from v to 1. So we we have the double integral of where integral of v =(0,1), integral of u = (v,1) sin(u)*cos(v) 2dudv.

    I know I have the right set up, but according to the solution, my error lies in setting up the integrals. Please help. I know this is like my 3rd question in two days =(
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 17, 2011 #2

    vela

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    The integral you set up will lead to the correct answer, so you're apparently just not calculating it correctly. Show us how you evaluated the uv integral.
     
  4. Jul 17, 2011 #3
    Attempt is attached
     

    Attached Files:

  5. Jul 17, 2011 #4

    vela

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    Sorry, you haven't made a mistake. You just don't recognize you have the right answer. You want to use the trig identity [itex]\sin 2\theta = 2 \sin \theta \cos \theta[/itex] to get
    [tex]1+\frac{\sin 2}{2}-2\sin 1\cos 1 = 1+\frac{\sin 2}{2} - \sin 2 = 1 - \frac{\sin 2}{2}[/tex]
     
  6. Jul 17, 2011 #5
    Thank you Vela so much!
     
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