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Double Integral over a triangular region

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f(x,y) = sin(∏*y^2). Let R be the triangular region on the x-y plane with the vertices at (0,0) (0,1) (.5,1). Consider the solid that is under z = f(x,y) and over the region R. Calculate the volume over that region using double integrals.

    2. Relevant equations



    3. The attempt at a solution
    ∫[itex]^{1}_{0}[/itex]∫[itex]^{1}_{.5}[/itex]sin(∏*y^2) dxdy
    Integrating with respect to x first gives
    .5∫[itex]^{1}_{0}[/itex]sin(∏*y^2)dy.
    After this i'm stuck because I know there is no simple anti-derivative for that function.
    If you reverse the order of integration then, sin(∏*y^2) would have to be integrated at first as well, so not sure what to do.
     
  2. jcsd
  3. Feb 28, 2012 #2

    SammyS

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    Hello mman014. Welcome to PF !

    You are integrating over a rectangular region.

    Write an equation for each line that is on the boundary of region R.
     
  4. Feb 28, 2012 #3
    Thanks.

    Anyways so I figured out that I had the limits wrong, and when I change it I am getting
    ∫[itex]^{1}_{0}[/itex]∫[itex]^{sin(∏*y^2)}_{0}[/itex]sin(∏*y^2)dxdy
    So from there I get
    ∫[itex]^{1}_{0}[/itex]sin(∏*y^2)[x][itex]^{sin(∏*y^2)}_{0}[/itex]
    = ∫[itex]^{1}_{0}[/itex] sin[itex]^{2}[/itex](∏*y^2) dy
    At this point Im stuck again, unless I am doing something wrong or I found the wrong limits again. From here the only way I can think of continuing is using trig substitution and then I would have
    the integral of 1/2 - (cos2(∏*y^2))/2. Which doesn't really help because I would still have to integrate the cos(2∏2y^2) which is the same problem as before.
     
    Last edited: Feb 28, 2012
  5. Feb 28, 2012 #4

    SammyS

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    You continue to have them wrong.

    As I suggested:
    "Write an equation for each line that is on the boundary of region R."​
     
  6. Feb 28, 2012 #5
    Wow so I just realized the mistake I made, wasn't thinking properly., equation of the line is just y = 2x so x = .5y, and I got it from there just using u sub and the rest is easy.
     
  7. Feb 28, 2012 #6

    SammyS

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    Excellent !!!
     
  8. Feb 28, 2012 #7
    Anyways thanks for making me realize my mistake haha. I was staring at the problem for too long, fresh eyes always help.
     
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