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Calculate double integer

  1. May 25, 2009 #1
    This was a problem on a final test I took this april in Reykjavík University and I whould be greatful if you could help me with it.

    1. The problem statement, all variables and given/known data

    Let f(x,y)=2x*cos(y^4) be a function and let D be area in R^2 defined by 0≤x≤1 and x^(2/3)≤y≤1.
    Calculate the double integer:
    ∫∫f(x,y)dA

    2. Relevant equations



    3. The attempt at a solution

    ∫dx∫2x*cos(y^4)dy
    I have tried to use substitution but that doesn´t lead me anywhere.
    I also tried to solve it this way...
    ∫dy∫2x*cos(y^4)dx which leads to...
    ∫dy*(x^2*cos(y^4)) and when I add in for x...
    ∫dy*cos(y^4) and if I use substitution now I will get...
    1/4*∫cos(u)du and the final answer isn´t sufficient...
    1/4*(sin(1)-sin(x^(8/3)))

    I would be very greatful if you could help me...
     
  2. jcsd
  3. May 25, 2009 #2

    Dick

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    Carefully draw a sketch of the region. Now when you integrate dx, what will be the upper and lower limits of the integration in terms of y?
     
  4. May 25, 2009 #3
    the upper limits are 1 and lower limits are x^(2/3) in terms of y
    and upper limits are 1 and lower limits are 1 in terms of x
     
  5. May 25, 2009 #4

    Dick

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    That's not what my picture looks like. The integration dx goes along a horizontal line through the region. It's the part above (above being the positive y direction) the curve x^(2/3)=y. Want to try again?
     
  6. May 26, 2009 #5
    I have no clue...:S
     
  7. May 26, 2009 #6

    Dick

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    The region inside of the square 0<=x<=1 and 0<=y<=1 above the curve y=x^(2/3). Pick a value of y and draw a horizontal line. Tell me what the x value is where it crosses the region. It looks to me like it will hit the y-axis first and then the curve, right?
     
  8. May 26, 2009 #7
    The problem requires you to change the order of integration. The limits that are given have you integrating with respect to y first.
     
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