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Triple Integral

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫∫∫z/(1+x^2)dxdydz with the range 0<=z<=y<=x^2<=1

    2. Relevant equations



    3. The attempt at a solution

    I have tried with the limits 0<=z<=1; z<=y<=1 and sqrt(y)<=x<=1, but it doesn't get me the right answer. Can you please help me and maybe give me a step-by-step solution.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 10, 2014 #2

    Simon Bridge

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    The notation does not seem to make any sense in this context.
    The only thing I can think of is:

    $$\int_0^1 dx \int_0^{x^2} dy \int_0^y dz \frac{z}{1+x^2}$$
     
  4. Mar 10, 2014 #3
    What I meant with 0<=z<=y<=x^2<=1, was that the area of integration is D, where D={(x,y,z)∈R^3 : 0<=z<=y<=x^2<=1}. Does that make more sense?

    Anyway, I tried to evaluate your integral and got (1/6)*(13/15 - pi/4), but it should be twice as much according to my key. Your limits seems correct, so I don't know where the error is. Can someone shed some light on this.

    Thanks for your reply by the way!
     
  5. Mar 10, 2014 #4

    vanhees71

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    Hm, it says [itex]x^2<1[/itex]. So what does that mean for the range of [itex]x[/itex]?
     
  6. Mar 10, 2014 #5

    Simon Bridge

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    "area of integration"?
    Suggests the limits are not as right as they look - maybe vanhhees is on to something?
    Perhaps you should tell us what the integral is supposed to find?
    Are you trying to find a volume?
     
  7. Mar 10, 2014 #6

    Range or volume of integration might be more correct, English is not my first language so I hope you understand anyway. The problem just says to evaluate the integral over the set D={(x,y,z)∈R^3 : 0<=z<=y<=x^2<=1}. The answer should be (1/3)*(13/15 - pi/4).
     
  8. Mar 10, 2014 #7
    So the range for x should be from -1 to 1? That makes sense :)
     
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