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Triple Integral Evaluation

  1. Dec 6, 2006 #1
    Triple Integral Evaluation (quick and easy)

    1. The problem statement, all variables and given/known data
    \int_{0}^{1} \int_{x^2}^{1} \int_{0}^{3y} ({y+2x^2z})dz dy dx

    2. Relevant equations


    3. The attempt at a solution

    Here is what I got at the end (the LaTeX takes too long to code in here, plus its not showing up):

    59/36 because after integrating the whole thing, and then putting in the very last limits (0 and 1), all of the x's go away leaving just the coefficients which I worked out to get 59/36...can someone please verify this for me? i've checked it twice and got the same solution..it takes about 3 minutes to do if you're a genious (unlike me, so i'm trying to appeal to the math geniuses)..

    Last edited: Dec 6, 2006
  2. jcsd
  3. Dec 7, 2006 #2


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    I get (actually Maple) [itex] \frac{32}{21} [/itex]

    ∫{0}^{3y}(y+2x²z)dz= 9x²y²+3y²

    ∫_{x²}^{1}( 9x²y²+3y²)dy= -3x^{8}-x^{6}+3x²+1

    ∫_{0}^{1}(-3x^{8}-x^{6}+3x²+1) dx= ((32)/(21))
  4. Dec 7, 2006 #3
    did it by hand and got 32/21
  5. Dec 8, 2006 #4


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    I did it by hand, too. :tongue2: I had to type everything and then use the "ctrl c-ctrl v" keys. It's not as easy as it seems...:biggrin:

  6. Dec 8, 2006 #5


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    [tex]\int_0^1\int_{x^2}^1\int_0^3y (y+ 2x^2z)dzdydx[/tex]
    [tex]\int_0^1\int_{x^2}^1\left[yz+ x^2z^2\right]_0^{3y}dydx[/tex]
    [tex]3\int_0^1\int_{x^2}^1(1+ x^2)y^2 dy dx[/tex]
    [tex]\int_0^1\left[(1+3x^2)y^3\right]_{x^2}^1 dx[/tex]
    [tex]\int_0^1(-3x^8- x^6+ 3x^2+ 1)dx[/tex]
    [tex]\left[-\frac{1}{3}x^9-\frac{1}{7}x^7+ x^3+ x\right]_0^1[/tex]
    [tex]-\frac{1}{3}-\frac{1}{7}+ 1+ 1[/tex]
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