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Math Help - Triple integral equality

  1. #1
    Member
    Joined
    May 2010
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    241

    Triple integral equality

    Well, Hi. I wanted to know if I did this exercise right. It gives a triple integral and ask me for the other five triple integrals equal to it.

    This is it: \displaystyle\int_{0}^{1}\displaystyle\int_{\sqrt[ ]{x}}^{1}\displaystyle\int_{0}^{1-y}dzdydx

    And this is what I did:

    z=1-y\rightarrow{y=1-z}, \sqrt[ ]{x}=1-z\longrightarrow{z=1-\sqrt[ ]{x}}
    x=(1-z)^2=(z-1)^2

    And the equalities.

    \displaystyle\int_{0}^{1}\displaystyle\int_{\sqrt[ ]{x}}^{1}\displaystyle\int_{0}^{1-y}dzdydx=\displaystyle\int_{0}^{1}\displaystyle\in  t_{0}^{1-\sqrt[ ]{x}}\displaystyle\int_{0}^{1-z}dydzdx=\displaystyle\int_{0}^{1}\displaystyle\in  t_{0}^{(z-1)^2}\displaystyle\int_{0}^{1-z}dydxdz=\displaystyle\int_{0}^{1}\displaystyle\in  t_{0}^{1-z}\displaystyle\int_{0}^{y^2}dxdydx=
    \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1-y}\displaystyle\int_{0}^{(z-1)^2}dxdzdy=\displaystyle\int_{0}^{1}\displaystyle  \int_{0}^{y^2}\displaystyle\int_{0}^{1-\sqrt[ ]{x}}dzdydx

    Is this right?

    Bye and thanks!
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  2. #2
    Banned
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    Oct 2009
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    I haven't checked all the math, but looking at the differentials I believe you're right.
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