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Math Help - Triple integral - volume

  1. #1
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    Triple integral - volume

    Hey.
    I need to find the volume of this integral.
    I'm pretty sure I did the second part fine, but I'm really not sure about the jacobian, can I do that?

    And another thing, I know I'm asking a lot of question these past few days, but it's only because I have a big test on Sunday
    Attached Thumbnails Attached Thumbnails Triple integral - volume-scan0004.jpg  
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  2. #2
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    I assume the volume you are trying to find is the set V=\{ (x,y,z)| (x+y+z)^2+(x-2y-z)^2+(2x-y+z)^2 \leq h^2 \} (where h>0).

    The Jacobian theorem says,
    \iiint_R f = \iiint_{\bold{g}^{-1}(R)} (f \circ \bold{g}) \cdot | \det D(\bold{g})|
    Here \bold{g}:\mathbb{R}^3 \to \mathbb{R}^3 is a differenciable vector function.
    And \det D (\bold g) is the determinant of the Frechet' derivative i.e. the Jacobian.
    Of course, \bold{g}^{-1}(R) is the set \{ (x,y,z)\in \mathbb{R}^3 | \bold{g}(x,y,z) \in R \}

    It is conventient to have \bold{g} be such a function so that \bold{g}(x+y+z,x-2y-z,2x-y+z)=(x,y,z).

    If we let A = \left( \begin{array}{ccc}1&1&1\\1&-2&-1\\2&-1&1 \end{array} \right) then A(x,y,z)^T = (x+y+z,x-2y-z,2x-y+z)^T.
    Thus, A^{-1} (x+y+z,x-2y-z,2x-y+z)^T = (x,y,z)^T.

    This means, the function \bold{g}(\bold{x}) = A^{-1}\bold{x} (thinking of vectors in \mathbb{R}^3 as coloums).

    Now, this is convenient because \bold{g}^{-1}(V) = \{ (x,y,z)| \bold{g}(x,y,z) \in V \} = \{ (x,y,z) | x^2+y^2+z^2 \leq h^2 \}

    The volume of V is given by,
    \iiint_V 1 = \iiint_{\bold{g}^{-1}(V)} |\det D(\bold{g})|

    But, \det D(\bold{g}) = \det A^{-1}  = (\det A)^{-1} = 3^{-1}

    Thus the volume is \tfrac{4}{9}\pi h^3 because the volume of a sphere is \tfrac{4}{3}\pi h^3
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    I assume the volume you are trying to find is the set V=\{ (x,y,z)| (x+y+z)^2+(x-2y-z)^2+(2x-y+z)^2 \leq h^2 \} (where h>0).

    The Jacobian theorem says,
    \iiint_R f = \iiint_{\bold{g}^{-1}(R)} (f \circ \bold{g}) \cdot | \det D(\bold{g})|
    Here \bold{g}:\mathbb{R}^3 \to \mathbb{R}^3 is a differenciable vector function.
    And \det D (\bold g) is the determinant of the Frechet' derivative i.e. the Jacobian.
    Of course, \bold{g}^{-1}(R) is the set \{ (x,y,z)\in \mathbb{R}^3 | \bold{g}(x,y,z) \in R \}

    It is conventient to have \bold{g} be such a function so that \bold{g}(x+y+z,x-2y-z,2x-y+z)=(x,y,z).

    If we let A = \left( \begin{array}{ccc}1&1&1\\1&-2&-1\\2&-1&1 \end{array} \right) then A(x,y,z)^T = (x+y+z,x-2y-z,2x-y+z)^T.
    Thus, A^{-1} (x+y+z,x-2y-z,2x-y+z)^T = (x,y,z)^T.

    This means, the function \bold{g}(\bold{x}) = A^{-1}\bold{x} (thinking of vectors in \mathbb{R}^3 as coloums).

    Now, this is convenient because \bold{g}^{-1}(V) = \{ (x,y,z)| \bold{g}(x,y,z) \in V \} = \{ (x,y,z) | x^2+y^2+z^2 \leq h^2 \}

    The volume of V is given by,
    \iiint_V 1 = \iiint_{\bold{g}^{-1}(V)} |\det D(\bold{g})|

    But, \det D(\bold{g}) = \det A^{-1}  = (\det A)^{-1} = 3^{-1}

    Thus the volume is \tfrac{4}{9}\pi h^3 because the volume of a sphere is \tfrac{4}{3}\pi h^3

    So I was suppose to take (Jacobian)^{-1}, right?
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  4. #4
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    Quote Originally Posted by asi123 View Post
    So I was suppose to take (Jacobian)^{-1}, right?
    I am not sure what you mean by this.
    You probably mean the reciprocal of the determinant you computed.
    Maybe you did not understand what I did - I know I do Jacobians my way so maybe it was confusing.
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  5. #5
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    I mean that I calculated that Jacobian with the first equation (in the pic) instead using the second one.
    I switched to u=u(x,y,z), v=(x,y,z) and R=(x,y,z) and as I was saying used the first one instead of the second.
    And my final answer would have also been \tfrac{4}{9}\pi h^3 if I would have taken Jacobian 3^{-1} so I think that was my mistake.
    Attached Thumbnails Attached Thumbnails Triple integral - volume-1.jpg  
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