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Math Help - [SOLVED] Volume (triple integral)

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Volume (triple integral)

    Find the volume of D= \left\{(x,y,z)\in R^3: z \geq y^2;\  y\geq x^2;\ x \geq z^2 \right\}

    I'm not quite sure where to start. Do I need to split D up into three different volumes? The body is symmetrical, so I would only need to calculate one of the three volumes in that case.

    One of the three volumes is: x^2 \leq y \leq \sqrt{z}, but that doesn't really help me without any boundaries for y.
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  2. #2
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    Quote Originally Posted by Spec View Post
    Find the volume of D= \left\{(x,y,z)\in R^3: z \geq y^2;\ y\geq x^2;\ x \geq z^2 \right\}

    I'm not quite sure where to start. Do I need to split D up into three different volumes? The body is symmetrical, so I would only need to calculate one of the three volumes in that case.

    One of the three volumes is: x^2 \leq y \leq \sqrt{z}, but that doesn't really help me without any boundaries for y.
    put: \frac{x^2}{y}=u, \ \frac{y^2}{z}=v, \ \frac{z^2}{x}=w. then your region will be transfered to the unit cube 0 \leq u \leq 1, \ 0 \leq v \leq 1, \ 0 \leq w \leq 1.

    a simple computation shows that \frac{1}{J}=\frac{\partial(u,v,w)}{\partial(x,y,z)  }=7. thus: \text{the volume of D} =\frac{1}{7} \int_0^1 \int_0^1 \int_0^1 du \ dv \ dw = \frac{1}{7}. \ \ \ \square
    Last edited by NonCommAlg; August 9th 2008 at 07:43 PM.
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