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Math Help - Volume under surface, change of variable

  1. #1
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    Volume under surface, change of variable

    (This was a question from my calculus final this semester.)

    Find the volume bounded by the surface \sqrt{x}+\sqrt{y}+\sqrt{z}=1 and the first octant.

    Using the change of variables u^2=x,\; v^2=y,\; w^2=z the Jacobian of the transformation is  \begin{vmatrix} 2u & 0 & 0 \\ 0 & 2v & 0 \\ 0 & 0 & 2w \end{vmatrix} = 8uvw .

    So now, I tried to find volume by calculating the triple integral over the constraints  V = \iiint\limits_E {dV} = \iiint\limits_{T(E)} 8uvw {dV}.

    What I'm not sure about is the bounds of the regions; here's what I have:

    For the original region E, <br />
\begin{array}{l}  \\ 0 \le x \le 1 \\ 0 \le y \le (1-\sqrt{x})^2 \\ 0 \le z \le (1 - \sqrt{x} - \sqrt{y})^2 \end{array} , and for the transformed region T(E), <br />
\begin{array}{l}  \\ -1 \le u \le 1 \\ -(1-u) \le v \le (1-u) \\ -(1-u-v) \le w \le (1 - u - v) \end{array} .

    Is that right? I'm confused by the positive/negative signs that result from squaring and square roots.
    Last edited by cubrikal; May 24th 2009 at 09:31 PM. Reason: typo
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  2. #2
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    Quote Originally Posted by cubrikal View Post
    (This was a question from my calculus final this semester.)

    Find the volume bounded by the surface \sqrt{x}+\sqrt{y}+\sqrt{z}=1 and the first octant.

    Using the change of variables u^2=x,\; v^2=y,\; w^2=z the Jacobian of the transformation is  \begin{vmatrix} 2u & 0 & 0 \\ 0 & 2v & 0 \\ 0 & 0 & 2w \end{vmatrix} = 8uvw .

    So now, I tried to find volume by calculating the triple integral over the constraints  V = \iiint\limits_E {dV} = \iiint\limits_{T(E)} 8uvw {dV}.

    What I'm not sure about is the bounds of the regions; here's what I have:

    For the original region E, <br />
\begin{array}{l} \\ 0 \le x \le 1 \\ 0 \le y \le (1-\sqrt{x})^2 \\ 0 \le z \le (1 - \sqrt{x} - \sqrt{y})^2 \end{array} , and for the transformed region T(E), <br />
\begin{array}{l} \\ -1 \le u \le 1 \\ -(1-u) \le v \le (1-u) \\ -(1-u-v) \le w \le (1 - u - v) \end{array} .

    Is that right? I'm confused by the positive/negative signs that result from squaring and square roots.
    the correct bounds are: 0 \leq u \leq 1, \ 0 \leq v \leq 1-u, \ 0 \leq w \leq 1-u-v. your substitution was actually \sqrt{x}=u, \ \sqrt{y}=v, \ \sqrt{z}=w, which implies that (but it's not equivalent to)

    x=u^2, \ y=v^2, \ z=w^2. it's clear now that u,v,w must be non-negative.
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