Originally Posted by

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

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

Using the change of variables $\displaystyle u^2=x,\; v^2=y,\; w^2=z$ the Jacobian of the transformation is $\displaystyle \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 $\displaystyle V = \iiint\limits_E {dV} = \iiint\limits_{T(E)} 8uvw {dV}$.

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

For the original region E, $\displaystyle

\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), $\displaystyle

\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.