# Thread: Volume under surface, change of variable

1. ## 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, $
\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), $
\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.

2. Originally Posted by cubrikal
(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, $
\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), $
\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.