Let

$\displaystyle W \rightarrow \Re^3$ be the tetrahedral region

{$\displaystyle (x, y, z) | x \ge 0, y \ge 0 , z \ge 0; x + y + z \le 1$}

Evaluate

$\displaystyle \iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z$

Printable View

- Mar 4th 2011, 02:16 AMmaximus101given tetrahedral region and evaluate triple integral
Let

$\displaystyle W \rightarrow \Re^3$ be the tetrahedral region

{$\displaystyle (x, y, z) | x \ge 0, y \ge 0 , z \ge 0; x + y + z \le 1$}

Evaluate

$\displaystyle \iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z$ - Mar 4th 2011, 02:32 AMCaptainBlack
- Mar 5th 2011, 10:24 AMmaximus101
- Mar 5th 2011, 11:10 AMAllanCuz
Well, if we know that x, y and z are all greater then zero and that they are also bounded by a function, in this case $\displaystyle x + y + z \le 1 $ that tells us something aboute the region.

Captain Black has decided to evaluate z first, y next and x last so we will keep that order for consistency (it actually doesn't matter).

Notice if we re-arrange the equation we can get,

$\displaystyle z \le 1 -(x+y) $

That physically means that the function $\displaystyle z = 1-(x+y) $ is our upper bound for z! And since we know that our lower bound is 0, we can say

$\displaystyle 0 \le z \le 1-(x+y) $

Certainly we can do the same for y realizing that in the x-y domain z = 0,

$\displaystyle x + y + z \le 1 \to x+y+0 \le 1$

And

$\displaystyle y \le 1-x $

Thus, $\displaystyle 0 \le y \le 1-x $

Our last variable, x, has to have constant bounds. Well, we know the lower bound is 0, but what about the upper bound? the upper bound happens when both y and z are 0, which from the above equation would yield 1. Thus,

$\displaystyle 0 \le x \le 1 $

All together this yields what Captain black had already posted. On how to evaluate this you evaluate the first integral (dz integral ) then the middle integral (dy integral) then the last integral (dx integral). - Mar 5th 2011, 10:08 PMCaptainBlack