# given tetrahedral region and evaluate triple integral

• Mar 4th 2011, 03:16 AM
maximus101
given tetrahedral region and evaluate triple integral
Let
$W \rightarrow \Re^3$ be the tetrahedral region
{ $(x, y, z) | x \ge 0, y \ge 0 , z \ge 0; x + y + z \le 1$}
Evaluate
$\iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z$
• Mar 4th 2011, 03:32 AM
CaptainBlack
Quote:

Originally Posted by maximus101
Let
$W \rightarrow \Re^3$ be the tetrahedral region
{ $(x, y, z) | x \ge 0, y \ge 0 , z \ge 0; x + y + z \le 1$}
Evaluate
$\iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z$

$\displaystyle \iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z =$ $\displaystyle \int_{x=0}^1 \int_{y=0}^{1-x}\int_{z=0}^{1-(x+y)} (x+y+2z)\ dz\;dy\;dx$

CB
• Mar 5th 2011, 11:24 AM
maximus101
Quote:

Originally Posted by CaptainBlack
$\displaystyle \iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z =$ $\displaystyle \int_{x=0}^1 \int_{y=0}^{1-x}\int_{z=0}^{1-(x+y)} (x+y+2z)\ dz\;dy\;dx$

CB

Hi thanks for this,

how do I solve that final integral?

and to obtain the limits you have in it?

• Mar 5th 2011, 12:10 PM
AllanCuz
Quote:

Originally Posted by maximus101
Hi thanks for this,

how do I solve that final integral?

and to obtain the limits you have in it?

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 $x + y + z \le 1$ 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,

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

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

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

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

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

And

$y \le 1-x$

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

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,

$0 \le x \le 1$

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, 11:08 PM
CaptainBlack
Quote:

Originally Posted by maximus101
Hi thanks for this,

how do I solve that final integral?

and to obtain the limits you have in it?