# given tetrahedral region and evaluate triple integral

• Mar 4th 2011, 02:16 AM
maximus101
given 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 AM
CaptainBlack
Quote:

Originally Posted by maximus101
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
$\displaystyle \iiint\limits_W \ (x + y + 2z) \mathrm{d}x\,\mathrm{d}y \mathrm{d}z$

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

CB
• Mar 5th 2011, 10:24 AM
maximus101
Quote:

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

CB

Hi thanks for this,

how do I solve that final integral?

and to obtain the limits you have in it?

Kind thanks for your help
• Mar 5th 2011, 11:10 AM
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?

Kind thanks for your help

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 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?

Kind thanks for your help

You start with the innermost integral, do that which then leaves you with a double integral. Now do the inner integral ...

CB