# integration

• Nov 19th 2008, 07:06 AM
koalamath
integration
I need help setting the bounds of this equation
Find the volume of the solid bounded by the planes x=0 y=0 z=0 and x+y+z=6
thank you
• Nov 19th 2008, 07:33 AM
Soroban
Hello, koalamath!

Quote:

Find the volume of the solid bounded by the planes: .$\displaystyle x+y+z\:=\:6,\;x=0,\;y=0,\;z=0$

The solid is in the first octant, bounded by the plane $\displaystyle x + y + z \:=\:6$
. . which has intercepts: (6,0,0), (0,6,0), (0,0,6).

We have: .$\displaystyle V \;=\;\int\int_A z\,dA$ . . . where $\displaystyle A$ is the region in the $\displaystyle x\text{-}y$ plane.

That region looks like this:
Code:

      |     6 *       |:*       |:::*       |:::::*       |:::::::*       |:::::::::*     - + - - - - - * - - -       |          6

We see that $\displaystyle y$ goes from $\displaystyle 0$ to $\displaystyle 6-x$

And $\displaystyle x$ goes from $\displaystyle 0$ to $\displaystyle 6.$

Therefore: .$\displaystyle V \;=\;\int^6_0\int^{6-x}_0 (6-x-y)\,dy\,dx$

• Nov 19th 2008, 10:35 AM
koalamath
Thank you so much
• Nov 19th 2008, 10:44 AM
koalamath
Wait i get 6 as the answer and that's wrong
• Nov 19th 2008, 12:36 PM
Krizalid
Your integration skills or arithmetic are wrong 'cause the answer is 36.