# setting up a triple integral to get center of mass.

• November 17th 2010, 02:34 PM
Evan.Kimia
setting up a triple integral to get center of mass.
I have this problem, but I didnt have the correct answer for the mass:
http://www.engr.uconn.edu/~emk09004/...30.31%20PM.png

the correct answer is 1/12 for the mass. Can someone tell me if my setup is correct?

I set it up as a triple integral, with 0 to 1 for my x bound, 0 to 2-x for my y bound, and 0 to 2 for my z bound. Im integrating x dz dy dx.
• November 18th 2010, 06:04 AM
TheEmptySet
Quote:

Originally Posted by Evan.Kimia
I have this problem, but I didnt have the correct answer for the mass:
http://www.engr.uconn.edu/~emk09004/...30.31%20PM.png

the correct answer is 1/12 for the mass. Can someone tell me if my setup is correct?

I set it up as a triple integral, with 0 to 1 for my x bound, 0 to 2-x for my y bound, and 0 to 2 for my z bound. Im integrating x dz dy dx.

I do not think so.

[IMG]19759[/IMG]

The front plane in the tetrahedron is given by
$z=2-2x-2y$
and bound by the coordinate axis. It is always helpful to sketch a graph of the region of integration. How will this affect you limits of integration?