# Center of Mass and Mass

## Homework Equations

triple integrals, center of mass

## The Attempt at a Solution

I set up the triple integral

x: lower limit 0 upper limit 1
y: lower limit 0 upper limit 2
z: lower limit 0 upper limit 1

∫∫∫ δ(x,y,z) dx dy dz
=> applied the limits for x, y and z, and δ(x,y,z) = 2 +xy -2z (given)

I got mass is equal to 1.

Then I tiried to find x bar (x coordinate of center of mass)
set up this way: ∫∫∫ (x * δ(x,y,z) dx) dz dy divide by Mass (Mass = 1 from previous result)

and got x bar = 1.667, which does not make sense, cuz 0≤ x ≤1 how come it is outside the x limits, or what was the mistake?

## Answers and Replies

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LCKurtz
Homework Helper
Gold Member
How are we to know what your mistake is without you showing us your work? Probably a mistake in algebra or integration.

Mark44
Mentor
Here's your integral, formatted using LaTeX. Click on it to see what I wrote.
$$\int_{z = 0}^1 \int_{y = 0}^2 \int_{x = 0}^1 2 + xy - 2z \ dx \ dy \ dz$$

LCKurtz
Homework Helper
Gold Member
Sorry. I won't click on unknown links.

SammyS
ehild
Homework Helper
∫∫∫ δ(x,y,z) dx dy dz
=> applied the limits for x, y and z, and δ(x,y,z) = 2 +xy -2z (given)

I got mass is equal to 1.
It is wrong. Check your work.

nysnacc
It is wrong. Check your work.
Realized the problem, should be 3, algebraic mistake :P

Thanks