# center mass of the solid

Find the center mass of the solid bounded by planes $x+y+z=1,x=0,y=0$ and $z=0$, assuming a mass density of $$\rho(x,y,z) = 10 \sqrt{z}.$$

I could not set up the integral!

• Typically, $$\text{mass} = \iiint_\Omega \rho(x,y,z) dV,$$ where $\Omega$ denotes your region. Please provide some thoughts on the problem. We will be glad to furnish more hints. – gt6989b Mar 19 '14 at 19:47

The mass of the volume is

$$M = \int_0^1 dx \, \int_0^{1-x} dy \, \int_0^{1-x-y} dz \, \rho$$

The $x$-coordinate of the COM is

$$\frac1{M} \int_0^1 dx\, x \, \int_0^{1-x} dy \, \int_0^{1-x-y} dz \, \rho$$

etc.

• I need to find (x,y,z) so I get what you mean by that but i still did not get the way to find those! – user131040 Mar 19 '14 at 19:58
• @user131040: Bill Cook's comment applies here as well, but what exactly do you think you are missing? You did ask for the integral to be set up, and here we are. – Ron Gordon Mar 19 '14 at 20:11

You can get bounds of integration by intersecting your various surfaces. For example: $x+y+z=1$ intersected with $z=0$ gives you $x+y+0=1$ so that $y=1-x$. Then intersect with $y=0$ and get $0=1-x$ so that $x=1$.

In the end, this region can be described as follows: $0 \leq z \leq 1-x-y$, $0 \leq y \leq 1-x$, $0 \leq x \leq 1$.

So to find the mass you'd need to integrate... $$\int_0^1 \int_0^{1-x} \int_0^{1-x-y} 10\sqrt{z}\,dz\,dy\,dx$$

Then the moment about the $yz$-plane is... $$\int_0^1 \int_0^{1-x} \int_0^{1-x-y} 10\sqrt{z} \cdot x\,dz\,dy\,dx$$

• I need to find (x,y,z) so I get what you mean by that but i still did not get the way to find those! – user131040 Mar 19 '14 at 19:58
• You already have what you need. I added the moment about the $yz$-plane above. That divided by the mass gives you the $x$-coordinate of the center of mass. You can get the $y$ and $z$ coordinates in a similar manner. – Bill Cook Mar 19 '14 at 20:08