# center of mass of a solid

• August 30th 2007, 02:03 PM
kittycat
center of mass of a solid
Find the center of mass of a solid of constant density rho bounded below by the disk R: x^2 +y^2 < or = 4 in the plane z=0 and above by the paraboloid z=4-x^2 -y^2 .

Thank you very much.
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• August 30th 2007, 04:03 PM
ThePerfectHacker
Quote:

Originally Posted by kittycat
Find the center of mass of a solid of constant density rho bounded below by the disk R: x^2 +y^2 < or = 4 in the plane z=0 and above by the paraboloid z=4-x^2 -y^2 .

Thank you very much.
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The mass is given by,
$\iiint_V \rho(x,y,z)dV = \iint_V \rho dV = \rho \iiint_V dV$
Thus,
$\rho \int_0^{2\pi} \int_0^2 \int_0^{4-r^2} r dz \ dr \ d\theta$
• August 30th 2007, 04:16 PM
Jhevon
Quote:

Originally Posted by kittycat
Find the center of mass of a solid of constant density rho bounded below by the disk R: x^2 +y^2 < or = 4 in the plane z=0 and above by the paraboloid z=4-x^2 -y^2 .

Thank you very much.
==================================

we are required to find the center of mass, thus we need to find the point $(\bar {x},\bar {y}, \bar {z})$

where $\bar {x} = \frac {M_{yz}}{m}$, $\bar {y} = \frac {M_{xz}}{m}$, and $\bar {z} = \frac {M_{xy}}{m}$

TPH gave you the formula for $m$, and using $\rho (x,y,z) = \rho$

recall that $M_{yz} = \iiint_E x \rho(x,y,z)~dV$, $M_{xz} = \iiint_E y \rho(x,y,z)~dV$, and $M_{xy} = \iiint_E z \rho(x,y,z)~dV$

By symmetry, $\bar {x} = \bar {y} = 0$, so don't bother finding those integrals

• August 30th 2007, 05:05 PM
kittycat
Thank you very much.

What is rho(x,y,z) for center of mass of a solid with CONSTANT DENSITY? Is rho(x,y,z)=1?
• August 30th 2007, 05:08 PM
Jhevon
Quote:

Originally Posted by kittycat
Thank you very much.

What is rho(x,y,z) for center of mass of a solid with CONSTANT DENSITY? Is rho(x,y,z)=1?

we just call it a variable, $\rho$ (see TPH's post), by convention. but i guess you could call it something else, say k

notwithstanding, everything TPH and i said so far could have been found in your text book, i don't think we shed anymore light on the subject than you should already be aware of. therefore, i would like you to try the problem and tell us your solution so we can make sure you get it. for instance, do you understand how TPH got his limits of integration?
• August 30th 2007, 05:16 PM
kittycat
hI Jhevon,

I am asking this because I know for a homogeneous lamina ( with constant density => rho(x,y)=1. But I don't know whether this goes to solid with constant density or not!

Note: center of mass of a solid G with constant density=> does this imply centroid of a solid G?

Also why x bar and y bar are equal to 0 by the symmetry? Can you explain this point more. Thank you very much.
• August 30th 2007, 05:55 PM
Jhevon
Quote:

Originally Posted by kittycat
hI Jhevon,

I am asking this because I know for a homogeneous lamina ( with constant density => rho(x,y)=1. But I don't know whether this goes to solid with constant density or not!

no, it's not 1, just call it $\rho$

Quote:

Note: center of mass of a solid G with constant density=> does this imply centroid of a solid G?
yes

Quote:

Also why x bar and y bar are equal to 0 by the symmetry? Can you explain this point more. Thank you very much.
the region in the xy-plane was a circle. the center of this circle is given by the origin, x = y = 0

in essence you are looking for symmetry in the region you're dealing with. for the x, you look for symmetry about any vertical line, for the y, you look for symmetry about any horizontal line.

so let's say the region was half a circle, in the upper half plane, with center the origin and radius 1. how can we get symmetry with this? well, we can cut it into two identical parts with a vertical line (namely, the y-axis), but vertical lines are represented by x = c for some constant c. the y-axis is the line x = 0, thus x bar = 0 here, since we found symmetry about a vertical line. if the circle was centered at (2,0), then we would have x bar = 2

there is no symmetry about any horizontal line in this example, so we would have to compute y bar

so for uniform objects, the center of mass is exactly what its name suggests, the center of the object.
• August 30th 2007, 06:59 PM
kittycat
Hi Jhevon,

I have already completed this question.

(0,0,4/3)

• August 30th 2007, 07:01 PM
Jhevon
Quote:

Originally Posted by kittycat
Hi Jhevon,

I have already completed this question.

(0,0,4/3)

good. that's correct :)
• December 6th 2008, 06:09 PM
aliksa79
Quote:

Originally Posted by Jhevon
we are required to find the center of mass, thus we need to find the point $(\bar {x},\bar {y}, \bar {z})$

where $\bar {x} = \frac {M_{yz}}{m}$, $\bar {y} = \frac {M_{xz}}{m}$, and $\bar {z} = \frac {M_{xy}}{m}$

TPH gave you the formula for $m$, and using $\rho (x,y,z) = \rho$

recall that $M_{yz} = \iiint_E x \rho(x,y,z)~dV$, $M_{xz} = \iiint_E y \rho(x,y,z)~dV$, and $M_{xy} = \iiint_E z \rho(x,y,z)~dV$

By symmetry, $\bar {x} = \bar {y} = 0$, so don't bother finding those integrals

Please can convert the Myz, Mxz, and Mxy to be polar with limit of the integral.
• December 8th 2008, 07:49 AM
Jhevon
Quote:

Originally Posted by aliksa79
Please can convert the Myz, Mxz, and Mxy to be polar with limit of the integral.

?? i am not exactly sure what you are asking. it seems that what you are asking TPH already did...
• December 8th 2008, 10:54 AM
aliksa79
Quote:

Originally Posted by Jhevon
?? i am not exactly sure what you are asking. it seems that what you are asking TPH already did...

HI Jhevon.
I'm asking how to implement the TPH "Myz, Mxz, and Mxy"as polar integration. Can you give of one of those include limit of the integral. However thank you for your reply.
• December 9th 2008, 06:39 AM
Jhevon
Quote:

Originally Posted by aliksa79
HI Jhevon.
I'm asking how to implement the TPH "Myz, Mxz, and Mxy"as polar integration. Can you give of one of those include limit of the integral. However thank you for your reply.

Ok, I gave the formulas for Myz, Mxz and Mxy. The limits of the integrals are exactly the same as those TPH gave in his definition for m. Thus, you reuse those limits, but integrate the formulas i gave.

Myz and Mxz will be zero, by the symmetry of the region, so don't waste your time calculating those. Just find the Mxy.

now, are you wondering how to get the limits? is that it? have you graphed the region to see what is going on? recall that the radius r is measured from the origin, and that $x = r \cos \theta$, $y = r \sin \theta$, and $z = z$, which in this case is $4 - (x^2 + y^2) = 4 - r^2$