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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we are required to find the center of mass, thus we need to find the point $\displaystyle (\bar {x},\bar {y}, \bar {z})$
where $\displaystyle \bar {x} = \frac {M_{yz}}{m}$, $\displaystyle \bar {y} = \frac {M_{xz}}{m}$, and $\displaystyle \bar {z} = \frac {M_{xy}}{m}$
TPH gave you the formula for $\displaystyle m$, and using $\displaystyle \rho (x,y,z) = \rho$
recall that $\displaystyle M_{yz} = \iiint_E x \rho(x,y,z)~dV$, $\displaystyle M_{xz} = \iiint_E y \rho(x,y,z)~dV$, and $\displaystyle M_{xy} = \iiint_E z \rho(x,y,z)~dV$
By symmetry, $\displaystyle \bar {x} = \bar {y} = 0$, so don't bother finding those integrals
if you have any questions about this stuff, don't be afraid to ask
we just call it a variable, $\displaystyle \rho$ (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?
hI Jhevon,
Thank you very much for your reply.
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.
no, it's not 1, just call it $\displaystyle \rho$
yesNote: center of mass of a solid G with constant density=> does this imply centroid of a solid G?
the region in the xy-plane was a circle. the center of this circle is given by the origin, x = y = 0Also why x bar and y bar are equal to 0 by the symmetry? Can you explain this point more. Thank you very much.
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.
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 $\displaystyle x = r \cos \theta$, $\displaystyle y = r \sin \theta$, and $\displaystyle z = z$, which in this case is $\displaystyle 4 - (x^2 + y^2) = 4 - r^2$