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The problem statement, all variables and given/known data

Find the center of mass of the solid of uniform density bounded by the graphs of the equations: Wedge: $x^2+y^2=a^2$ $z=cy\qquad(c>0),\;y \geq 0, z \geq 0$.

Relevant equations

$$M_x= \int y \cdot \rho(x,y) dA$$ $$M_y= \int x \cdot \rho(x,y) dA$$ $$\overline{x} = \frac{M_y}{m}$$ $$\overline{y} = \frac{M_x}{m}$$

The attempt at a solution

I set up all the equations for $M_x, M_y, \overline{x}, \overline{y}$ but I cant seem to realize what the limits of integration are. I can't see how the $z=cy$ comes into play at all. Does it imply a 3 dimensional figure? Do I just integrate with respect to the limits of $x^2+y^2=a^2$?

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Why are $M_x$ and $M_y$ not called $M_y$ and $M_x$, respectively? And why doesn't $\overline{z}$ appear? Finding that would also be part of finding the centre of mass. –  joriki Apr 20 '11 at 5:55
    
Ah, I think see now -- it seems you're thinking in terms of a two-dimensional problem. A "solid" is usually a three-dimensional solid. The answer to your first question is, yes, $z=cy$ bounds the figure in the third dimension (since the other equation describes a cylinder, which is unbounded in $z$). Your symbols for the moments $M_x$ and $M_y$ make sense in a two-dimensional setting, but not in three dimensions, where $x$ and $y$ are not distinguished with respect to $z$, so I'd switch them around to avoid confusion, and add a corresponding $M_z$ and $\overline{z}$. –  joriki Apr 20 '11 at 6:02
    
understood. But still I would like to know how to find the limits of the integration. Is 'c' the same as 'a' ? please attempt to answer if time permits. –  Virtuoso Apr 20 '11 at 6:07
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1 Answer 1

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No, there's no reason for $c$ to be the same as $a$. In a sense, the limits of integration are exactly the equations as written, though some of them are in appropriate form for Cartesian coordinates and other for cylindrical coordinates, so you need to rewrite some of them depending on which coordinates you want to use -- the integration can be performed in either.

If you want to use Cartesian coordinates, you need to solve $x^2+y^2=a^2$ for $x$ or $y$. In this case $y$ is perhaps slightly easier since you have $y\ge0$ but not $x\ge0$. If you want to use cylindrical coordinates $(x,y,z)=(\rho\cos\phi,\rho\sin\phi,z)$, you need to rewrite $z \le cy$ as $z\le c\rho\sin\phi$ and $y\ge0$ as $0\le\phi\le\pi$.

In the Cartesian case you also need limits for $x$; you can get those by considering for which values of $x$ the limits for $y$ make sense.

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in other words this is my integral for solving Mx... tinyurl.com/3vu92gn –  Virtuoso Apr 20 '11 at 6:27
    
a) You forgot the factor of $x$ (or $y$ in your notation). b) You omitted the restriction $y\ge0$. –  joriki Apr 20 '11 at 6:30
    
Corrected: tinyurl.com/42t78q6 –  Virtuoso Apr 20 '11 at 6:36
    
If $k$ is the constant mass density, that looks OK now. –  joriki Apr 20 '11 at 7:09
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