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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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    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.2011-04-20

1 Answers 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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    If $k$ is the constant mass density, that looks OK now.2011-04-20