The motion of a solid object can be analyzed by thinking of the mass as concentrated at a single point, the center of mass. If the object has density enter image description here at the point (x,y,z) and occupies a region W, then the coordinates enter image description here of the center of mass are given by enter image description here Assume x, y, z are in cm. Let C be a solid cone with both height and radius 2 and contained between the surfaces enter image description here and enter image description here If C has constant mass density of 5 g/cm^3, find the z-coordinate of C's center of mass.

I can not find the right limits for and the integral function, so please help me to find it right


Work in cylindrical coordinates because of the symmetry of the problem.

This is an upside-down cone, with the point at the origin, with surface $z = r$, and "base" of radius $r=2$ at height $z=2$.

I think of the $z$ integration as going from the surface $z=r$ up to the base at $z=2$. The radius goes from $0$ to $2$. The volume element can be taken as $2\pi r dr dz$. So we have $$\overline{z} = \frac{\int_{r=0}^2{\int_{z=r}^2 z\,2\pi r dr dz}}{\int_{r=0}^2{\int_{z=r}^2 2\pi r dr dz}} = \frac{4\pi}{8\pi/3}= {3\over 2}\,.$$

(The integral in the denominator is just the total volume. Note that the uniform density of the material cancels out.)

The $8\pi/3$ in the denominator is what you would get from the usual cone volume formula ${1\over 3}\pi R^2h$. The final answer is a point one-fourth of the way from the base to the point. You can check this on Wikipedia for example.

  • $\begingroup$ @user131040 By the way, I should have noted that I like to put the mass integral in the denominator, even when I know a stock formula for the shape in question, because doing so can help me diagnose an error in my limits of integration, volume element, etc. By not using the stock formula, I can check for consistency between the result of the mass integration and the result of the stock formula. That gives me extra confidence in the overall result. (It doesn't prove it right, of course. But to me it's worth the extra effort.) $\endgroup$ – Jason Zimba Mar 28 '14 at 13:55

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