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The Problem

Let $C$ be the cone $$C:= \{ (x,y,z) \in \mathbb{R^3}: \sqrt{x^2+y^2}\leq z \leq 1$$ Assume that $C$ has a constant mass density and find the z coordinate of the center of mass.

The work I have done:

\begin{align*} M_{xy}&=\iint_{R}\int_{\sqrt{x^2+y^2}}^{1} z \,\delta \,dx \,dy \,dx\\ &=\iint_{R} \delta \,\frac{z^2}{2} \Bigg|_{\sqrt{x^2+y^2}}^{1} \,dy \,dx\\ &=\iint_{R} \delta \,\Big(\frac{1}{2}-\frac{\sqrt{x^2+y^2}}{2} \Big) \,dy \,dx \end{align*}

Switching to polar coordinates now and pulling out the constants: \begin{align*} &=\frac{\delta}{2}\int_{0}^{2\pi}\int_{0}^{1} \big(r-r^3 \big) \,dr \,d\theta\\ &=\frac{\delta}{2} (\frac{\pi}{2})=\frac{\pi}{4}\,\delta\,.\\ \\ M&=\iint_{R}\int_{\sqrt{x^2+y^2}}^{1} \,\delta \,dx \,dy \,dx\\ &=\iint_{R} \delta (1-\sqrt{x^2+y^2}) \,dy \,dx \end{align*}

Switching again to polar: \begin{align*} &=\delta \int_{0}^{2\pi} \int_{0}^{1} (1-r) \,r \,dr \,d\theta\\ &=\delta \int_{0}^{2\pi} \frac{1}{6} d\theta = \frac{\pi}{3}\,\delta \end{align*} Therefore:

$$\bar{z}=\frac{M_{xy}}{M} = \frac{\pi\delta}{4}\cdot\frac{3}{\pi\delta}=\frac{3}{4}$$

Is my answer correct?

Edit: I also know that I could have switched to cylindrical from the beginning. If I did, would the integral I need to evaluate be $$M=\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} \,r \,dz \,dr\,d\theta$$

and

$$M_{xy}=\int_{0}^{2\pi}\int_{0}^{1}\int_{r}^{1} \,rz \,dz \,dr\,d\theta\,?$$

math-physicist
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user8290579
  • 886

2 Answers2

1

This is a cone of base radius $1$, and height $1$. The center of mass formula is

$ \overline{z} = \displaystyle \dfrac{ \iiint z dV } {\iiint dV} $

The numerator is

$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^3)dz = \dfrac{\pi}{2} $

The denominator is

$ \displaystyle \iiint z dV = \int_{z=0}^1 (2 \pi z^2)dz = \dfrac{2\pi}{3} $

Hence,

$ \overline{z} = \dfrac{ \dfrac{1}{2} }{ \dfrac{ 2}{3}} = \dfrac{3}{4} $

Hosam Hajeer
  • 21,978
0

The cylindrical integral is slightly off.

You are computing the center of mass for a Cone of height 1 and base radius 1. So the cylindrical integral will be $$ M=\int_{0}^{1}\,dz\int_{0}^{2\pi}d\theta\int_{0}^{z} \,r \,dr\, $$ You could integrate each $z$ plane by eye. $$ M=\int_{0}^{1}\,dz \pi z^2 =\frac 13\pi\\ M_z=\int_{0}^{1}\,dz \pi z^2 z=\frac 14\pi $$ $$ \bar z = M_z / M=3/4 $$

Binxu Wang 王彬旭
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