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Find the centre of mass $\overline{P}=(\overline{x},\overline{y},\overline{z}) $ of a unconstrained body $0\le z \le e^{-(x^2+y^2)}$. The density $\delta(x,y,z)$ of the body is constant.

I think we should use cylindricals.

$0\le z \le e^{-r^2}$ and then $r\in[0,+\infty]$, $\phi\in[0,2\pi]$ and $z\in[0,e^{-r^2}]$.

\begin{align*} V(D)&=\int_0^{2\pi}\int_0^{+\infty}\int_0^{e^{-r^2}}rdzdrd\phi= \int_0^{2\pi}\int_0^{+\infty}\Biggl[zr\Biggr]_0^{e^{-r^2}}drd\phi = \int_0^{2\pi}\int_0^{+\infty}re^{-r^2}drd\phi \\ &=\int_0^{2\pi}\Biggl[\frac{-e^{-r^2}}{2}\Biggr]_0^{+\infty}d\phi= -\frac{2\pi}{2}\left(e^{-\infty}-e^0\right)=-\pi(0-1)=\pi \end{align*} now we can calculate $\overline{z}$ \begin{align*} \overline{z}&=\frac{1}{V(D)}\int_0^{2\pi}\int_0^{+\infty}\int_0^{e^{-r^2}}zrdzdrd\phi= \frac{1}{\pi}\int_0^{2\pi}\int_0^{+\infty}\Biggl[\frac{rz^2}{2}\Biggr]_0^{e^{-r^2}}drd\phi =\frac{1}{\pi} \int_0^{2\pi}\int_0^{+\infty}re^{-2r^2}drd\phi \\ &=\frac{1}{\pi}\int_0^{2\pi}\Biggl[\frac{-e^{-2r^2}}{4}\Biggr]_0^{+\infty}d\phi= -\frac{2\pi}{4\pi}\left( e^{-\infty}-e^0\right)=-\frac 12 (0-1)=\frac 12 \end{align*} so $\overline{P}=(0,0, \frac 12)$?

ELEC
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1 Answers1

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Obviously $\bar x=\bar y=0$. Let's call the domain $D$. Then $$\int_Dzdxdydz=\int_{\mathbb R^2} dxdy\int_0^{e^{-(x^2+y^2)}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!zdz=\int_{\mathbb R^2}\frac{e^{-2(x^2+y^2)}}{2}dz=\int_o^{2\pi}d\varphi\int_0^{+\infty}\frac{e^{-2\rho^2}}{2}\rho d\rho=-\frac{\pi}{2}[e^{-2\rho^2}]_0^{+\infty}=\frac{\pi}{2}$$ and $$\int_Ddxdydz=\int_{\mathbb R^2} dxdy\int_0^{e^{-(x^2+y^2)}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!dz=\int_{\mathbb R^2}e^{-2(x^2+y^2)}dz=\int_o^{2\pi}d\varphi\int_0^{+\infty}e^{-2\rho^2}\rho d\rho=-\pi[e^{-2\rho^2}]_0^{+\infty}=\pi$$ so $\bar z = \frac 1 2.$

alex
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