# Center of mass on an unbounded interval [closed]

I am having a hard time locating the center of mass of $$y=\frac{1}{x}$$ on $$(1,\infty)$$. I used the formula but I get a value of infinity for the mass of the lamina alone. I was able to get an improper integral after using the formula but then it just gave me an infinity as value at the end. I tried to check it with the graph and it seems that it isn't at infinity. How do we solve center of mass when the interval is unbounded? Any idea is greatly appreciated. Thank you so much.

• Where did this question come from? Jul 31, 2020 at 21:24

If your lamina is assumed to have finite mass $$M$$, then it is not possible that its linear density $$\lambda$$ is constant, as you have verified. However, it can have e.g. density $$\lambda(x) = \frac{Mxe^{1-x}}{2\sqrt{1+\frac1{x^4}}}.$$ Then parametrizing your lamina with $$\gamma : [1,+\infty\rangle \to \Bbb{R}^2$$, $$\gamma(t)=\left(t,\frac1t\right)$$ gives that the mass is indeed $$M = \int_\gamma dm = \int_1^\infty \lambda(t)\gamma'(t)\,dt = \frac{M}2\int_1^\infty te^{1-t}\,dt= M$$ and the center of mass is at \begin{align} (x,y) &= \left(\int_\gamma x\,dm, \int_\gamma y\,dm\right) \\ &= \left(\int_\gamma t\lambda(t)\gamma'(t)\,dt, \int_\gamma \frac1t\lambda(t)\gamma'(t)\,dt\right) \\ &= \frac{M}2\left(\int_1^\infty t^2e^{1-t}\,dt, \int_1^\infty e^{1-t}\,dt\right)\\ &= \frac{M}2(5,1). \end{align}