gabriels horn centre of mass

Question

I am trying to find the centre of mass of Gabriel's horn with constant density. I have chosen to orient it about the x axis so that its centre of mass lies on the x axis. I am therefore only interested in finding X bar.

It is straight forward to find the mass of the horn to be π and with that you can find X bar to be 2 by taking the integral from 1 to x of dt/t^2 and setting that equal to 1/2. this works because this gives the value of x for which their is equal mass before and after the point.

However, when you apply the normal methods of finding the centre of mass things get strange. Usually I would multiply the integrand by x to weight it and divide the result by the mass but this results in the integral of 1/x which does not converge. Either this formula breaks down in this case or I have done something wrong. I suspect it is me so please explain where I have gone wrong. Thanks.

formula: y = 1/x (rotated about the x axis) 1≤x

Answer 1

The center of mass is infinitely far away. $$V=\iiint \mathrm{d}V=\int_1^{\infty}\int_{-\frac1x}^{\frac 1x}\int_{-\sqrt{\frac{1}{x^2}-y^2}}^{\sqrt{\frac{1}{x^2}-y^2}}\mathrm{d}z\mathrm{d}y\mathrm{d}x=\pi$$ Implying the mass is $\pi$ (if density is equal to one). Then, for the $y$, and $z$ coordinates of the center of mass we do get $$\bar {y} =\iiint \rho y\mathrm{d}V=\int_1^{\infty}\int_{-\frac1x}^{\frac 1x}\int_{-\sqrt{\frac{1}{x^2}-y^2}}^{\sqrt{\frac{1}{x^2}-y^2}}y\mathrm{d}z\mathrm{d}y\mathrm{d}x=0$$ $$\bar {z} =\iiint \rho z\mathrm{d}V=\int_1^{\infty}\int_{-\frac1x}^{\frac 1x}\int_{-\sqrt{\frac{1}{x^2}-y^2}}^{\sqrt{\frac{1}{x^2}-y^2}}z\mathrm{d}z\mathrm{d}y\mathrm{d}x=0$$ However, for $x$ the integral $$\bar {x} =\iiint \rho x\mathrm{d}V=\int_1^{\infty}\int_{-\frac1x}^{\frac 1x}\int_{-\sqrt{\frac{1}{x^2}-y^2}}^{\sqrt{\frac{1}{x^2}-y^2}}x\mathrm{d}z\mathrm{d}y\mathrm{d}x$$ Diverges, meaning the center of mass is indeed on the $x$ axis but infinitely far away.