- #1
bdforbes
- 152
- 0
I have a cylinder of height 2a and radius R centered at the origin, and I want to integrate 1/r over it. Ie:
[tex] I=\int_{V}\frac{1}{\left|\texbf{x}\right|}d^{3}\texbf{x}[/tex]
I know how to do it:
[tex] I=\int^{}_{}\int^{}_{}\int^{}_{}\frac{\rho d\phi d\rho dz}{\sqrt{\rho^2+z^2}}[/tex]
But I'm wondering about why the integral is convergent. I understand that the volume elements for a given cylindrical partition depend on rho and thus get very small at the origin, compensating for the unbounded integrand, but if we think about it in cartesian coordinates it doesn't work. In cartesian coords the volume elements for a given partition are the same size everywhere, so as we make the partition finer, there is nothing to compensate for the unbounded integral.
Even if we can't use cartesian coordinates to calculate the integral exactly, in theory shouldn't we be able to numerically calculate the limiting value by using Riemann sums in cartesian coords? I haven't attempted to do this yet, but as far as I can tell, it will diverge.
Am I wrong about that? If so, can anyone give me a intuitive reason for why I should expect the cartesian Riemann sums to converge?
[tex] I=\int_{V}\frac{1}{\left|\texbf{x}\right|}d^{3}\texbf{x}[/tex]
I know how to do it:
[tex] I=\int^{}_{}\int^{}_{}\int^{}_{}\frac{\rho d\phi d\rho dz}{\sqrt{\rho^2+z^2}}[/tex]
But I'm wondering about why the integral is convergent. I understand that the volume elements for a given cylindrical partition depend on rho and thus get very small at the origin, compensating for the unbounded integrand, but if we think about it in cartesian coordinates it doesn't work. In cartesian coords the volume elements for a given partition are the same size everywhere, so as we make the partition finer, there is nothing to compensate for the unbounded integral.
Even if we can't use cartesian coordinates to calculate the integral exactly, in theory shouldn't we be able to numerically calculate the limiting value by using Riemann sums in cartesian coords? I haven't attempted to do this yet, but as far as I can tell, it will diverge.
Am I wrong about that? If so, can anyone give me a intuitive reason for why I should expect the cartesian Riemann sums to converge?