Recent content by cotufa

THanks for the help

Ok so 1 works for z0 If I factor a z-1 how would the equation look?

Homework Statement Solve z^3 - 3z^2 + 6z - 4 = 0 The Attempt at a Solution I tried factoring a z and quadratic equation but went nowhere Input apreciated

Thank you

I can't seem to see how this simplifies, having an algebra brain fart 2(2-x)^2-x(2-x)^2- 2/3 (2-x)^3 (2-x)(2-x)^2- 2/3 (2-x)^3 1/3(2-x)^3

Yes to the volume part, but I have been to every class and this is the first time I even found rho in a problem I guess this would be a physics two problem? Thanks everyone in the thread for the input

so \rho h \pi=M/R^2? Those equations weren't even given to us :|

Sure, just getting used to putting in the tex code Btw I have no idea if I am even on the right track \int_0^{2\pi} \int_0^R \int_0^h \rho r^2 \,dz\,dr\,d \theta = \rho * h \int_0^{2\pi} \int_0^R r^2 \,dr\,d \theta = (\rho * h)/3 \int_0^{2\pi} r^3 \,d \theta = 1/3...

height, so it has to be from 0 to something, just don't know what the height should be to get the 1/2 * MR^2

So the density function for the cylinder is the same as for the sphere? r^2\rho dV What would the limit of integration be for the dz part of the triple integral? Sorry if the questions seem trivial, I need more calc 3 The K part was just a constant

for the density function of a triple integral of the cylinder in cylindrical coordinates, what would be the limits of integration for the dz part? \int_0^{2\pi} \int_0^R \int_0^? (Kr) r \,dz\,dr\,d \theta And would the density function be K\sqrt {x^2 + y^2} = Kr ?

I just need the first part, I understand now the triple integration and the coordinate systems, I just don't seem to know where to start from eg to derive the moment of inertia of a solid sphere of uniform density and radius R. dI = r_{\perp}^{2} dm = r_{\perp}^{2} \rho dV I need this...

Can you post the derivation of it in cylindrical coordinates?

I think I found the derivation in spherical coordinates https://www.physicsforums.com/showthread.php?t=159293 I got to look for the cylindrical coordinates, and then I got to study everything about the two systems :DEdit the cylinder would end in 1/2 *MR^2

Thanks for the quick reply I guess it is not really a proof, but the question was worded vague so I really cannot improve it The moment of inertia of a solid sphere is 2/5 MR^2 But how would I go at calculating it using the cylindrical coordinates or spherical coordinates? I been...