So I was trying to do this using triple integrals (I know there are simpler ways) when I encountered something I can't really justify..here it is
the mass is uniformly distributed so $dm=\rho dv$ and the equation of the cone is $$ x^2 + y^2 = \frac{R^2}{h^2}z^2$$ where R is the radius of circular top of the cone and h is it's height , now the square of the distance from the z axis is $x^2 + y^2 $ and I evaluate the integral using cylindrical coordinates and the equation becomes $r=\frac{R}{h}z$ (taking the positive root)
$$\iiint_v r^3\rho\,dz\,dr\,d\theta$$ where the limits of z are $z=\frac{h}{R}r$ and h , r from zero to R and $\theta$ from zero to $2\pi$
I do the integral and get the correct answer which $\frac{3MR^2}{10}$
here's my question..when I replace the $r^3$ term in the integrand with $r^3=\frac{R^3}{h^3}z^3$ and do the integral (using the same limits and order of integration) I get an incorrect result and I can't explain why this happens
Yes the correct set up is as follows
$$\int_0^{2\pi}\,d\theta\int_0^h dz \int_0^{\frac{R}{h}z} \rho r^3\,dr$$
we can't substitute $r^3=\frac{R^3}{h^3}z^3$ because the $r$ indicated here represent the maximum radius to be considered for any fixed $z$, that is $$r_{MAX}^3=\frac{R^3}{h^3}z^3$$ but for any $z$ we have $0 \le r \le r_{MAX}$.