Moment of inertia of hollow paraboloid

Question

A hollow paraboloid of length $l$ and radius $r$ with mass $m$ and uniform density $\rho$ has the origin at the base of its circular cross section, and at the center of its cross section. I am trying to find the moment of inertia for the axis that cuts through the paraboloid's center of gravity (its x axis) and the remaining axis y and z which should yield the same values of $I$ since it is axisymmetrical about the y and z axes. I have worked out the solution for what I believe is a filled paraboloid, but I am unsure how to calculate it for a hollow paraboloid, whether it be an infinitesimally small thickness or a uniform thickness $t$. I consider $x$ to be the distance from the origin of the paraboloid along the x axis towards the tip of the paraboloid.

$dm=\rho \pi r^2 dx$

$dI_x=\frac{1}{2}r^2dm$

$I_x=\int_0^l \frac{1}{2}r^2 \rho \pi r^2 dx$

$ = \frac{1}{2}r^4\rho \pi x|_0^l$

$I_y=I_z=\int_0^l (\frac{1}{4}r^2+x^2)\rho \pi r^2 dx$

$=(\frac{1}{4}\rho \pi r^4 x+\frac{1}{3}x^3\rho \pi r^2)|_0^l$

Revised solution (thanks to Math Lover)

$a$ is the constant that determines the shape of the nose cone of the paraboloid

$dI_x=\frac{1}{2}axdm$

$I_x=\int_0^l \frac{1}{2}a^2x^2 \rho \pi dx$

$I_x=\frac{1}{6}a^2x^3 \rho \pi$

$I_y=I_z=\int_0^l (\frac{1}{4}ax+x^2)\rho \pi ax dx$

$I_y=I_z=\int_0^l (\frac{1}{4}a^2x^2\rho \pi+ax^3\rho \pi ) dx$

$I_y=I_z=(\frac{1}{12}a^2x^3\rho\pi+\frac{1}{4}ax^4\rho \pi )$

Second revised solution (once again thanks to Math Lover)

Being a hollow paraboloid, we need to take the surface integral. For a hollow paraboloid we consider the surface area of everything but the base. This results in the equation:

$A=\frac{\pi b}{6h^2} [(b^2+4h^2)^{\frac{3}{2}}-b^3]$

Substituting base radius $b$ and height $h$ with our equivalent variables:

$A=\frac{\pi \sqrt{ax}}{6x^2} [(ax+4x^2)^{\frac{3}{2}}-(ax)^{\frac{3}{2}}]$

Answer 1

We have a hollow paraboloid of length $L$ and radius $R$ along $x-$axis. If the equation of the paraboloid is $y^2 + z^2 = r^2 = ax$,

Then $R^2 = aL \implies a = \frac{R^2}{L}$. We will continue our working with $a$ and substitute this value of $a$ in the end.

As we know, moment of inertia of mass $dm$ around an axis = $dm \cdot r^2$

For a hollow paraboloid, $dm = \rho \ dS$ (where $\rho$ is the uniform density represented as mass per unit surface area)

$dS = \sqrt{1 + (\frac{\partial x}{\partial y})^2 + (\frac{\partial x}{\partial z})^2} \ dA = \sqrt{1 + (\frac{2y}{a})^2 + (\frac{2z}{a})^2} \ dA$

where $dA$ is the area of the projection of $dS$ in $YZ$ plane.

$dS = \frac{1}{a} \sqrt{a^2 + 4y^2 + 4z^2} \ dy \ dz = \frac{1}{a} \sqrt{a^2 + 4r^2} \ r \ dr \ d\theta$

So, $\displaystyle I_x = \frac{2 \pi \rho}{a} \int_0^R r^3 \ \sqrt{a^2 + 4r^2} \ dr$

For moment of inertia around $y$ or $z$ axis, we note that the radius is $\sqrt{r^2 + x^2}$.

So $\displaystyle I_y = I_z = \frac{2 \pi \rho}{a} \int_0^R r \big(r^2 + \frac{r^4}{a^2}\big) \ \sqrt{a^2 + 4r^2} \ dr$