I have done a bunch of work and simply wish to check that it makes sense.
I have a hollow parabola of height b and base radius b ($ z = \frac{x^2 + y^2}{b}$ bounded by z = b)
1) surface area of paraboloid
$da = 2\pi rds$ (Is an infinitesimally small lampshade)
$r = \sqrt{x^2 + y^2} \implies z = \frac{r^2}{b}$
$ds = \sqrt{dr^2 + dz^2)} = \sqrt{1 + (\frac{dz}{dr})^2)}dr = \sqrt{1 + \frac{4r^2}{b^2}}dr$
$$\therefore Area = \int da = 2 \pi \int_0^br \sqrt{1 + \frac{4r^2}{b^2}}dr = \frac{\pi b^2 (5 \sqrt{5} – 1)}{6}$$
$$\therefore \sigma = Mass/ Area$$
2) Center of Mass (assuming uniform mass density)
Clearly by symmetry that the cross sections are circular, the center of mass will reside at the point (0,0, Zcm)
$Z_{cm} = \frac{1}{M}\int_0^m z dm$
$$dm = \sigma da = \sigma 2\pi r ds = \sigma 2\pi r \sqrt{1 + \frac{4r^2}{b^2}}dr$$
$$\therefore z_{cm} = \frac{1}{M }\int zdm = \frac{1}{M }\int \frac{x^2 + y^2}{b} dm = \frac{\sigma 2 \pi}{M b }\int_0^b r^3 \sqrt{1 + \frac{4r^2}{b^2}}dr$$ Using the identity $cosh^2 - sinh^2 = 1$ (and a little help from wolfram }:p ) I get this integral to come out to be :
$$\frac{\sigma \pi b^3( \frac{2}{15} + \frac{10 \sqrt{5}} {3})}{8M} = b \frac{3(\frac{2}{15} + \frac{10 \sqrt{5}} {3})}{4(5 \sqrt{5} – 1)} = Z_{cm}$$
3) Moment of Inertia about the Z axis
The axis of rotation passes right through the parabola's center of mass, so I don't need to worry about the old parallel axis theorem.
$I_z = \int_0^m r^2 dm$
But $dm = \sigma 2\pi r \sqrt{1 + \frac{4r^2}{b^2}}dr$
$$\therefore I_z = 2 \pi \sigma \int_0^b r^3 \sqrt{1 + \frac{4r^2}{b^2}} dr$$
I am a little confused because this is the same integral as before, and I am not sure if it is supposed to be like that. anyways, I already know the answer to this one now, so I get:
$$I_z = \frac{ \sigma 2 \pi b^4 (1 + 25 \sqrt{5}}{120} = \frac{M b^2 (1 + 25 \sqrt{5})}{10 (5 \sqrt{5} – 1)} $$
I think this result is just, $mb^2$ multiplied by $z_{cm}$ Im not sure if it is because I did something wrong, it is coincidence, or because it is supposed to be like that.
Best Answer
Your calculations seem to be correct. It's not that suprising that the two calculations are closely related, since the integrands $z$ and $r^2$ are related by the equation of the paraboloid (which, by the way, is not a parabola). Some potential improvements:
You don't need hyperbolic functions for the integral; you can integrate by parts, differentiating $r^2$ and integrating the rest to reduce the integral to the previous one that you'd already solved.
You can simplify your expression for $z_{\text{cm}}$; the $3$ and a factor of $2$ cancel (as you seem to have noticed yourself, since your expression for the same integral later on is much simpler); and you can multiply through by $5\sqrt5+1$ to get rid of the square root in the denominator and leave only one root in the numerator. (Also you could take a bit more care balancing parentheses.)