# Surface area , center of mass, and moment of inertia of paraboloid

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.

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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.)