Moment of Inertia around z axis

Hello I am having difficulty with the following;

I am wanting to find I, the moment of inertia about the z axis of the region that is bounded by the paraboloid $z=x^{2}+y^{2}$ and the $z=1$ plane, where the density is proportional to the distance from the z axis.

Here is what I have tried:

I thought maybe I could use the formula

$I= \iiint_{D}(x^2+y^2)\rho(x,y,z)dV$

and use that since $\rho(x,y,z)$ is proportional to distance from the $z$ axis, then for some constant $K$ we have $\rho(x,y,z)=K$(distance from z axis)

But I am not sure which distance from the z axis to use, or which formula to use.

Anyway, assuming the above is correct, then I would have

$I= \iiint_{D}K(x^2+y^2)(distance from z -axis)dV$

and the region D would be determined by knowing we are bound by the paraboloid and z=1 plane.

But I am stuck on it so far, can anyone please help me? Is it the correct approach? and how should I find distance to z axis? Maybe I don't need triple integrals? Anyways, I have tried my best to work on it, but I am not making any progress. I think if someone helped to explain then I could understand it for the future

Thank you

Update:

I am still working on this problem days after. I really wish I could just get some help so I can study it and move on!

Here is what else I have tried;

If we try cylindrical coordinates,

then $z=r^{2}$ and z goes to the plane ie $z =1$

\theta, would go from $0$ to $2pi$, and $r$ from $0$ to $1$

(however I still don't know how I could represent $K\rho $

but I would then have $\iiint_{D'} (r^3)(K(\rho))dzdrd\theta$ my only guess would be that K(\pho)=K(r)

giving $$K\iiint_{D'}r^{4}dzdrd\theta$$ But I don't know if so far it is the right approach? Please anyone?

1 Answer

Work in cylindrical coordinates $(r,\theta,z)$. The element of volume is $r\,dr\,d\theta\,dz$. The distance of a point to the $z$ axis is just $r$, and the density $\lambda r$. The paraboloid is $r^2=z$ or $r=\sqrt z$.

So the mass is

$$M=\int_{z=0}^1\int_{\theta=0}^{2\pi}\int_{r=0}^{\sqrt z}\lambda r\,r\,dr\,d\theta\,dz=\lambda2\pi\int_{z=0}^1\left.\frac{r^3}3\right|_0^{\sqrt z}dz=\lambda2\pi\left.\frac2{3\cdot5}z^{5/2}\right|_0^1=\lambda\frac{4\pi}{15}.$$

And the moment of inertia around $z$

$$I_{zz}=\int_{z=0}^1\int_{\theta=0}^{2\pi}\int_{r=0}^{\sqrt z}r^2\lambda r\,r\,dr\,d\theta\,dz=\lambda2\pi\int_{z=0}^1\left.\frac{r^5}5\right|_0^{\sqrt z}dz=\lambda2\pi\left.\frac2{5\cdot7}z^{7/2}\right|_0^1=\lambda\frac{4\pi}{35}.$$

Then

$$I_{zz}=\frac37M.$$