# Moment of inertia of a solid of rotation

1. Jul 1, 2011

### Grand

1. The problem statement, all variables and given/known data
The first quadrant area bounded by the curve
$$x^3+y^3=8$$
is rotated around y-axis to give a solid of rotation. The question asks for an integral which represents the solid's moment of inertia around the axis.

$$I_y=M\frac{\int_0^2x^2ydx}{\int_0^2x^2dx}$$

but seems like I'm wrong. Could someone help me out, please?

2. Jul 1, 2011

### tiny-tim

Last edited by a moderator: Apr 26, 2017
3. Jul 1, 2011

### SammyS

Staff Emeritus
The Moment of Inertial of a cylindrical shell of mass, m, and radius, R, is: m(R2).

4. Jul 1, 2011

### Grand

We need to integrate over the height, that is from 0 to 2, and for each y we need to account for a disk of radius
$$x=\sqrt[3]{(8-y^3)}$$
which moment of inertia is:
$$dI=\rho x^2 dy=\rho (8-y^3)^{2/3}dy$$
For the whole body this is:
$$I=\int_0^2\rho (8-y^3)^{2/3} dy$$
$$I=\frac{m}{V}\int_0^2 (8-y^3)^{2/3} dy$$

I think this is wrong.

5. Jul 1, 2011

### SammyS

Staff Emeritus
The Moment of Inertial of a disk of mass, m, and radius, R, is: (½)m(R2).

However, I think it would be better to use cylindrical shells as tiny-tim recommended.

The last integral you give, $\displaystyle \int_0^2 (8-y^3)^{2/3} dy$ , gives the volume of your solid of revolution.

Last edited: Jul 1, 2011
6. Jul 1, 2011

### tiny-tim

Hi Grand!
This is wrong for two reasons:

i] the moment of inertia of a disc is 1/2 mr2, not mr2

ii] you can't just use ρ instead of m … you need m for the whole disc, which is ρr2dy, so the whole formula for moment of inertia of the disc is 1/2 ρr4dy

But why are you making things so difficult for yourself? As SammyS says, cylindrical shells would be much easier.

Try again.