# Calculating the moment of Interia

1. Nov 9, 2009

### PenTrik

1. The problem statement, all variables and given/known data
Prove that the moment of inertia of the same region about the x-axis, Ixx, is given by bh^3/21

where y = h(x/b)^2

2. Relevant equations

$$\int(y^2*dA)$$

3. The attempt at a solution

so I first figure that dA has to equal
$$h(x/b)^2 * dx$$
and that y has to equal
$$h^2 * (x/b)^4$$

so when I put it into the intergral, I get something along the lines of
$$\int(h(\frac{x}{b})^2 * h(\frac{x}{b}) ^ 2 * h(\frac{x}{b})*dx)$$
from 0 to b

except when I calculate it out, I get a solution of
$$\frac{h^3*b}{7}$$

$$\frac{h^3*b}{21}$$

somehow I am missing a 1/3.

if it at all helps, I've already calculated the centroids to be <0.75b, .3h>

Last edited: Nov 9, 2009
2. Nov 10, 2009

### phyzmatix

Set dA = dydx and evaluate a double integral over the given region (you have the limits of x as well as y). Remember to first evaluate with respect to the variable limits (your "inner integral") and then with respect to constants (your "outer integral" should have constant limits). That will give you the right answer.

3. Nov 10, 2009

### PenTrik

Ah uh, would my limit for dx be from 0 to b, and for dy to go from 0 to y=h(x/b)^2?

4. Nov 10, 2009

### phyzmatix

Think you made a typo (should be h(x/b)^2) but that is exactly right

EDIT: As I was replying, you fixed it

5. Nov 10, 2009

### PenTrik

Holy ****, I get this now.
Thank you much.

6. Nov 10, 2009

### phyzmatix

HAHAHA!!! Glad to be of help! You see, you can do it