# Inertia of a solid vertical disk about it's diameter

## Homework Statement

Derive the moment of inertia of a solid, uniform density, disk rotating about a diameter using vertical rods.
Diagram: http://i.imgur.com/TQIjz.png

Only the radius of the circle and the inertia of a thin rod is given

## Homework Equations

Inertia of thin vertical rod = m*r^2
σ= area density
M= Total mass
I= inertia
dm= infinitesimal amount of mass
dI= infinitesimal amount of inertia
A= total area

## The Attempt at a Solution

This is the method given in my class. Because calculus 1 is the only prerequisite we can not use double integrals.

I of rod = m*x^2
dI= dm
*x^2
dm=σ*da
da= 2y*dx
σ= M/A
A=pi*R^2
σ= M/(pi*R^2)
dm=M/(pi*R^2) * 2y*dx
dI= M/(pi*R^2) * 2y*x^2 * dx

From here we have to put y in terms of x which I think must be done using the Pythagorean theorem so:
y=(R^2-x^2)^(1/2)

subbing in we get:

dI= M/(pi*R^2) * 2(R^2-x^2)^(1/2) * x^2 * dx

This is why I get stuck because I am not quite sure how to integrate that. What is should come out to be is I = 1/4 MR^2. If anyone can offer an explanation it would be very appreciated.