Proving parallel axis theorem for equilateral triangle moment of inertia

[Starwatcher16]

Imagine we have a Equilateral triangle in the xy plane with one vertex at the origin. The triangle will be rotated about the z axis. The depth of the triangle, d, will be constant.$$B_r=B(r)$$

From the picture I see that $$B_r=\frac{r}{h}B , V=\frac{B_rhd}{2}=\frac{Bdr}{2}$$

First, I need to find I.

$$I=\sum M_i*r_i^2=\int r^2 dm. p=M/V, so, dp=p dv$$

$$I=p\int r^2 dv$$ Substituting using$$dp=p dv and dv=\frac{Bd}{2}dr$$, we get:

$$I=\frac{pBd}{2}*\int r^2 dr=\frac{pBd}{2}*\frac{r^3}{3}.$$ Using p=M/V:
$$I=\frac{Mr^3}{3h}=\frac{Mr^2}{3}.$$ R_total/h=1.

Now, let's say I want to find I for the same shape, but now there is a distance d from the vertex to the z axis.

Everything is the same as the above, except, $$B_r$$ is now a piecewise function, where B(r)=0 for r<d and B(r)=$$B_r=\frac{r-d}{h}B for r>d.$$

Reworking it with the new B_r, I can get the new I. After that I just solve for x in $$I_2=I_1+x$$ to get the parallel axis theorem.

The thing is, I don't know how to generalize this one example to work for all shapes. Any help is appreciated.

 

[Starwatcher16]

Lets pretend that whoever answers my question get free cookies.

Also, now that I think about it, I am not sure this method is valid.

 

[Cyosis]

Yes picking a shape will not help you prove the theorem. What is the definition of the moment of inertia? Note that it is a vector quantity. Now displace the axis a distance d away from the original axis, note that d is a vector too.