Thermodynamics- thermal expansion coefficient

[Fellowroot]

Homework Statement

Prove that the volume thermal expansion coefficient of a solid is equal to the sum of its linear expansion coefficients in the three directions. $\beta$=$\alpha$x +$\alpha$y+$\alpha$z

For isotopic solid when $\beta$ = 3$\alpha$

Homework Equations

$\beta$=[1/v][dv/dt]p= $\alpha$x +$\alpha$y+$\alpha$z

The Attempt at a Solution

I have looked at several websites but can't seem to get it.
http://www.ami.ac.uk/courses/topics/0197_cte/index.html
http://en.wikipedia.org/wiki/Coefficient_of_thermal_expansion

I thought about just plugging in the $\beta$ but I don't think that is correct.

 

[Mapes]

Consider a cube, and calculate β in terms of changing edge lengths, and then in terms of α. What do you get? What approximations can you make if α is very small?

 

[Fellowroot]

Thanks for your help.

I know see what you mean.

I did of course only understand it until I found this solution online, I wanted to post it here again just in case someone else wants it.
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Let's start with that cube of side L.

If we heat it up, the length of each side will change from L to (L+ΔL), where ΔL = αLΔT.

In other words, the new length is:

L′ = L + αLΔT = L(1 + αΔT)

So the new volume is:

Vo′ = (L′)^3
= [L(1 + αΔT)]^3
= (L^3)(1 + 3αΔT + 3(αΔT)² + (αΔT)^3)
= Vo(1 + 3αΔT + 3(αΔT)² + (αΔT)^3)

Now we make an approximation. Since αΔT is quite small compared to 1, we can safely say that the "3(αΔT)²" term and the "(αΔT)^3" term are negligible compared to the "3αΔT" term. Therefore, to a good approximation,

Vo′ = Vo(1 + 3αΔT)

But we already know, from the definition of β, that:

Vo′ = Vo(1 + βΔT)

So this means that β = 3α.