Why the coefficient of volume expansion is three times the linear coefficient expansion?

Question

In volume expansion $\beta\approx3\alpha$, let's suppose we have a rectangular solid with height $h_o$, width $w_o$, and length $l_o$, then $V_o=h_ow_ol_o$

Now, after heating, each side increases by a factor of $\alpha\Delta T$,

$V=h_o(1+\alpha\Delta T)w_o(1+\alpha\Delta T)l_o(1+\alpha\Delta T)$

$\Delta V= V-V_o=h_ow_ol_o(1+\alpha\Delta T)^3-V_o$

We will substitute first equation, since $V_o=h_ow_ol_o$

$\Delta V=V_o(1+\alpha\Delta T)^3-V_o$

$\Delta V= V_o((1+\alpha\Delta T)^3-1)$

$\Delta V= V_o(1+\alpha\Delta T)(1+\alpha\Delta T)(1+\alpha\Delta T)-1)$

$\Delta V=V_o(3\alpha\Delta T+3(\alpha \Delta T)^2+(\alpha \Delta T)^3)$

Now, the last part is the part that I don't get, the last two parts cancel out because if $\alpha$ was small, then we will cancel $\alpha^2 $ because $\alpha^2 $ will get even smaller, and $\alpha^3 $ will be even much smaller, thus we get rid of them, and we are left with $\Delta V=V_o3\alpha \Delta T$.

I don't understand why are we making this assumption here? What if $\alpha$ was big, the last two terms won't really tend to zero... it just doesn't make sense to me in which to why we cancel them out.

Please excuse my ignorance, I would really appreciate it if someone would kindly explain it in a way that it would make sense to me.

Answer 1

The more advanced way (and more precise way) of treating this is to express the linear expansion as $$\frac{dL}{L}=\alpha dT$$ where dL and dT are differentials. Then by the product rule, $$dV=d(hwl)=wldh+hldw+hwdl=hwl\left(\frac{dh}{h}+\frac{dw}{w}+\frac{dl}{l}\right)=hwl(3\alpha dT)$$So,$$\frac{dV}{V}=3\alpha dT$$