formula:

$V(l,r,R)=l(R^2-r^2).$

According $to$ your measurements, the volume $of$ toilet paper $is(to$ the nearest cubic $mm)$:

$V(93,21,56)=,m{m}^3$

However, since your initial measurements were only accurate $to$ the nearest $mm,$ the true volume $of$ toilet paper may

$be$ different.

The total differential approximation can $be$ used $to$ estimate how the measured volume $V(93,21,56)$ differs from the

true value $ofV(l,r,R)as$:

$V(l,r,R)$~~$V(93,21,56)+\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{l}}(93,21,56)(l-93)+\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{r}}(93,21,56)(r-21)+\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{R}}(93,21,56)(R-56)$

$We$ calculate the partial derivatives $(to$ the nearest integer$)$:

$\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{l}}(93,21,56)=$

$\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{r}}(93,21,56)=$

$\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{R}}(93,21,56)=$

Since $we$ measured $l,r$ and $Rto$ the nearest $mm,$

and

Using the integer approximations above together with the triangle inequality,

$|V(l,r,R)-V(93,21,56)||\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{l}}(93,21,56)||(l-93)|+|\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{r}}(93,21,56)||(r-21)|+|\frac{\text{d}\text{e}\text{l}\text{V}}{\text{d}\text{e}\text{l}\text{R}}(93,21,56)||(R-56)|$

And $so$ our measured volume $V(93,21,56)$ differs from the true volume $by$

$|V(l,r,R)-V(93,21,56)|,m{m}^3$