Answer
The estimate value of the maximum error in $V$ is $8.948$ ${m^3}$
Work Step by Step
The volume $V$ of a right-circular cylinder of radius $r$ and height $h$ is given by $V = \pi {r^2}h$.
So, the partial derivatives with respect to r and h are
$\frac{{\partial V}}{{\partial r}} = 2\pi rh$ ${\ \ }$ and ${\ \ }$ $\frac{{\partial V}}{{\partial h}} = \pi {r^2}$,
respectively.
Using the linear approximation, Eq. (5) the change in $V$ is estimated to be
$\Delta V \approx \frac{{\partial V}}{{\partial r}}\Delta r + \frac{{\partial V}}{{\partial h}}\Delta h$
Substituting $\frac{{\partial V}}{{\partial r}}$ and $\frac{{\partial V}}{{\partial h}}$ in $\Delta V$ gives
$\Delta V \approx 2\pi rh\Delta r + \pi {r^2}\Delta h$
Since $V = \pi {r^2}h$, so $2\pi rh = \frac{{2V}}{r}$ and $\pi {r^2} = \frac{V}{h}$. Therefore,
$\Delta V \approx \frac{{2V}}{r}\Delta r + \frac{V}{h}\Delta h$
Hence,
$\frac{{\Delta V}}{V} \approx \frac{{2\Delta r}}{r} + \frac{{\Delta h}}{h}$
If $r$ and $h$ each has a possible error of at most $5\%$, then
$\frac{{\Delta r}}{r} = \frac{{\Delta h}}{h} = 0.05$
The estimate of the maximum error in $V$ is
$\frac{{\Delta V}}{V} \approx 2\cdot0.05 + 0.05 = 0.15$
Thus, the estimate percentage of the maximum error in $V$ is $15\%$.
If the diameter is $3.5$ m and the height $6.2$ m, then $r=1.75$ and $h=6.2$. The volume is
$V = \pi {r^2}h = \pi {\left( {1.75} \right)^2}\cdot6.2 \simeq 59.651$
Thus, the estimate value of the maximum error in $V$ is
$0.15\cdot59.651 = 8.948$ ${m^3}$
