Answer
a. $0.5\%$
b. $5\%$
Work Step by Step
a. Let $D$ be the interior diameter, $h=10m$; we need to have for the volume $\frac{dV}{V}\leq1\%$. Since $V=\pi(D/2)^2h=\frac{5\pi}{2}D^2$, we have $dV=5\pi DdD$ and $\frac{dV}{V}=\frac{5\pi DdD}{\frac{5\pi}{2}D^2}=\frac{2dD}{D}\leq1\%$. Thus $\frac{dD}{D}\leq0.5\%$
b. Let $P$ be the exterior diameter, $h=10m$; we need to have for the side area $\frac{dA}{A}\leq5\%$. Since $A=\pi Ph=10\pi P$, we have $dA=10\pi dP$ and $\frac{dA}{A}=\frac{10\pi dP}{10\pi P}=\frac{dP}{P}\leq5\%$
