A closed circular cylinder has height \(16\) in. and radius \(r\) in. The total surface area is \(A\) sq. in.

Prove that \[\begin{equation*} \frac{dA}{dr} = 4\pi(r+8). \end{equation*}\]

The surface of a closed circular cylinder can be split into three parts, the two discs at the end, and the rectangular central section.

The two discs each have area \(\pi r^2\).

The central section is a curled up rectangle whose length is the height of the cylinder (\(16\)), and whose width is the circumference of the circle (\(2\pi r\)).

Hence the area of the rectangle is \(32 \pi r\), so that \[\begin{equation*} A = 2 \pi r^2 + 32 \pi r. \end{equation*}\] As a result, \[\begin{equation*} \frac{dA}{dr} = 4\pi r + 32\pi = 4\pi(r+8). \end{equation*}\]

Use this result to calculate an approximation for the increase in area when the radius increases from \(4\) to \(4.02\) in., the height remaining constant. You may leave the answer in terms of \(\pi\).

If we increase \(r\) by a small amount \(\delta r\), we increase the surface area \(A\) by a small amount \(\delta A\).

Moreover, we can say \(\dfrac{\delta A}{\delta r} \approx \dfrac{dA}{dr}\) evaluated at that point, and when \(r = 4, \dfrac{dA}{dr} = 48\pi\).

So \(\dfrac{\delta A}{\delta r} =\dfrac{\delta A}{0.02} \approx 48 \pi\), which tells us that \(\delta A \approx 0.96\pi\).