The area function is A(r) = πr^{2}, so

\begin{aligned}dA & = A′(r) dr \\ & = 2πr dr. \end{aligned}

(For small values of dr, this will be a good approximation to ΔA, the true change in A.) Under the conditions of the problem, we get

dA = 2π × 5 × 0.2 ≈ 6.28 cm^{2}.