Who likes 2D donuts?

The area of the chocolate part is:

$A = \pi(3r)^2 - \pi r^2$ $\implies A = 8\pi r^2$

$r = 6 \implies A = 288 \pi$

Now, when the $r$ increases by $p$ , then:

$A_n = 8\pi(6+p)^2$

The increase in area is therefore:

$\Delta A = A_n - A = 8\pi(6+p)^2 - 288 \pi$ $\Delta A = 288 \pi + 96p \pi + 8 \pi p^2 - 288 \pi$ $\Delta A = 96p \pi + 8 \pi p^2$

Neglecting the $p^2$ term as $p$ is considered small leads to: $\boxed{\Delta A \approx 96p \pi}$

First we calculate the area of the chocolate part of the donut:

$A = 9πr^{2} - πr^{2} = 8πr^{2}$

Differentiating gives:

$\dfrac {dA} {dr} = 16πr$

when $r = 6$ , $\dfrac {dA} {dr} = 96π$

Since $p$ is small we can say that

$\dfrac {δA} {δr} ≈ \dfrac {dA} {dr}$

$\dfrac {δA} {p} ≈ 96π$

Therefore $δA ≈ 96πp$