Find the approximate volume of metal in a hollow spherical shell whose internal and external radii are $3 \text{ cm}$ and $3.0005 \text{ cm}$,respectively.

(N/A) Let internal radius $= r = 3 \text{ cm}$ and external radius $= R = 3.0005 \text{ cm}$.
The volume of a hollow spherical shell is given by $V = \frac{4}{3} \pi (R^3 - r^3)$.
Let $f(x) = x^3$. Then $R^3 = f(3.0005)$ and $r^3 = f(3) = 3^3 = 27$.
Using the differential approximation,$\Delta V \approx dV = \frac{dV}{dR} \Delta R$.
Here,$V = \frac{4}{3} \pi R^3$,so $\frac{dV}{dR} = 4 \pi R^2$.
Given $R = 3$ and $\Delta R = 0.0005$,we have:
$dV = 4 \pi (3)^2 \times 0.0005$
$dV = 4 \pi \times 9 \times 0.0005$
$dV = 36 \pi \times 0.0005 = 0.018 \pi \text{ cm}^3$.
Thus,the approximate volume of metal is $0.018 \pi \text{ cm}^3$.