Step 1

Volume of an hollow spherical shell = $ \large\frac{4}{3}\pi (r_2^3-r_1^3)$

External radii $r_2=3.005 cm$

Internal radii $r_1=3cm$

$ v=\large\frac{4}{3}\pi (r_2^3-r_1^3)$

Let us find the value of $(3.005)^3$

Let y = $f(r)=(3.005)^3$

Let $x=3\: and \: x+\Delta x=3.005$

$ \therefore \Delta x = 3.005-3=0.005$

Hence $ x = 3\: and \: \Delta x=0.005$

for $x=3$ we have

$y=(3)^3$

$=27$

$ dx=\Delta x = 0.005$

Step 2

$y=x^3$

Differentiating on both sides. w.r.t $x$,

$ \large\frac{dy}{dx}=3x^2$

$ = dy=3x^2.dx$

Substituting for $x$ and $dx$ we get

$ dy= \Delta y=3(3)^2 \times 0.005$

$=27 \times 0.005$

$ \therefore dy=\Delta y = 0.135$

Hence $(3.005)^3=y+ \Delta y$

$ = 27+.135$

$ 27.135$

Hence the volume of the hollow spherical shell is

$ v = \large\frac{4}{3} \times \pi \times (27.135-27)$

$ = \large\frac{4}{3} \times \pi \times 0.135$

$ = 0.018 \pi cm^2$