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

Volume of sphere having radius r is given by

The volume is in terms of r hence

We get volume of sphere of some radius by putting value of r in f(r)

The volume of metal can be found by volume of complete sphere having external radius minus the volume of hollow sphere(hollow space) having internal radius

Let r_e = 3.0005 and r_i =3 be external and internal radius respectively

Hence the volume of metal will be given by

⇒ volume of metal = f(r_e) – f(r_i)

⇒ volume of metal = f(3.0005) – f(3)

⇒ volume of metal = f(3.0005) – 36π …(i)

We will find the approximate value of f(3.0005) using derivatives

Let

Let r = 3 and as we have to find for r_e = 3.0005(that is f(3.0005)) hence the small change dr will be 3.0005 – 3 = 0.0005

Using dy = f(r + dr) – f(r)

⇒ dy = f(3 + 0.0005) – f(3)

⇒ dy = f(3.0005) – 36π …since

⇒ f(3.0005) = 36π + dy …(a)

Now we have to find dy

Differentiate y with respect to r

⇒ dy = 4πr²dr

As r = 3 and dr = 0.0005

⇒ dy = 4π(3²)(0.0005)

⇒ dy = (0.0005)36π

Put this value of dy in (a)

⇒ f(3.0005) = 36π + (0.0005)36π

Put this value of f(3.0005) in (i)

⇒ volume of metal = 36π + (0.0005)36π – 36π

⇒ volume of metal = 0.0005 × 36π

⇒ volume of metal = 0.018π

Hence approximate volume of metal in hollow sphere is 0.018π