Given: a hollow spherical shell with internal radii 3cm and external radii 3.0005 cm

To find: the approximate volume of the metal in the hollow spherical shell

Explanation: Let the internal and external radii of the hollow spherical shell be r and R, respectively.

So it is given,

R = 3.0005 and r = 3

And let the volume of the hollow spherical shell be V.

Then we know,

Now substituting the values of R and r, we get

Now using the differentiation to get the approximate value of (3.0005)³.

But the integer nearest to 3.0005 is 3,

So 3.0005 = 3+0.0005

So let a = 3 and h = 0.0005

Hence, (3.0005)³ = (3+0.0005)³

Let the function becomes,

f(x) = x³………(ii)

Now applying first derivative, we get

f’(x) = 2x²……….(iii)

Now let f(a+h) = (3.0005)³

Now we know,

f(a+h) = f(a)+hf’(a)

Now substituting the function from (ii) and (iii), we get

f(a+h) = a³+h(3a²)

Substituting the values of a and h, we get

f(3+0.0005) = 3³+(0.0005) (3(3²))

⇒ f(3.0005) = 27+(0.0005)(3(9))

⇒ (3.0005)³ = 27+(0.0005)(27)

⇒ (3.0005)³ = 27+0.0135

⇒ (3.0005)³ = 27.0135

Hence the approximate value of (3.0005)³ = 27.0135.

Now substituting this in equation (i), we get

V = 4π(0.0045)

V = 0.018π cm³

Hence the approximate volume of the metal in the hollow spherical shell is 0.018π cm³.