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

Given: A hollow spherical shell having an internal radius of 3cm and external radii of 3.0005 cm

To find: The amount of metal used in the formation of the spherical shell.

Explanation: Let the r and R be the internal and external radii respectively.

So, it is given,

R = 3.0005 and r = 3

Let V be the volume of the hollow shell.

So according to question,

\begin{aligned} &\mathrm{V}=\frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right)\\ &\mathrm{R} \text { and } \mathrm{r} \text { values are substituted, we get }\\ &\mathrm{V}=\frac{4}{3} \pi\left((3.0005)^{3}-3^{3}\right) \ldots (i) \end{aligned}

To get the approximate value of $(3.0005)^3$ , differentiation is carried out

But the integer nearest to 3.0005 is 3,

So 3.0005 = 3+0.0005

So let a = 3 and h = 0.0005

Hence, $(3.0005)\textsuperscript{3} = (3+0.0005)\textsuperscript{3}$

Let the function becomes,

$f(x) = x\textsuperscript{3} \ldots \ldots \ldots (ii)$

Now applying first derivative, we get

$f'(x) = 2x\textsuperscript{2} \ldots \ldots \ldots .(iii)$

Now let  $f(a+h) = (3.0005)\textsuperscript{3}$

Now we know,

$f(a+h) = f(a)+hf'(a)$

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

$f(a+h) = a\textsuperscript{3}+h(3a\textsuperscript{2})$

Substituting the values of a and h, we get

\begin{aligned} &f(3+0.0005)=3^{3}+(0.0005)\left(3\left(3^{2}\right)\right)\\ &\Rightarrow f(3.0005)=27+(0.0005)(3(9))\\ &\Rightarrow(3.0005)^{3}=27+(0.0005)(27)\\ &\Rightarrow(3.0005)^{3}=27+0.0135\\ &\Rightarrow(3.0005)^{3}=27.0135\\ &\text { Thus, the approximate value of }(3.0005)^{3}=27.0135 \text { . } \end{aligned}

So, substituting of the value in equation (i),

\begin{aligned} \mathrm{V} &=\frac{4}{3} \pi(27.0135-27) \\ \mathrm{V} &=\frac{4}{3} \pi(0.0135) \\ \mathrm{V} &=4 \pi(0.0045) \\ \mathrm{V} &=0.018 \pi \mathrm{cm}^{3} \end{aligned}

Hence, the approximate volume of the metal in the hollow spherical shell is $0.018\pi cm^3$.