Surface Area and Volume of Sphere
Determining the expressions for the surface area and volume of a sphere requires a more detailed analysis, so we will not discuss those here. Instead, for now you are expected to simply memorize these expressions:

SA of sphere = \(4\pi {r^2}\)

Curved SA of hemisphere = \(2\pi {r^2}\)

Flat SA of hemisphere = \(\pi {r^2}\)

Total SA of hemisphere= \(\left\{ \begin{array}{l} 2\pi {r^2} + \pi {r^2}\\ 3\pi {r^2} \end{array} \right.\)

Volume of sphere = \(\frac{4}{3}\pi {r^3}\)

Volume of hemisphere = \(\frac{2}{3}\pi {r^3}\)
Example 1: Find the surface area and volume of a sphere of radius 5 cm.
Solution: We have:
\[\begin{align}&SA = 4\pi {r^2} \approx 314\,{\rm{c}}{{\rm{m}}^2}\\&V = \frac{4}{3}\pi {r^3} \approx 523.3\,{\rm{c}}{{\rm{m}}^3}\end{align}\]
Example 2: A spherical shell has an inner radius of 5 cm and a uniform thickness of 1 cm. The material to construct this shell costs INR 500 / cm^{3}. What is the cost of building this shell?
Solution: First, we need to determine the volume of the material used in building the shell. Note that the outer radius of the shell will be 5 + 1 = 6 cm. Now, the total volume of the outer sphere (shell + inner hollow) is
\[{V_{outer}} = \frac{4}{3}\pi {\left( 6 \right)^3} \approx 904.3\,{\rm{c}}{{\rm{m}}^3}\]
The volume of the inner hollow is
\[{V_{inner}} = \frac{4}{3}\pi {\left( 5 \right)^3} \approx 523.3\,{\rm{c}}{{\rm{m}}^3}\]
Thus, the volume of the shell alone is
\[\begin{align}&{V_{shell}} = 904.3\,{\rm{c}}{{\rm{m}}^3}  523.3\,{\rm{c}}{{\rm{m}}^3}\\&\;\,\,\,\,\,\,\,\,\,\, = 381\,{\rm{c}}{{\rm{m}}^3}\end{align}\]
Finally, the cost of building the shell will be
\[\begin{align}&\left( {381\,{\rm{c}}{{\rm{m}}^3}} \right) \times \left( {{\rm{INR}}\,\,500/{\rm{c}}{{\rm{m}}^3}} \right)\\ &= {\rm{INR}}\,\,1,90,500\end{align}\]
Example 3: A copper cylinder of base radius 10 cm and height 50 cm is melted to form small hemispherical objects known as pellets, each having a radius of 2 cm. How many pellets will be formed?
Solution: The volume of the copper cylinder is
\[\pi {R^2}H = \pi {\left( {10} \right)^2}\left( {50} \right) \approx 15,700\,{\rm{c}}{{\rm{m}}^3}\]
The volume of each pellet is
\[\frac{2}{3}\pi {r^3} = \frac{2}{3}\pi {\left( 2 \right)^3} \approx 16.747\,{\rm{c}}{{\rm{m}}^3}\]
Thus, the number of pellets formed will be
\[\frac{{15,700\,{\rm{c}}{{\rm{m}}^3}}}{{16.747\,{\rm{c}}{{\rm{m}}^3}}} \approx 937\]