# Surface Area and Volume of Sphere

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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 / cm3. 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

$\frac{{15,700\,{\rm{c}}{{\rm{m}}^3}}}{{16.747\,{\rm{c}}{{\rm{m}}^3}}} \approx 937$

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