[tex] \large\begin{gathered} {\underline{\boxed{ \rm {\red{Volume \: \: of \: \: sphere \: = \: \frac{4}{3} \: \pi \: {r}^{3} }}}}}\end{gathered}[/tex]
- r denotes radius of sphere
[tex]\bf \large \hookrightarrow \: \: r \: = \: \frac{Diameter}{2} [/tex]
[tex]\bf \large \hookrightarrow \: \: r \: = \: \cancel\frac{23}{2} [/tex]
[tex]\bf \large \hookrightarrow \: \: r \: = \: 11.5 \: ft[/tex]
Now , substuting the values in formula
[tex]\bf \large \rightarrow \: \: \frac{4}{3} \: \times \pi \: \times \: {(11.5)}^{3} \\ [/tex]
[tex]\bf \large \rightarrow \: \: \frac{4}{3} \: \times \: \pi \: \times 1520.875 \\ [/tex]
[tex]\bf \large \rightarrow \: \: \frac{4}{ \cancel3} \: \times \: \pi \: \times \cancel{1520.875} \: \: \: ^{506.9583} \\ [/tex]
[tex]\bf \large \rightarrow \: \: 4 \: \times \: \pi \: \times \: 506.9583[/tex]
[tex]\bf \large \rightarrow \: \:2027.83 \: \: \pi \: \: ft \: ^{3} [/tex]
Hence , the volume of sphere is 2027.83 π ft³.
