Answer:
Volume of the rolling pin = 246.09 [tex]in^{3}[/tex]
Step-by-step explanation:
Length of larger cylinder L = 12 in
Diameter of larger cylinder D = 5 in
Diameter of two handles = 1.5 in
Length of two handles = 3 in
Volume of the rolling pin = Volume of two handles + Volume of larger cylinder
Volume of two handles = 2 ([tex]\pi r^{2}L[/tex])
Volume of two handles = 2 (3.14 × [tex]0.75^{2}[/tex] × 3 )
Volume of two handles = 10.59 [tex]in^{3}[/tex]
Volume of larger cylinder = ([tex]\pi r^{2}L[/tex])
Volume of larger cylinder = (3.14 × [tex]2.5^{2}[/tex] × 12 )
Volume of larger cylinder = 235.5 [tex]in^{3}[/tex]
Volume of the rolling pin = 10.59 + 235.5
Volume of the rolling pin = 246.09 [tex]in^{3}[/tex]
