A can of StarKist tuna has a volume LaTeX: 18\pi\:cm^318 π c m 3 and a height of 2 cm. Find the area of the StarKist label below the wraps around the entire can and does not overlap. Write your answers in terms of LaTeX: \piπ.

Respuesta :

Answer:

Area of the StarKist label around the can in terms of π = 12π cm²

Step-by-step explanation:

Given;

the volume of a can of StarKist tuna, V = 18 π cm³

height of the can of StarKist tuna, h =  2 cm

To determine the area of the StarKist label that wraps around the entire can and does not overlap, we assume the can to have a shape of a cylinder.

Volume of the can = πr²h

where;

r is radius of the can

h is height of the can

πr² x 2 = 18 π

2r² = 18

r² = 18/2

r² = 9

r = 3

Area around the can = curved surface area of the can (cylinder)

Curved surface area of the can = 2πr × h = 2πrh

Curved surface area of the can = 2πrh = 2π x 3 x 2 = 12π cm²

Area of the StarKist label around the can in terms of π = 12π cm²