Question

A soup can have a volume of 54 π in³ and a height of 6 in.

What is the area for the label needed to wrap around the can with no overlaps? Express your answer in terms of π.

Answer 1

18π in ²

Explanation:

This problem bothers on the mensuration of solid shapes, a cylinder.

Given data

volume v = 54 π in³

Height h = 6 in

In this problem we are considering the surface area of the cylinder

(not the total though ) but the curved surface area

Let us solve for the radius of the shape

Volume = πr²h

54π= π*r²*6

54/6= r²

r²= 9

r= √ 9

r= 3in

Area of curved surface = 2πrh

Area = 2πrh

Area =2π*3*6

Area = 18π in ²

Answer 2

The area of the label is 36*pi in²

Explanation:

A can of soup has a shape of a cylinder, therefore it's volume is equal to:

Volume = pi*r²*h

We need to use the formula above to find the radius of the base, because we need to calculate the lateral area of the cylinder, which is given by:

Lateral area = 2*pi*r*h

Applying the data from the problem we have:

54*pi = pi*r²*6

pi*r²*6 = 54*pi

r² = 54*pi/(pi*6)

r² = 54/6 = 9

r = sqrt(9) = 3 in

Therefore the lateral area is:

lateral area = 2*pi*(3)*(6) = 36*pi in²