how much aluminum is need to fit a can and the radius is 0.8 inches and the height is 3,2 inch. including the top and bottom show the work.?

I understand the British English word "fit" to mean "install".

In my non-British English, I would use the word "construct".

The major work involves:

a = π r²

and

c = π d

Not taking into account the thickness of the metal or whether you are giving inside dimensions or outside dimensions

(2 (3.1416 x .64)) + (3.1416 x 1.6 x 3.2)

4.021248 + 16.084992 = 20.10624 in²

or 20,10624 in² in British notation.

(See picture below)

You will need 6.4π in² of aluminum ≈ 20.11 in².............ANS

There are three parts to a can. The top and bottom are circles. The amount of aluminum is equal to the area of these two circles. The side of the can is a rectangle. The length of the rectangle is equal to the circumference of the circle. Its height is equal to the height of the can. The area of the side is product of the circumference of the circle and the height of the can.

For each circle, area = π * 0.8^2 = 0.64 * π

For two circles, area = 1.28 * π

This is approximately 4 in^2.

For the side, area = 2 * π * 0.8 * 3.2 = 5.12 * π

This is approximately 16 in^2.

Total area = 1.28 * π + 5.12 * π = 6.4 * π

This is approximately 20 in^2. I hope this is helpful for you.

Surface area.

Area of 2 ends + area of sides

2*pi*r^2 + 2*pi*r*h

= 2*pi*r*(r+h)

= 2*pi*0.8*(0.8+3.2)

= 6.4*pi square inches

~ 20.1 sq in

Without thickness you can not compute the weight of material.

Total surface area of a cylinder = 2πr(r+h)

you mean form a can? Do you mean area of sheet aluminum?

is that 32 inches? or 3.2 ?

guessing at 3.2

area of a cylinder = 2πr² + 2πrh = 2π0.8² + 2π0.8•3.2 = 4.02 + 16.08 = 20.1 square inches