A covered water tank is in the shape of a cylinder. The tank has a 28-inch diameter and a height of 66 inches. To the nearest square inch, what is the surface area of the water tank

Answer 1

The radius of the cylinder is 1/2 the diameter, so 14 inches.

Assuming the problem is asking for total surface area (i.e. the top and bottom are included), the top and bottom are congruent circles,. So area of one is πr² or 14²π = 196π in². Since there are two such circles the area of the top and bottom is twice that: 392π in².

The lateral surface area is calculated as 2πrh. If you envision cutting the surface of the cylinder along a line perpendicular to the top and bottom, then laying that surface flat, perhaps you can picture the result as a rectangle. One of the sides of the rectangle will actually be the circumference of the top (or bottom). The other side will be the height of the cylinder. Since the circumference is given by 2πr and the height by h, the area of that rectangle is 2πrh= 2π(14)(66) = 1848π. Adding the area of the top and bottom to the lateral surface area: 392π + 1848π = 2240π in². Since you are asked to give the surface area to the nearest square inch, use 3.14159 for π, then round to the nearest square inch. Total Surface Area ≈ 7037 in².

You might have a formula in your text for the total surface area of a right circular cylinder with radius r and height h as:

S.A. = 2πr² + 2πrh. The first term represents the combined area of top and bottom, and the 2nd term the lateral surface area.