Home › Geometry - solid figures

Find the curved and visible surface area of cylinders

Count only the surfaces named in the question

A cylinder's curved surface has area $2 π r h = π d h$ , and each circular base has area $π r^{2}$ . For a composite solid, do not count faces hidden by contact with another solid or a table. A covered circular patch must be subtracted from the exposed base.

Problem

Example 1. An open cylindrical pipe has diameter $10 cm$ and length $80 cm$ . Find the area of its outside curved surface to be painted. Do not include the inner surface or end rims.

Example 2. An upright cylinder with radius $3 cm$ and height $4 cm$ stands on top of a larger cylinder with radius $6 cm$ and height $5 cm$ . The smaller cylinder's entire bottom disk rests on the larger top. The larger cylinder sits on a table. Find the total exposed surface area.

Answer

Example 1:

800 π cm^{2} \approx 2513.27 cm^{2}

. Example 2:

120 π cm^{2} \approx 376.99 cm^{2}

For the open pipe, there are no circular end caps to paint. For the stacked solid, both contact faces and the bottom face on the table are excluded.

Step-by-step solution

1. Unroll the pipe's curved surface

The outside curved surface unrolls into a rectangle. Its length is the circumference $π d$ , and its other dimension is the pipe length $h$ . Thus

A = π d h = π \cdot 10 \cdot 80 = 800 π cm^{2} .

Rounding only at the end gives approximately $2513.27 cm^{2}$ . The value $10 cm$ is a diameter, not a radius.

2. List the exposed faces of the stacked solid

The smaller cylinder contributes its curved wall and top disk. Its bottom disk is hidden. The larger cylinder contributes its curved wall and the uncovered part of its top disk. Its bottom is hidden by the table.

3. Calculate the smaller cylinder's area

With $r = 3$ and $h = 4$ ,

A_{small} = 2 π (3) (4) + π (3)^{2} = 24 π + 9 π = 33 π cm^{2} .

4. Calculate the larger cylinder's area

Its curved wall has area $2 π (6) (5) = 60 π$ . Its whole top disk has area $36 π$ , but the smaller cylinder covers $9 π$ . Therefore

A_{large} = 60 π + (36 π - 9 π) = 87 π cm^{2} .

5. Add and check

The total exposed area is

A = 33 π + 87 π = 120 π cm^{2} .

As a check, the two exposed horizontal pieces together have area $9 π + (36 π - 9 π) = 36 π$ . Adding the curved walls gives $(24 + 60 + 36) π = 120 π$ . If the solid were lifted off the table and its bottom also counted, that would be a different question with an additional $36 π cm^{2}$ .

Add to Chrome - solve your own

Screenshot any problem on your screen and get the steps. Chrome on a computer; free.