Section 4.2 Surface Area

In this section the Surface Area we will address will relate to the measure of surfaces of 3 dimensional objects. One way to look at it is in terms of determining how much “area” you would be “painting” or “coating” in some way if you covered all (or sometimes part) of the surfaces of a three dimension object.

Example: In the box show on the right we need to consider the following:

How accurate I need to be depends on the application.

If I’m using a cheap paint, it won’t matter.

If I’m using gold leaf, every square inch matters.

Homework: Problems 1-8.

Before leaving class today, it is important that we review every homework problem to lay out a possible approach for solving.

What are we “covering?”InsideOutsideBottomTop

What shapes do you see?TrapezoidsTriangleRectanglesRounded edges

Is the bottom flush or not? What is the thickness of the material?

In order to find the requested surface area, you will more than likely have to divide the objects in shapes that you know.

Example 1

Find the surface area of the plate.

• 38 cm diameter• 4 cm thick• 6 cm diameter holes drilled through

Example 2

Find the weight of the 6mm thick steel tank.

Example 3

Find the weight of the 3/16-inch thick steel plate. The original rectangular shape was 10” x 14”. The corners were rounded off with a 2 inch radius. The inner rectangular cut-out is 4” x 6”.

Example 4

Find the surface area of the concrete wall. Do not include the bottom of the wall since it is against the ground.

1. 17 gallons2. 64,296 mm2

3. 1,435 ft2

4. 854.5 mm2

5. 502.7 m2

6. 21,038 kg7. 18,169 lbs8. 1,412 kg