Solution:

Sphere: [latex]SA = 4 \pi r^2[/latex]

Rectangular Prism: SA = 2 ( LW + WH + LH)

Solution:

Cube: [latex]SA = 6s^2[/latex]

Cylinder: [latex]SA = 2 \pi r^2 + 2 \pi r h[/latex]

1. 
   A cylindrical jam jar has a radius of 2.5 inches and a height of 5 inches. What is the surface area of the jar? Round to two decimal places at the end.

Solution:

[latex]SA = 2 \pi(2.5)^2 + 2\pi(2.5)(50) \approx 98.17 \text{in}^2[/latex]

2.

The Earth is not perfectly spherical, but we will assume it is for this problem. The diameter of Earth is approximately 12,756 kilometers. What is the surface area of Earth in kilometers squared and miles squared? Hint: There are 0.621 miles in one kilometer.

Solution:

Now the radius is half of the diameter, so the radius of the earth is approximately 6378 kilometers. This leads to the following computation.

[latex]SA = 4 \pi (6378)^2 \approx 511,185,932.5 \text{km}^2[/latex]

Since 1 kilometer is about 0.62.1 miles, we can make the conversion:

[latex]511,185,932.5 \text{km} (\frac{1km^2}{(0.621)^2\text{mi}^2}) \approx 197369991.95478 \text{mile}^2[/latex]

3.

There are two toy boxes in the nursery. One box is 3 feet long, 2 feet tall, and 1.5 feet wide. The other box is 2.5 feet long, 4 feet wide, and 1 foot tall. Which box has a larger surface area?

Solution:

SA for Box 1 = [latex]2(3\cdot 1.5 + 1.5\cdot 2 + 3\cdot 2) = 27 \text{ft}^2[/latex]

SA for Box 2 = [latex]2(2.5\cdot 1 + 1\cdot 4 + 2.5 \cdot 4) = 33 \text{ft}^2[/latex]

So the second box has a larger surface area.

Find the surface area of the right regular pyramid above if b = 5 and a = 6. Note that b in the figure is only half the length of the base.

Solution:

Since b = 5, the length of each aside is 2 • 5 = 10 and P = 4 • 10 = 40. B is the area of the base which is [latex]10^2 = 100[/latex].

SA = [latex]\frac{1}{2} (10) \cdot 6 + 100 = 30 + 100 = 130[/latex]

• To find the surface area of irregular figures, we can divide the given figure into shapes that we can find the surface area of and add those values together to get the entire surface area. Be sure to recognize two sides of individual solids that the together as are not part of the surface area.

A 60-foot grain silo is in the shape of a cylinder with a semi-sphere as the roof. The radius of the sphere is 7 feet. Find the surface area to the nearest square foot.

Solution:

First, we need the surface area of the cylinder, but we do not need the top or the bottom of the cylinder, so we only use the portion of the formula for the curved surface which is , [latex]2 \pi r h[/latex]. We also need half of the surface area of the sphere for the semi-sphere roof, or [latex]2 \pi r^2[/latex].

SA = [latex]2 \pi r h + 2 \pi r^2 = 2\pi \cdot 7 \cdot 60 + 2 \pi 7^2 \approx 2793 \text{ft}^2[/latex]

Find the surface area of the right circular cone shown, using the formula

Use the radius 12 cm and the slant height, L, of 18 cm.

Solution: