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# 8.8: Surface Area and Volume of Spheres

Created by: CK-12

## Introduction

The Pinata

The Olympics at Montgomery Middle School was a huge success. The students and guests had a terrific time and all decided that this event would be an annual event from now on.

For the final celebration, the students hosted a huge party in the gymnasium. Joe took on creating a piñata that students could enjoy. In fact, he made four of them for the party. When finished with the basic shape, Joe began decorating the sphere shaped piñata with green construction paper.

Each sphere has a radius of 2.4 feet and Joe made four piñatas. How much construction paper will he need to cover each piñata completely?

Figuring out the surface area and volume of a sphere can be a tricky thing to figure out. For this problem, you will need to know how to solve for the surface area of a sphere. Pay close attention throughout this lesson and you will know how to solve the problem by the end of the lesson.

What You Will Learn

In this lesson, you will learn how to complete the following skills.

• Recognize and apply the formula for the surface area of a sphere.
• Recognize and apply the formula for the volume of a sphere.
• Compare surface area and volume of spheres, cylinders and cones.
• Solve real – world problems involving surface area and volume of spheres.

Teaching Time

I. Recognize and Apply the Formula for the Surface Area of a Sphere

In this lesson we will learn to find the surface area and volume of spheres. A sphere is a solid figure that exists in three-dimensional space. Spheres consist of all the points that are equidistant from a center point. Every point on the sphere is the distance of the radius from the center. Spheres are perfectly round.

When thinking about spheres, we will be working with surface area and volume. In this chapter, you have been working with the surface area and volume of other three-dimensional figures. Let’s review the definitions of surface area and volume.

Surface area is the total of the areas of each face in a solid figure. Spheres do not have any faces because they are round. Still, we can think of its surface as a flat plane that we can unroll. Imagine you could wrap a sphere in wrapping paper, like a present. The amount of wrapping paper needed to cover the figure exactly represents its surface area.

The volume of a solid figure is the measure of how much three-dimensional space it takes up or holds. Imagine a filling a balloon with water. The size of the balloon determines how much water it can hold. The amount of water tells the volume of the balloon. We measure volume in cubic units, because we are measuring in three dimensions: length, width, and height.

We can calculate the surface area and volume of a sphere by using formulas. Let’s look at finding surface area first.

As we’ve said, surface area is the total area for the surface of a sphere. Imagine you need to wrap a present in the shape of a sphere. How much wrapping paper would you need to cut to cover the sphere?

If we could “unroll” the sphere and show it as a rectangle, the rectangle would have a width that is equivalent to the diameter of the sphere. Its length would be the same as the circumference of the sphere (recall that circumference is the distance around a circle). Now this gives us something we can work with, because we can use the area formula for rectangles to find the area of the “unrolled” sphere.

The formula for the area of rectangles is $A = lw$. In other words, we multiply the length and the width. Now let’s think about this in terms of the sphere. We have said that the length is the same as the circumference and that the width is the same as the diameter. This gives us $A = Cd$. Now let’s substitute the formula for circumference in for $C$, which is $2 \pi r$, where $r$ is the radius. Now we have $A = 2 \pi rd$. There’s one more substitution we can make to make the calculations easier. Remember, the diameter of a circle or sphere is exactly twice the radius. Let’s us $2r$ instead of $d$. Now we have $A = 2 \pi r \times 2r$. Finally, we can simplify this by multiplying to get $4 \pi r^2$. Let’s sum all of this up:

$A & = l \times w\\A & = C \times d\\A & = 2 \pi r \times 2r\\SA & = 4 \pi r^2$

The formula for finding the surface area of a sphere is $4 \pi r^2$.

Write this formula down in your notebook.

All we need to do is substitute the measure of the radius for $r$ in the formula and solve for $SA$, the surface area. Let’s give it a try.

Example

What is the surface area of the sphere below?

We can see that the radius of the sphere is 8, so we put this into the formula and solve.

$SA & = 4 \pi r\\SA & = 4 \pi (8^2)\\SA & = 4 \pi (64)\\SA & = 256 \pi$

Calculating numbers with pi is a bit complicated, because pi is actually a decimal number that goes on forever.

The most exact measure of the sphere’s surface area is to leave it as $256 \pi$. However, we often round the decimal to 3.14 to represent pi. Then we would multiply 256 by 3.14 to get a surface area of 803.84 square centimeters for this sphere. Remember, we always use square units to measure area, because we measure area in two dimensions.

Now let’s look at how we can calculate the volume of a sphere.

II. Recognize and Apply the Formula for the Volume of a Sphere

Volume, as we have said, is the measure of three-dimensional space a figure takes up. We can also think of it as how much space the figure “holds.” For spheres, finding volume is a bit complicated because it doesn’t have any flat surfaces.

To find the volume of spheres, we can use pyramids. Imagine a pyramid with its base on the surface of the sphere and its point as the center of the sphere. The radius of the sphere would be the height of the pyramid.

The pyramid makes up a portion of the sphere’s volume. If we can fill the whole sphere with pyramids like this, we would know the volume of the sphere. It would be equal to the volumes of all the pyramids put together. How many pyramids would it take to fill a sphere? That depends on the surface area of the sphere. We can combine the surface area of a sphere with the volume formula for a pyramid to calculate the volume of all the pyramids contained within the sphere.

Let’s look at how this information can give us the formula for finding the volume of a sphere.

$V = \frac{1}{3} Bh$

Volume formula for a pyramid, where $B$ represents the area of its base

$V = \frac{1}{3} \times \text{surface area of sphere} \times r$

The surface area of the sphere is equal to the area of the bases of all the pyramids. The height of the pyramid is equal to the radius of the sphere, so we substitute $r$ for $h$.

$V = \frac{1}{3} \times 4 \pi r^2 \times r$

We can simplify the formula by combining like terms.

$V = \frac{4}{3} r^3$

The formula for finding the volume of a sphere is $V = \frac{4}{3} r^3$.

Write this formula down in your notebook.

Again, all we need to know is the radius of the sphere. We put the value in for $r$ in the formula and solve for $V$, the volume. Let’s try it.

Example

Find the volume of the sphere below.

We know that the radius of the sphere is 6 meters, so we put this value in for $r$ and solve.

$V & = \frac{4}{3} \pi r^3\\V & = \frac{4}{3} \pi (6^3)\\V & = \frac{4}{3}(216) \pi\\V & = 288 \pi$

We can leave the volume as $288 \pi$, or we can use 3.14 to approximate an answer. This gives us $288 \times 3.14 = 904.32 \ cubic \ meters$. Remember, we measure volume in three dimensions, so we use cubic units.

The volume and surface areas of different solid figures can be compared because they have a relationship. Let’s take a look at how we can compare these figures.

III. Compare Surface Area and Volume of Spheres, Cylinders and Cones

We can compare the surface area and volume of spheres, cylinders and cones with the same radii and height. There is a relationship between the measurements of each of these. Let’s look at this picture.

You can see from this picture that there is a definite relationship between these figures. This is an illustration of how we can compare the volume of the solids. The volume of a cone is $\frac{1}{3}$ the volume of the cylinder. We can think about this because we could slide that cone right inside the cylinder. Because it has the same radius and height, one will fit inside the other.

The volume of a sphere is $\frac{2}{3}$ the volume of the cylinder. Once again, we can see how the sphere would fit inside the cylinder given the same radius and height.

We know that surface area refers to the outer covering on a solid figure. We can compare the surface areas of cylinders, cones and spheres too.

You can see from the diagram that the surface area of cylinders and spheres with the same radii and height have the exact same surface area. Think about unrolling a cylinder into a rectangle, well, we can do the same thing with a sphere. Think about the wrapping paper example from the last section-now you can see how the surface areas are the same.

The surface area of a cone is one-half the surface area of a cylinder if they have the same height and radius. Let’s look at an example.

Cone = Radius of 10 cm, height of 6 cm

$SA & = 3.14(10^2)+3.14(10)(6) \\SA & = 502.4 \ sq.cm$

Cylinder with Radius of 10 cm and height of 6 cm

$SA & = 2(3.14)(10^2)+2(3.14)(10)(6) \\SA & = 1004.8 \ sq.cm$

You can see the relationship between these two solids.

IV. Solve Real – World Problems Involving Surface Area and Volume of Spheres

We can use the formulas we have learned to solve real-world problems involving spheres. First, be sure you understand what the question is asking. Do you need to find the surface area or the volume? Second, make sure you know what the radius of the sphere is. If you have been given the diameter, divide it in half to find the radius. Let’s practice with a few problems.

Example

Maria has a paperweight that is a glass sphere. The sphere is full of red, sparkly liquid. If the diameter of the paperweight is 6 inches, how much red liquid does it contain?

First of all, what is the problem asking us to find? We need to find how much liquid the paperweight contains. This amount will be the volume of the paperweight, so we’ll need to use the volume formula. Now let’s see if we know the radius of the paperweight. We know that the diameter is 6 inches. Therefore the radius is $6 \div 2 = 3 \ inches$. Now we can put this into the volume formula and solve.

$V & = \frac{4}{3} \pi r^3\\V & = \frac{4}{3} \pi (3^3)\\V & = \frac{4}{3} (27) \pi\\V & = 36 \pi$

The sphere has a volume of $36 \pi$. We can approximate a numeric value if we use 3.14 for pi. This gives us a volume of $36 \times 3.14 = 113.04 \ cubic \ inches$.

Now let’s go back and apply what we have learned to the problem from the introduction.

## Real-Life Example Completed

The Pinata

Here is the original problem once again. Reread it and then solve the problem.

The Olympics at Montgomery Middle School was a huge success. The students and guests had a terrific time and all decided that this event would be an annual event from now on.

For the final celebration, the students hosted a huge party in the gymnasium. Joe took on creating a piñata that students could enjoy. In fact, he made four of them for the party. When finished with the basic shape, Joe began decorating the sphere shaped piñata with green construction paper.

Each sphere has a radius of 2.4 feet and Joe made four piñatas. How much construction paper will he need to cover each piñata completely?

Now solve for the surface area of each piñata.

Solution to Real – Life Example

This problem is asking us about covering the surface of a piñata, so we will need to calculate its surface area. The problem tells us that the radius of the piñata is 2.4 feet, so we can put this into the surface area formula and solve.

$SA & = 4 \pi r^2\\SA & = 4 \pi (2.4^2)\\SA & = 4 (5.76) \pi\\SA & = 23.04 \pi\\SA & = 72.35 \ ft^2$

The piñata has a surface area of 72.35 square feet when we approximate pi as 3.14, so this is how much paper Joe will need to cover one.

But Joe made four piñatas, so we need to multiply this answer by 4.

$72.35(4) = 289.4 \ sq. feet$

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Sphere
a perfectly round solid figure where all of the points of it are equidistant from a center point.
Surface Area
the measurement of the outer covering of a solid figure.
Volume
the amount of capacity contained inside a solid figure.

## Time for Practice

Directions: Find the surface area of each sphere. Use 3.14 to approximate pi.

1. A sphere with a radius of 4 in.
2. A sphere with a radius of 2 in.
3. A sphere with a radius of 3.5 ft.
4. A sphere with a radius of 6.7 in.
5. A sphere with a radius of 12 cm.
6. A sphere with a radius of 1.6 ft.
7. A sphere with a radius of 9 m.

Directions: Find the volume of each sphere.

1. A sphere with a radius of 3 m.
2. A sphere with a radius of 5 in.
3. A sphere with a radius of 6 in.
4. A sphere with a radius of 2.5 m.
5. A sphere with a radius of 3 mm.
6. A sphere with a diameter of 8 in.
7. A sphere with a diameter of 10 ft.

Directions: Use what you have learned to solve each problem.

1. What is the surface area of a sphere whose diameter is 22 centimeters?
2. A sphere has a diameter of 12 feet. What is its volume?
3. Kelly has a perfume bottle in the shape of a sphere. The diameter of the bottle is 6 inches. How much perfume does Kelly have left if the bottle is only half full?
4. Bruce is making a sculpture in his art class that is made of 3 spheres. Each sphere has a radius of 2.3 feet. He will paint them all with blue poster paint. If each bottle of paint covers 20 square feet, how many bottles will Bruce need to buy?

Jan 14, 2013