Surface Area and Volume of Cylinders, Cones, and Spheres

Volume of Cylinders

When we find the volume of a 3D shape, we are finding how much space is inside it. You can think of volume as the amount a container can hold.

A cylinder is a solid shape with two matching circular bases and one curved side. Examples of cylinders include soup cans, water bottles, and paper towel rolls.

To understand the volume of a cylinder, it helps to remember a big idea from earlier: for any prism-like solid, volume can be found by multiplying the area of the base by the height.

That means:

$$V = B \cdot h$$

Here, \(V\) is volume, \(B\) is the area of the base, and \(h\) is the height.

For a cylinder, the base is a circle. The area of a circle is:

$$B = \pi r^2$$

So the volume formula for a cylinder becomes:

$$V = \pi r^2 h$$

This is the main formula you will use for this topic.

What each part means:

• \(V\) = volume
• \(\pi\) = about \(3.14\)
• \(r\) = radius of the circular base
• \(h\) = height of the cylinder

Important: The radius is the distance from the center of the circle to the edge. If you are given the diameter instead, remember that:

$$r = \frac{d}{2}$$

Also, volume is measured in cubic units, such as:

• cubic centimeters: \(cm^3\)
• cubic meters: \(m^3\)
• cubic inches: \(in^3\)

Why the formula makes sense

Imagine stacking many thin circles on top of each other until you build a cylinder. Each layer has the same circular area. The number of layers depends on the height. So multiplying the area of one circular base by the height gives the total space inside.

This is the same idea as the prism formula:

$$\text{Volume} = \text{base area} \times \text{height}$$

For cylinders, the base just happens to be a circle.

Steps for finding the volume of a cylinder

1. Identify the radius \(r\) and height \(h\).
2. If needed, change diameter to radius.
3. Use the formula \(V = \pi r^2 h\).
4. Square the radius.
5. Multiply by the height.
6. Multiply by \(\pi\), or use \(3.14\) if asked for a decimal answer.
7. Write the answer in cubic units.

Worked Example 1: Whole-number radius and height

Find the volume of a cylinder with radius \(4\,cm\) and height \(7\,cm\).

Use the formula:

$$V = \pi r^2 h$$

Substitute the values:

$$V = \pi (4)^2(7)$$

Square the radius:

$$V = \pi (16)(7)$$

Multiply:

$$V = 112\pi$$

Exact answer:

$$112\pi\,cm^3$$

Approximate decimal answer:

$$V \approx 112(3.14) = 351.68\,cm^3$$

So the volume is \(112\pi\,cm^3\) or about \(351.68\,cm^3\).

Worked Example 2: Given diameter instead of radius

Find the volume of a cylinder with diameter \(10\,m\) and height \(8\,m\).

First find the radius:

$$r = \frac{10}{2} = 5\,m$$

Now use the formula:

$$V = \pi r^2 h$$ $$V = \pi (5)^2(8)$$

Square the radius:

$$V = \pi (25)(8)$$

Multiply:

$$V = 200\pi$$

Approximate:

$$V \approx 200(3.14) = 628\,m^3$$

So the volume is \(200\pi\,m^3\) or about \(628\,m^3\).

Worked Example 3: Decimal measurements

A can has radius \(2.5\,in\) and height \(12\,in\). What is its volume?

Use the formula:

$$V = \pi r^2 h$$ $$V = \pi (2.5)^2(12)$$

Square the radius:

$$2.5^2 = 6.25$$

Now multiply:

$$V = \pi (6.25)(12) = 75\pi$$

Approximate:

$$V \approx 75(3.14) = 235.5\,in^3$$

So the volume is \(75\pi\,in^3\) or about \(235.5\,in^3\).

Worked Example 4: Finding a missing measurement

A cylinder has a volume of \(314\,cm^3\) and a radius of \(5\,cm\). Find the height. Use \(\pi \approx 3.14\).

Start with the formula:

$$V = \pi r^2 h$$

Substitute the known values:

$$314 = 3.14(5)^2h$$

Square the radius:

$$314 = 3.14(25)h$$ $$314 = 78.5h$$

Now divide both sides by \(78.5\):

$$h = \frac{314}{78.5} = 4$$

So the height is \(4\,cm\).

Common mistakes to avoid

• Using diameter instead of radius: If the problem gives diameter, divide by 2 first.
• Forgetting to square the radius: In the formula \(V = \pi r^2 h\), only the radius is squared.
• Mixing units: Radius and height should be in the same unit before calculating.
• Using square units for volume: Volume must be written in cubic units, like \(cm^3\), not \(cm^2\).

Helpful tip

If you ever forget the cylinder formula, remember this idea:

$$\text{Volume} = \text{base area} \times \text{height}$$

Then ask yourself: what is the base of a cylinder? It is a circle. So use the area of a circle, \(\pi r^2\), as the base area.

Quick check questions

• If a cylinder has radius \(3\) and height \(10\), what formula would you use?
• If the diameter is \(14\), what is the radius?
• Why is the answer written in cubic units?

Summary

The volume of a cylinder is found by multiplying the area of its circular base by its height. Since the area of a circle is \(\pi r^2\), the formula for cylinder volume is:

$$V = \pi r^2 h$$

Always make sure you use the radius, square it, multiply by the height, and write your answer in cubic units. This connects directly to the general volume rule for prisms: base area times height.

Surface Area of Cylinders

In this lesson, you will learn how to find the surface area of a cylinder. Surface area means the total area on the outside of a 3D shape.

A cylinder is a solid shape with:

• two matching circular bases, and
• one curved surface that wraps around the sides.

Common examples of cylinders are soup cans, water bottles, and paper towel rolls.

To find the surface area of a cylinder, it helps to break the shape apart into simpler pieces.

A cylinder can be thought of as:

• 2 circles for the top and bottom, and
• 1 rectangle for the curved side when it is unwrapped.

Step 1: Find the area of the two circular bases

The area of one circle is:

$$A=\pi r^2$$

Since a cylinder has two circular bases, the total area of both circles is:

$$2\pi r^2$$

Here, \(r\) is the radius of the base.

Step 2: Find the area of the side

If you cut the curved surface and lay it flat, it becomes a rectangle.

The height of this rectangle is the height of the cylinder, which is \(h\).

The length of this rectangle is the distance around the circle, which is the circumference:

$$2\pi r$$

So the area of the rectangle is:

$$2\pi r \cdot h = 2\pi rh$$

Step 3: Add all the parts together

Total surface area:

$$SA = 2\pi r^2 + 2\pi rh$$

This is the formula for the surface area of a cylinder.

You may also see it written as:

$$SA = 2\pi r(r+h)$$

Both formulas mean the same thing.

What do the variables mean?

• \(SA\) = surface area
• \(r\) = radius of the circular base
• \(h\) = height of the cylinder
• \(\pi\) is about \(3.14\)

Important idea: radius vs. diameter

Sometimes a problem gives the diameter instead of the radius.

Remember:

$$d = 2r$$

So:

$$r = \frac{d}{2}$$

Always make sure you are using the radius in the formula.

Units

Because surface area measures area, the answer must be in square units.

• square centimeters: \(cm^2\)
• square meters: \(m^2\)
• square inches: \(in^2\)

Worked Example 1: Find the surface area when radius and height are given

A cylinder has radius \(3\,cm\) and height \(5\,cm\). Find its surface area.

Use the formula:

$$SA = 2\pi r^2 + 2\pi rh$$

Substitute \(r=3\) and \(h=5\):

$$SA = 2\pi(3^2) + 2\pi(3)(5)$$

$$SA = 2\pi(9) + 30\pi$$

$$SA = 18\pi + 30\pi$$

$$SA = 48\pi$$

Approximate using \(\pi \approx 3.14\):

$$SA \approx 48(3.14) = 150.72$$

Answer: The surface area is \(48\pi\,cm^2\), or about \(150.72\,cm^2\).

Worked Example 2: Given the diameter instead of the radius

A cylinder has diameter \(10\,in\) and height \(8\,in\). Find its surface area.

First find the radius:

$$r = \frac{10}{2} = 5$$

Now use the formula:

$$SA = 2\pi r^2 + 2\pi rh$$

$$SA = 2\pi(5^2) + 2\pi(5)(8)$$

$$SA = 2\pi(25) + 80\pi$$

$$SA = 50\pi + 80\pi$$

$$SA = 130\pi$$

Approximate:

$$SA \approx 130(3.14) = 408.2$$

Answer: The surface area is \(130\pi\,in^2\), or about \(408.2\,in^2\).

Worked Example 3: Finding the parts separately

A cylinder has radius \(4\,m\) and height \(7\,m\). Find the surface area by adding the areas of the parts.

Area of the two circles:

$$2\pi r^2 = 2\pi(4^2)=2\pi(16)=32\pi$$

Area of the rectangle (side):

$$2\pi rh = 2\pi(4)(7)=56\pi$$

Add them:

$$SA = 32\pi + 56\pi = 88\pi$$

Approximate:

$$SA \approx 88(3.14)=276.32$$

Answer: The surface area is \(88\pi\,m^2\), or about \(276.32\,m^2\).

Worked Example 4: A word problem

A can is shaped like a cylinder. It has a radius of \(6\,cm\) and a height of \(12\,cm\). How much metal is needed to make the whole outside of the can?

Since the whole outside includes the top, bottom, and side, we find the total surface area.

Use the formula:

$$SA = 2\pi r^2 + 2\pi rh$$

$$SA = 2\pi(6^2) + 2\pi(6)(12)$$

$$SA = 2\pi(36) + 144\pi$$

$$SA = 72\pi + 144\pi$$

$$SA = 216\pi$$

Approximate:

$$SA \approx 216(3.14)=678.24$$

Answer: The can needs \(216\pi\,cm^2\) of metal, or about \(678.24\,cm^2\).

When might you not use the whole formula?

Sometimes a problem asks only for the lateral surface area. That means just the curved side, not the top and bottom.

The formula for lateral surface area is:

$$LSA = 2\pi rh$$

If a problem asks for the total surface area, include all parts:

$$SA = 2\pi r^2 + 2\pi rh$$

Read each question carefully to see whether it wants:

• just the side, or
• the entire outside surface.

Common mistakes to avoid

• Using diameter when the formula needs radius.
• Forgetting there are two circular bases.
• Forgetting to include the side rectangle.
• Writing units without squaring them.
• Mixing up surface area with volume.

Quick steps for solving

1. Identify the radius \(r\) and height \(h\).
2. If given the diameter, divide by 2 to get the radius.
3. Use $$SA = 2\pi r^2 + 2\pi rh$$
4. Simplify.
5. If needed, use \(\pi \approx 3.14\) for a decimal answer.
6. Write the correct square units.

Summary

A cylinder’s surface area comes from adding the areas of two circles and one rectangle. The formula is:

$$SA = 2\pi r^2 + 2\pi rh$$

This works because the top and bottom are circles, and the curved side unwraps into a rectangle. If you remember to use the radius, include all parts, and write square units, you can find the surface area of any cylinder.

Volume of Cones

Have you ever seen an ice cream cone or a party hat? Both are shaped like a cone. In math, we often want to know how much space is inside a 3D shape. That amount of space is called volume.

In this lesson, you will learn how to find the volume of a cone and why the formula works. You will also see how cones are connected to cylinders.

What is a cone?

A cone is a three-dimensional shape with:

• one circular base,
• a curved surface, and
• a point at the top called the vertex.

To find the volume of a cone, you need to know two measurements:

• the radius of the circular base, written as \(r\),
• the height of the cone, written as \(h\).

The height is the straight distance from the center of the base to the vertex. It is not the slanted side.

The formula for the volume of a cone

The formula is:

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

Here is what each part means:

• \(V\) = volume
• \(\pi\) = about \(3.14\)
• \(r\) = radius of the base
• \(h\) = height

Why is there a \(\frac{1}{3}\) in the formula?

A cone is closely related to a cylinder. If a cone and a cylinder have the same base radius and the same height, then the cone holds exactly one-third of the volume of the cylinder.

The volume of a cylinder is:

$$V=\pi r^2 h$$

So the volume of a cone is one-third of that:

$$V=\frac{1}{3}(\pi r^2 h)$$

That is why the cone formula includes \(\frac{1}{3}\).

Steps for finding the volume of a cone

1. Identify the radius \(r\) and height \(h\).
2. Square the radius: \(r^2\).
3. Multiply by \(\pi\).
4. Multiply by the height.
5. Multiply by \(\frac{1}{3}\), or divide by 3.
6. Write the answer in cubic units, such as \(\text{cm}^3\), \(\text{m}^3\), or \(\text{in}^3\).

Important note about units

Because volume measures space inside a 3D shape, the units are always cubic units. For example:

• centimeters become \(\text{cm}^3\)
• meters become \(\text{m}^3\)
• inches become \(\text{in}^3\)

Worked Example 1: Find the volume of a cone

A cone has radius \(3\) cm and height \(8\) cm. Find its volume.

Use the formula:

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

Substitute the values:

$$V=\frac{1}{3}\pi (3)^2(8)$$

Square the radius:

$$V=\frac{1}{3}\pi (9)(8)$$

Multiply:

$$V=\frac{1}{3}\pi (72)$$ $$V=24\pi$$

Now approximate using \(\pi \approx 3.14\):

$$V \approx 24(3.14)=75.36$$

Answer: The volume is \(24\pi\text{ cm}^3\), or about \(75.36\text{ cm}^3\).

Worked Example 2: Diameter is given instead of radius

A cone has a diameter of \(10\) m and a height of \(12\) m. Find its volume.

Be careful: the formula uses radius, not diameter.

Find the radius:

$$r=\frac{10}{2}=5$$

Now use the formula:

$$V=\frac{1}{3}\pi r^2 h$$ $$V=\frac{1}{3}\pi (5)^2(12)$$

Square the radius:

$$V=\frac{1}{3}\pi (25)(12)$$

Multiply:

$$V=\frac{1}{3}\pi (300)$$ $$V=100\pi$$

Approximate:

$$V\approx 100(3.14)=314$$

Answer: The volume is \(100\pi\text{ m}^3\), or about \(314\text{ m}^3\).

Worked Example 3: Compare a cone and a cylinder

A cone and a cylinder have the same radius, \(4\) in, and the same height, \(9\) in. Find both volumes.

First, find the volume of the cylinder:

$$V_{\text{cylinder}}=\pi r^2 h$$ $$V_{\text{cylinder}}=\pi (4)^2(9)$$ $$V_{\text{cylinder}}=\pi (16)(9)=144\pi$$

So the cylinder has volume \(144\pi\text{ in}^3\).

Now find the cone volume:

$$V_{\text{cone}}=\frac{1}{3}\pi r^2 h$$ $$V_{\text{cone}}=\frac{1}{3}(144\pi)=48\pi$$

Answer:

• Cylinder: \(144\pi\text{ in}^3\)
• Cone: \(48\pi\text{ in}^3\)

This shows that the cone has one-third the volume of the cylinder.

Worked Example 4: A word problem

A paper cone cup has radius \(6\) cm and height \(15\) cm. How much liquid can it hold?

Use the volume formula:

$$V=\frac{1}{3}\pi r^2 h$$ $$V=\frac{1}{3}\pi (6)^2(15)$$

Square the radius:

$$V=\frac{1}{3}\pi (36)(15)$$

Multiply:

$$V=\frac{1}{3}\pi (540)$$ $$V=180\pi$$

Approximate:

$$V\approx 180(3.14)=565.2$$

Answer: The cone cup can hold about \(565.2\text{ cm}^3\) of liquid.

Common mistakes to avoid

• Using diameter instead of radius: Always divide the diameter by 2 first.
• Forgetting the \(\frac{1}{3}\): This is a very common mistake. Without it, you are finding the volume of a cylinder, not a cone.
• Not squaring the radius: The formula uses \(r^2\), not just \(r\).
• Using the slanted side instead of the height: The formula needs the straight up-and-down height.
• Forgetting cubic units: Volume answers must be in cubic units.

Quick check

• If \(r=2\) cm and \(h=6\) cm, then $$V=\frac{1}{3}\pi (2)^2(6)=\frac{1}{3}\pi (24)=8\pi\text{ cm}^3$$
• If a cone and cylinder have the same radius and height, the cone volume is one-third of the cylinder volume.

Summary

The volume of a cone tells how much space is inside it. To find it, use the formula $$V=\frac{1}{3}\pi r^2 h$$ where \(r\) is the radius and \(h\) is the height.

A cone with the same base and height as a cylinder has one-third the cylinder’s volume. Remember to use the radius, square it, include the \(\frac{1}{3}\), and write your answer in cubic units.

Surface Area of Cones

When we talk about surface area, we mean the total area covering the outside of a 3D shape.

A cone has two outside parts:

• a circular base
• a curved side

To find the total surface area of a cone, we add the area of the base and the area of the curved side.

This lesson will show you how to:

• identify the parts of a cone
• use the formula for surface area
• find a missing slant height using the Pythagorean theorem
• solve real examples step by step

1. Parts of a Cone

A cone has three important measurements:

• radius \\(r\\): the distance from the center of the circular base to the edge
• height \\(h\\): the straight up-and-down distance from the center of the base to the tip
• slant height \\(l\\): the distance from the edge of the base to the tip along the side of the cone

The slant height is not the same as the height. The height goes straight up. The slant height goes along the slanted side.

2. Formula for Surface Area of a Cone

The total surface area of a cone is:

$$SA = \pi r^2 + \pi r l$$

Here is what each part means:

• \(\pi r^2\) is the area of the circular base
• \(\pi r l\) is the area of the curved surface

So you can also think of it as:

$$\text{Surface Area} = \text{base area} + \text{curved area}$$

or

$$SA = \pi r^2 + \pi r l$$

You can factor this formula if you want:

$$SA = \pi r(r+l)$$

Both forms mean the same thing.

3. Finding the Slant Height

Sometimes a problem gives you the radius and the height, but not the slant height.

To find the slant height, use the Pythagorean theorem. The radius, height, and slant height make a right triangle.

$$l^2 = r^2 + h^2$$

So:

$$l = \sqrt{r^2 + h^2}$$

This works because:

• the radius is one leg of the triangle
• the height is the other leg
• the slant height is the longest side, or hypotenuse

4. Steps for Solving Surface Area of a Cone

1. Find the radius \\(r\\).
2. Find the slant height \\(l\\).
3. If \\(l\\) is missing, use \\(l = \sqrt{r^2+h^2}\\).
4. Substitute into the formula \\(SA = \pi r^2 + \pi r l\\).
5. Simplify. Leave your answer in terms of \\(\pi\\) or use a decimal approximation.
6. Write the correct square units, such as \\(cm^2\\), \\(m^2\\), or \\(in^2\\).

Worked Example 1: Surface Area When Radius and Slant Height Are Given

A cone has radius \\(r=4\\) cm and slant height \\(l=7\\) cm. Find its total surface area.

Use the formula:

$$SA = \pi r^2 + \pi r l$$

Substitute the values:

$$SA = \pi(4^2) + \pi(4)(7)$$ $$SA = 16\pi + 28\pi$$ $$SA = 44\pi$$

Approximate with \\(\pi \approx 3.14\\):

$$SA \approx 44(3.14) = 138.16$$

Answer: The total surface area is \\(44\pi \text{ cm}^2\\), or about \\(138.16 \text{ cm}^2\\).

Worked Example 2: Find Slant Height First

A cone has radius \\(r=3\\) m and height \\(h=4\\) m. Find its total surface area.

First find the slant height:

$$l = \sqrt{r^2+h^2}$$ $$l = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$$

Now use the surface area formula:

$$SA = \pi r^2 + \pi r l$$ $$SA = \pi(3^2) + \pi(3)(5)$$ $$SA = 9\pi + 15\pi$$ $$SA = 24\pi$$

Approximate:

$$SA \approx 24(3.14) = 75.36$$

Answer: The total surface area is \\(24\pi \text{ m}^2\\), or about \\(75.36 \text{ m}^2\\).

Worked Example 3: A Larger Cone

A cone has radius \\(5\\) inches and height \\(12\\) inches. Find the total surface area.

Step 1: Find the slant height.

$$l = \sqrt{5^2 + 12^2}$$ $$l = \sqrt{25+144} = \sqrt{169} = 13$$

Step 2: Use the formula.

$$SA = \pi r^2 + \pi r l$$ $$SA = \pi(5^2) + \pi(5)(13)$$ $$SA = 25\pi + 65\pi$$ $$SA = 90\pi$$

Approximate:

$$SA \approx 90(3.14) = 282.6$$

Answer: The total surface area is \\(90\pi \text{ in}^2\\), or about \\(282.6 \text{ in}^2\\).

Worked Example 4: Real-World Style Question

A party hat is shaped like a cone with radius \\(6\\) cm and slant height \\(10\\) cm. If you want to cover the entire outside of the hat, including the base, what is the surface area?

Use the formula:

$$SA = \pi r^2 + \pi r l$$ $$SA = \pi(6^2) + \pi(6)(10)$$ $$SA = 36\pi + 60\pi$$ $$SA = 96\pi$$

Approximate:

$$SA \approx 96(3.14) = 301.44$$

Answer: The surface area is \\(96\pi \text{ cm}^2\\), or about \\(301.44 \text{ cm}^2\\).

5. Common Mistakes to Avoid

• Using height instead of slant height: In the formula \\(SA = \pi r^2 + \pi r l\\), you must use slant height \\(l\\), not height \\(h\\).
• Forgetting the base: Total surface area includes both the curved surface and the circular base.
• Using diameter instead of radius: If the problem gives the diameter, divide by 2 to get the radius.
• Forgetting square units: Surface area is always measured in square units.

6. Quick Check

Try these on your own:

• A cone has \\(r=2\\) cm and \\(l=5\\) cm. Find the surface area.
• A cone has \\(r=8\\) ft and \\(h=15\\) ft. First find \\(l\\), then find the surface area.

Answers:

• \(SA = \pi(2^2)+\pi(2)(5)=4\pi+10\pi=14\pi \text{ cm}^2\)
• \(l=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\), so \\(SA=\pi(8^2)+\pi(8)(17)=64\pi+136\pi=200\pi \text{ ft}^2\\)

Summary

The total surface area of a cone is the area of its circular base plus the area of its curved side.

Use the formula:

$$SA = \pi r^2 + \pi r l$$

If the slant height is missing, find it with:

$$l = \sqrt{r^2+h^2}$$

Always make sure you are using the radius and slant height, and remember to include square units in your final answer.

Volume of Spheres

In this lesson, you will learn how to find the volume of a sphere. Volume tells us how much space a 3-dimensional object takes up.

A sphere is a perfectly round solid, like a basketball, marble, or globe. When we find the volume of a sphere, we are finding how much space is inside it.

You will also learn about a hemisphere, which is half of a sphere, like half of a ball cut straight through the middle.

1. Important vocabulary

• Volume: the amount of space inside a solid figure
• Sphere: a round 3D shape where every point on the surface is the same distance from the center
• Hemisphere: half of a sphere
• Radius ((r)): the distance from the center of the sphere to the outside
• Diameter ((d)): the distance across the sphere through the center

Remember the relationship between radius and diameter:

\(d = 2r\) and \(r = \frac{d}{2}\)

2. Formula for the volume of a sphere

The formula for the volume of a sphere is:

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

In this formula:

• \(V\) is the volume
• \(r\) is the radius
• \(\pi\) is about \(3.14\)
• \(r^3\) means \(r \times r \times r\)

This means you:

1. Find the radius
2. Cube the radius
3. Multiply by \(\pi\)
4. Multiply by \(\frac{4}{3}\)

3. Formula for the volume of a hemisphere

A hemisphere is half of a sphere, so its volume is half the sphere formula.

$$V = \frac{1}{2}\left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3$$

So the volume of a hemisphere is:

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

4. Units for volume

Since volume measures space inside a solid, the units are always cubic units.

• cubic centimeters: \(cm^3\)
• cubic inches: \(in^3\)
• cubic meters: \(m^3\)

If you forget to use cubic units, your answer is not complete.

5. Worked Example 1: Find the volume when the radius is given

Find the volume of a sphere with radius \(r = 3\) cm.

Use the formula:

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

Substitute \(r = 3\):

$$V = \frac{4}{3}\pi (3)^3$$

Cube the radius:

$$3^3 = 27$$

Now multiply:

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

Using \(\pi \approx 3.14\):

$$V \approx 36(3.14) = 113.04$$

Answer: The volume is \(36\pi\, cm^3\), or about \(113.04\, cm^3\).

6. Worked Example 2: Find the volume when the diameter is given

Find the volume of a sphere with diameter \(10\) inches.

First, find the radius:

$$r = \frac{d}{2} = \frac{10}{2} = 5$$

Now use the formula:

$$V = \frac{4}{3}\pi r^3$$ $$V = \frac{4}{3}\pi (5)^3$$

Cube the radius:

$$5^3 = 125$$

Substitute:

$$V = \frac{4}{3}\pi (125)$$ $$V = \frac{500}{3}\pi$$

Now approximate with \(\pi \approx 3.14\):

$$V \approx \frac{500}{3}(3.14) \approx 523.33$$

Answer: The volume is \(\frac{500}{3}\pi\, in^3\), or about \(523.33\, in^3\).

7. Worked Example 3: Volume of a hemisphere

Find the volume of a hemisphere with radius \(6\) m.

Use the hemisphere formula:

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

Substitute \(r = 6\):

$$V = \frac{2}{3}\pi (6)^3$$

Cube the radius:

$$6^3 = 216$$

Multiply:

$$V = \frac{2}{3}\pi (216)$$ $$V = 144\pi$$

Approximate:

$$V \approx 144(3.14) = 452.16$$

Answer: The volume of the hemisphere is \(144\pi\, m^3\), or about \(452.16\, m^3\).

8. Worked Example 4: Solve a word problem

A toy ball has a radius of \(4\) cm. How much space is inside the ball?

This is asking for the volume of a sphere.

Use the formula:

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

Substitute \(r = 4\):

$$V = \frac{4}{3}\pi (4)^3$$

Cube the radius:

$$4^3 = 64$$

Multiply:

$$V = \frac{4}{3}\pi (64) = \frac{256}{3}\pi$$

Approximate:

$$V \approx \frac{256}{3}(3.14) \approx 267.95$$

Answer: The ball holds about \(267.95\, cm^3\) of space.

9. Common mistakes to avoid

• Using the diameter instead of the radius: Always check whether the problem gives radius or diameter.
• Forgetting to cube the radius: \(r^3\) means multiply the radius by itself 3 times.
• Forgetting the \(\frac{4}{3}\): It is an important part of the sphere formula.
• Using square units instead of cubic units: Volume must be written in units like \(cm^3\), not \(cm^2\).
• Not halving for a hemisphere: A hemisphere is half a sphere, so use \(\frac{2}{3}\pi r^3\).

10. Tips for success

• Write the formula before substituting numbers.
• If the diameter is given, divide by 2 first.
• Cube carefully: \(2^3 = 8\), \(3^3 = 27\), \(4^3 = 64\), \(5^3 = 125\).
• If your teacher wants an exact answer, leave \(\pi\) in the answer.
• If your teacher wants an approximate answer, use \(\pi \approx 3.14\).

11. Quick check

Ask yourself these questions when solving:

1. Is this a sphere or a hemisphere?
2. Do I have the radius, or do I need to find it from the diameter?
3. Did I cube the radius?
4. Did I use cubic units in my answer?

Summary

To find the volume of a sphere, use $$V = \frac{4}{3}\pi r^3$$. The most important value is the radius, so make sure you do not confuse it with the diameter.

To find the volume of a hemisphere, use $$V = \frac{2}{3}\pi r^3$$ because a hemisphere is half of a sphere. Always write your final answer in cubic units.

Lesson: Surface Area of Pyramids and Cones

When we find the surface area of a 3D shape, we are finding the total area of all the outside surfaces that cover it.

In this lesson, you will learn how to find the surface area of regular pyramids and right cones. You will also learn the difference between lateral surface area and total surface area.

Lateral surface area means the area of the side surfaces only. It does not include the base.

Total surface area means the area of all outside surfaces, including the base.

These shapes often use a measurement called slant height. The slant height is the distance from the top of the shape down along its side, not straight up and down.

Part 1: Surface Area of a Regular Pyramid

A regular pyramid has a base that is a regular polygon, such as a square, and triangular faces that are all the same size.

To find the surface area of a regular pyramid, we often use:

• the perimeter of the base, written as \(P\)
• the slant height, written as \(l\)
• the area of the base, written as \(B\)

The formula for the lateral surface area of a regular pyramid is:

$$L = \frac{1}{2}Pl$$

This works because the side faces are triangles, and their total area can be added together using half the base perimeter times the slant height.

The formula for the total surface area of a regular pyramid is:

$$SA = \frac{1}{2}Pl + B$$

So for pyramids:

• First find the lateral area: \(\frac{1}{2}Pl\)
• Then add the area of the base

Part 2: Surface Area of a Right Cone

A right cone has a circular base and a point directly above the center of the base.

For a cone, we use:

• the radius of the base, written as \(r\)
• the slant height, written as \(l\)

The formula for the lateral surface area of a cone is:

$$L = \pi rl$$

The formula for the total surface area of a cone is:

$$SA = \pi rl + \pi r^2$$

Here, \(\pi r^2\) is the area of the circular base.

Important idea: In both pyramids and cones, the slant height is used in the lateral area formula, not the vertical height.

How to Solve Surface Area Problems

1. Decide whether the problem asks for lateral surface area or total surface area.
2. Identify the measurements you need.
3. Choose the correct formula.
4. Substitute the numbers into the formula.
5. Solve carefully and include square units, such as \(cm^2\) or \(m^2\).

Worked Example 1: Lateral Surface Area of a Square Pyramid

A regular square pyramid has a base perimeter of \(24\) cm and a slant height of \(7\) cm. Find the lateral surface area.

Use the pyramid lateral area formula:

$$L = \frac{1}{2}Pl$$

Substitute the values:

$$L = \frac{1}{2}(24)(7)$$ $$L = 12 \times 7$$ $$L = 84$$

The lateral surface area is \(84\text{ cm}^2\).

Worked Example 2: Total Surface Area of a Square Pyramid

A regular square pyramid has a base side length of \(6\) m and a slant height of \(5\) m. Find the total surface area.

Step 1: Find the perimeter of the square base.

$$P = 4 \times 6 = 24\text{ m}$$

Step 2: Find the lateral surface area.

$$L = \frac{1}{2}Pl$$ $$L = \frac{1}{2}(24)(5) = 12 \times 5 = 60\text{ m}^2$$

Step 3: Find the area of the base.

The base is a square, so:

$$B = 6^2 = 36\text{ m}^2$$

Step 4: Add them together.

$$SA = L + B = 60 + 36 = 96\text{ m}^2$$

The total surface area is \(96\text{ m}^2\).

Worked Example 3: Lateral Surface Area of a Cone

A right cone has radius \(4\) cm and slant height \(9\) cm. Find the lateral surface area.

Use the cone lateral area formula:

$$L = \pi rl$$

Substitute the values:

$$L = \pi (4)(9)$$ $$L = 36\pi$$

So the exact answer is \(36\pi\text{ cm}^2\).

If you use \(\pi \approx 3.14\), then:

$$L \approx 36(3.14) = 113.04$$

The lateral surface area is about \(113.04\text{ cm}^2\).

Worked Example 4: Total Surface Area of a Cone

A right cone has radius \(3\) in and slant height \(8\) in. Find the total surface area.

Use the formula:

$$SA = \pi rl + \pi r^2$$

Substitute the values:

$$SA = \pi(3)(8) + \pi(3^2)$$ $$SA = 24\pi + 9\pi$$ $$SA = 33\pi$$

So the exact answer is \(33\pi\text{ in}^2\).

Using \(\pi \approx 3.14\):

$$SA \approx 33(3.14) = 103.62$$

The total surface area is about \(103.62\text{ in}^2\).

Common Mistakes to Avoid

• Mixing up lateral and total surface area. Lateral area does not include the base. Total surface area does.
• Using height instead of slant height. The formulas here use slant height.
• Forgetting the base area. This only matters when finding total surface area.
• Using the wrong base measurement. For a pyramid, you may need the base perimeter. For a cone, you need the radius.
• Forgetting square units. Surface area is always in square units.

Quick Check

Use what you learned to think about these questions:

• If a pyramid has \(P = 30\) cm and \(l = 6\) cm, what is its lateral surface area?
• If a cone has \(r = 5\) m and \(l = 7\) m, what is its lateral surface area?
• What extra piece of information do you need to find the total surface area of a pyramid?

Answers:

• Pyramid lateral area: $$L = \frac{1}{2}(30)(6) = 90\text{ cm}^2$$
• Cone lateral area: $$L = \pi(5)(7) = 35\pi\text{ m}^2$$
• You need the area of the base.

Lesson Summary

Surface area tells us how much space covers the outside of a 3D shape. For pyramids and cones, we often separate the side area from the base area.

For a regular pyramid:

$$L = \frac{1}{2}Pl$$ $$SA = \frac{1}{2}Pl + B$$

For a right cone:

$$L = \pi rl$$ $$SA = \pi rl + \pi r^2$$

Always check whether the question wants lateral surface area or total surface area, and make sure you use the slant height.

Surface Area of Spheres

Have you ever wondered how much material it would take to cover a basketball, a globe, or a marble? That amount is called the surface area.

The surface area of a sphere is the total area on the outside of the sphere. A sphere is a perfectly round 3D shape, like a ball.

In this lesson, you will learn the formula for the surface area of a sphere, how to use it correctly, and how to solve problems step by step.

1. What is a sphere?

A sphere is a three-dimensional shape where every point on the outside is the same distance from the center.

• A basketball is shaped like a sphere.
• A tennis ball is shaped like a sphere.
• A globe is shaped like a sphere.

The distance from the center of the sphere to the outside edge is called the radius.

The distance all the way across the sphere, through the center, is called the diameter.

Remember:

• \(\text{diameter} = 2 \times \text{radius}\)
• \(\text{radius} = \frac{\text{diameter}}{2}\)

2. Formula for surface area of a sphere

The formula is:

$$SA = 4\pi r^2$$

• \(SA\) means surface area
• \(r\) means radius
• \(\pi\) is pi, which is about \(3.14\)

This formula tells us how many square units cover the outside of the sphere.

Important: Surface area is always measured in square units, such as:

• square inches \((in^2)\)
• square centimeters \((cm^2)\)
• square meters \((m^2)\)

3. Steps for finding surface area

1. Find the radius.
2. Square the radius: \(r^2\).
3. Multiply by \(4\pi\).
4. Write the answer in square units.

If the problem gives you the diameter instead of the radius, divide by 2 first.

4. Worked Examples

Example 1: Radius is given

Find the surface area of a sphere with radius \(5\) cm.

Use the formula:

$$SA = 4\pi r^2$$

Substitute \(r = 5\):

$$SA = 4\pi(5^2)$$ $$SA = 4\pi(25)$$ $$SA = 100\pi$$

Using \(\pi \approx 3.14\):

$$SA \approx 100(3.14) = 314$$

Answer: The surface area is \(100\pi\, cm^2\), or about \(314\, cm^2\).

Example 2: Diameter is given

A sphere has a diameter of \(12\) inches. Find its surface area.

First find the radius:

$$r = \frac{12}{2} = 6$$

Now use the formula:

$$SA = 4\pi r^2$$ $$SA = 4\pi(6^2)$$ $$SA = 4\pi(36)$$ $$SA = 144\pi$$

Using \(\pi \approx 3.14\):

$$SA \approx 144(3.14) = 452.16$$

Answer: The surface area is \(144\pi\, in^2\), or about \(452.16\, in^2\).

Example 3: Word problem

A toy ball has a radius of \(3.5\) cm. How much surface area does it have?

Use the formula:

$$SA = 4\pi r^2$$

Substitute \(r = 3.5\):

$$SA = 4\pi(3.5^2)$$ $$SA = 4\pi(12.25)$$ $$SA = 49\pi$$

Using \(\pi \approx 3.14\):

$$SA \approx 49(3.14) = 153.86$$

Answer: The surface area is \(49\pi\, cm^2\), or about \(153.86\, cm^2\).

Example 4: Comparing two spheres

Sphere A has radius \(2\) m. Sphere B has radius \(4\) m. Find the surface area of each sphere.

Sphere A:

$$SA = 4\pi(2^2) = 4\pi(4) = 16\pi$$ $$SA \approx 16(3.14) = 50.24$$

So Sphere A has surface area \(16\pi\, m^2\), or about \(50.24\, m^2\).

Sphere B:

$$SA = 4\pi(4^2) = 4\pi(16) = 64\pi$$ $$SA \approx 64(3.14) = 200.96$$

So Sphere B has surface area \(64\pi\, m^2\), or about \(200.96\, m^2\).

Notice that when the radius doubled from \(2\) to \(4\), the surface area became 4 times as large.

5. Common mistakes to avoid

• Using the diameter instead of the radius
  If the formula needs \(r\), make sure you do not plug in the diameter by mistake.
• Forgetting to square the radius
  The formula uses \(r^2\), not just \(r\).
• Forgetting square units
  Surface area is measured in units like \(cm^2\) or \(in^2\).
• Not using parentheses carefully
  When substituting, write values clearly, like \(4\pi(5^2)\).

6. Quick check: Which formula do I use?

If the question asks for the area covering the outside of a sphere, use:

$$SA = 4\pi r^2$$

If you see words like cover, wrap, paint the outside, or outside surface, that means you need surface area.

7. Helpful tip

It can help to remember the formula like this:

Surface Area of a Sphere = 4 times pi times radius squared

Say it aloud:

"4 pi r squared"

8. Practice thinking

Ask yourself these questions when solving:

• Did I use the radius?
• Did I square the radius?
• Did I multiply by \(4\pi\)?
• Did I write square units?

Summary

The surface area of a sphere is the total area on the outside of the sphere. To find it, use the formula $$SA = 4\pi r^2$$ where \(r\) is the radius.

Always check whether the problem gives the radius or diameter. If you are given the diameter, divide by 2 first. Then square the radius, multiply by \(4\pi\), and write your answer in square units.

Volume of Composite Solids

Sometimes a 3D shape is made by joining two or more solids together. These are called composite solids.

In this lesson, you will learn how to find the total volume of composite solids made from cylinders, cones, and hemispheres.

Remember: volume tells us how much space is inside a solid. Volume is measured in cubic units, such as cubic centimeters \,\(cm^3\) or cubic meters \,\(m^3\).

1. The big idea

To find the volume of a composite solid, you usually:

• Break the solid into simpler parts.
• Find the volume of each part.
• Add the volumes if the parts are joined together.
• Subtract a volume if a piece is missing or cut out.

So the main strategy is:

$$\text{Volume of composite solid} = \text{sum of volumes of parts}$$

or, if there is a hole or missing piece,

$$\text{Volume of composite solid} = \text{whole} - \text{missing part}$$

2. Formulas you need

Here are the volume formulas for the solids in this topic.

Cylinder

$$V = \pi r^2 h$$

where \(r\) is the radius and \(h\) is the height.

Cone

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

where \(r\) is the radius and \(h\) is the height.

Sphere

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

Hemisphere

A hemisphere is half of a sphere, so its volume is:

$$V = \frac{1}{2}\left(\frac{4}{3}\pi r^3\right) = \frac{2}{3}\pi r^3$$

3. Important things to watch for

• Use the same units for every measurement before calculating.
• Make sure you know the difference between radius and diameter.
• If you are given the diameter, then the radius is half of it: \(r = \frac{d}{2}\).
• Check whether shapes are being added together or whether one part is removed.
• Write the final answer in cubic units.

4. How to solve volume of composite solids

1. Look at the solid carefully.
2. Decide which basic solids it is made from.
3. Label the radius and height of each part.
4. Use the correct formula for each part.
5. Add or subtract the volumes.
6. Round only if the problem asks you to.

Worked Example 1: Cylinder with a cone on top

A solid is made from a cylinder with a cone on top. Both parts have radius \(3\,cm\). The cylinder has height \(8\,cm\), and the cone has height \(4\,cm\). Find the total volume.

Step 1: Find the volume of the cylinder.

$$V_{\text{cyl}} = \pi r^2 h$$

$$V_{\text{cyl}} = \pi(3)^2(8) = \pi(9)(8) = 72\pi$$

So the cylinder volume is \(72\pi\,cm^3\).

Step 2: Find the volume of the cone.

$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h$$

$$V_{\text{cone}} = \frac{1}{3}\pi(3)^2(4) = \frac{1}{3}\pi(9)(4) = 12\pi$$

So the cone volume is \(12\pi\,cm^3\).

Step 3: Add the volumes.

$$V_{\text{total}} = 72\pi + 12\pi = 84\pi$$

$$V_{\text{total}} = 84\pi\,cm^3$$

Using \(\pi \approx 3.14\),

$$84\pi \approx 84(3.14) = 263.76$$

Total volume: \(84\pi\,cm^3\) or about \(263.76\,cm^3\).

Worked Example 2: Cylinder with a hemisphere on top

A solid is made from a cylinder and a hemisphere. The radius of both parts is \(5\,m\). The cylinder has height \(10\,m\). Find the total volume.

Step 1: Find the cylinder volume.

$$V_{\text{cyl}} = \pi r^2 h$$

$$V_{\text{cyl}} = \pi(5)^2(10) = \pi(25)(10) = 250\pi$$

Step 2: Find the hemisphere volume.

$$V_{\text{hemi}} = \frac{2}{3}\pi r^3$$

$$V_{\text{hemi}} = \frac{2}{3}\pi(5)^3 = \frac{2}{3}\pi(125) = \frac{250}{3}\pi$$

Step 3: Add the volumes.

$$V_{\text{total}} = 250\pi + \frac{250}{3}\pi$$

Write \(250\pi\) with denominator 3:

$$250\pi = \frac{750}{3}\pi$$

So,

$$V_{\text{total}} = \frac{750}{3}\pi + \frac{250}{3}\pi = \frac{1000}{3}\pi$$

Using \(\pi \approx 3.14\),

$$V_{\text{total}} \approx \frac{1000}{3}(3.14) \approx 1046.67$$

Total volume: \(\frac{1000}{3}\pi\,m^3\) or about \(1046.67\,m^3\).

Worked Example 3: A cone-shaped hole removed from a cylinder

A cylinder has radius \(4\,cm\) and height \(12\,cm\). A cone-shaped hole is removed from the top. The cone has the same radius, \(4\,cm\), and height \(6\,cm\). Find the volume left.

This time, we subtract because part of the solid is removed.

Step 1: Find the cylinder volume.

$$V_{\text{cyl}} = \pi r^2 h$$

$$V_{\text{cyl}} = \pi(4)^2(12) = \pi(16)(12) = 192\pi$$

Step 2: Find the cone volume.

$$V_{\text{cone}} = \frac{1}{3}\pi r^2 h$$

$$V_{\text{cone}} = \frac{1}{3}\pi(4)^2(6) = \frac{1}{3}\pi(16)(6) = 32\pi$$

Step 3: Subtract.

$$V_{\text{left}} = 192\pi - 32\pi = 160\pi$$

Using \(\pi \approx 3.14\),

$$160\pi \approx 160(3.14) = 502.4$$

Volume left: \(160\pi\,cm^3\) or about \(502.4\,cm^3\).

Worked Example 4: Watch out for diameter

A solid is made from a cylinder and a hemisphere. The diameter is \(10\,in\), so the radius is \(5\,in\). The cylinder height is \(7\,in\). Find the total volume.

Step 1: Find the radius.

$$r = \frac{10}{2} = 5\,in$$

Step 2: Find the cylinder volume.

$$V_{\text{cyl}} = \pi r^2 h = \pi(5)^2(7) = \pi(25)(7) = 175\pi$$

Step 3: Find the hemisphere volume.

$$V_{\text{hemi}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(5)^3 = \frac{2}{3}\pi(125) = \frac{250}{3}\pi$$

Step 4: Add.

$$V_{\text{total}} = 175\pi + \frac{250}{3}\pi$$

Convert \(175\pi\) to thirds:

$$175\pi = \frac{525}{3}\pi$$

Then,

$$V_{\text{total}} = \frac{525}{3}\pi + \frac{250}{3}\pi = \frac{775}{3}\pi$$

Using \(\pi \approx 3.14\),

$$V_{\text{total}} \approx \frac{775}{3}(3.14) \approx 811.17$$

Total volume: \(\frac{775}{3}\pi\,in^3\) or about \(811.17\,in^3\).

5. Common mistakes

• Using the diameter as the radius by mistake.
• Forgetting that a cone has the factor \(\frac{1}{3}\).
• Forgetting that a hemisphere is half a sphere.
• Adding volumes when you should subtract, or subtracting when you should add.
• Writing square units like \(cm^2\) instead of cubic units like \(cm^3\).

6. Quick check questions

Try these on your own:

• A cylinder of radius \(2\,cm\) and height \(9\,cm\) has a cone on top with the same radius and height \(3\,cm\). What is the total volume?
• A hemisphere of radius \(6\,m\) sits on a cylinder of the same radius and height \(4\,m\). What is the total volume?
• A cylinder of radius \(3\,in\) and height \(10\,in\) has a cone-shaped hole with the same radius and height \(5\,in\). What volume remains?

7. Final summary

Composite solids are 3D figures made from simpler solids joined together or with parts removed.

To find volume, split the solid into familiar shapes like cylinders, cones, and hemispheres. Then use the correct formula for each part and add or subtract the volumes.

The most important skills are identifying the parts, using the correct radius and height, and remembering whether the pieces are being combined or taken away.

Missing Dimensions in Solids means finding a measurement that is not given, such as a radius, diameter, or height, by using a formula for volume or surface area.

In this lesson, we will focus on cylinders, cones, and spheres. Instead of always plugging numbers into a formula, we will sometimes work backward. That means we will be given the volume or surface area and use inverse operations to solve for the missing dimension.

This is an important skill because many real-world problems do not give every measurement directly. You may know how much a container holds or how much material covers an object, and from that information you need to figure out its size.

Before we begin, remember these words:

• Radius: the distance from the center of a circle to the edge
• Diameter: the distance across a circle through the center; it is twice the radius, so \(d = 2r\)
• Height: how tall a solid is
• Inverse operations: operations that undo each other, like multiplying and dividing, or squaring and taking a square root

Step 1: Know the formulas

To find a missing dimension, you must start with the correct formula.

Volume of a cylinder:

$$V = \pi r^2 h$$

Volume of a cone:

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

Volume of a sphere:

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

Surface area of a cylinder:

$$SA = 2\pi r^2 + 2\pi rh$$

Surface area of a sphere:

$$SA = 4\pi r^2$$

At this grade level, you will most often solve for a missing dimension in these formulas by isolating the variable. That means getting the unknown by itself.

Step 2: Solve by undoing operations

When solving for a missing dimension, follow these steps:

1. Write the correct formula.
2. Substitute the values you know.
3. Use inverse operations to isolate the unknown.
4. If needed, take a square root or cube root.
5. Check whether the question asks for radius or diameter.

Important idea: If the formula has \(r^2\), you will usually need a square root. If the formula has \(r^3\), you will usually need a cube root.

Working with cylinders

A cylinder has two circular bases and a height. Its volume formula is:

$$V = \pi r^2 h$$

If the height is missing, divide by \(\pi r^2\):

$$h = \frac{V}{\pi r^2}$$

If the radius is missing, divide by \(\pi h\) first:

$$r^2 = \frac{V}{\pi h}$$

Then take the square root:

$$r = \sqrt{\frac{V}{\pi h}}$$

Working with cones

A cone is similar to a cylinder, but its volume is one-third as much when the radius and height are the same.

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

To solve for height:

$$h = \frac{3V}{\pi r^2}$$

To solve for radius:

$$r = \sqrt{\frac{3V}{\pi h}}$$

Working with spheres

A sphere has no height like a cylinder or cone. Its size is determined by its radius.

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

To solve for radius, first isolate \(r^3\):

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

Then take the cube root:

$$r = \sqrt[3]{\frac{3V}{4\pi}}$$

For the surface area of a sphere:

$$SA = 4\pi r^2$$

To solve for radius:

$$r^2 = \frac{SA}{4\pi}$$

$$r = \sqrt{\frac{SA}{4\pi}}$$

Worked Example 1: Find the height of a cylinder

A cylinder has volume \(144\pi\text{ cm}^3\) and radius \(4\text{ cm}\). Find its height.

Step 1: Use the cylinder volume formula.

$$V = \pi r^2 h$$

Step 2: Substitute the known values.

$$144\pi = \pi(4)^2h$$

Step 3: Simplify.

$$144\pi = 16\pi h$$

Step 4: Divide both sides by \(16\pi\).

$$h = \frac{144\pi}{16\pi} = 9$$

Answer: The height is \(9\text{ cm}\).

Worked Example 2: Find the radius of a cone

A cone has volume \(75\pi\text{ in}^3\) and height \(9\text{ in}\). Find the radius.

Step 1: Use the cone volume formula.

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

Step 2: Substitute the known values.

$$75\pi = \frac{1}{3}\pi r^2(9)$$

Step 3: Simplify \(\frac{1}{3} \cdot 9\).

$$75\pi = 3\pi r^2$$

Step 4: Divide both sides by \(3\pi\).

$$r^2 = \frac{75\pi}{3\pi} = 25$$

Step 5: Take the square root.

$$r = 5$$

Answer: The radius is \(5\text{ in}\).

Worked Example 3: Find the radius of a sphere from volume

A sphere has volume \(\frac{500}{3}\pi\text{ m}^3\). Find its radius.

Step 1: Use the sphere volume formula.

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

Step 2: Substitute the known value.

$$\frac{500}{3}\pi = \frac{4}{3}\pi r^3$$

Step 3: Multiply both sides by 3 to clear the fraction.

$$500\pi = 4\pi r^3$$

Step 4: Divide both sides by \(4\pi\).

$$r^3 = \frac{500\pi}{4\pi} = 125$$

Step 5: Take the cube root.

$$r = \sqrt[3]{125} = 5$$

Answer: The radius is \(5\text{ m}\).

Worked Example 4: Find the diameter of a sphere from surface area

A sphere has surface area \(196\pi\text{ cm}^2\). Find its diameter.

Step 1: Use the sphere surface area formula.

$$SA = 4\pi r^2$$

Step 2: Substitute the known value.

$$196\pi = 4\pi r^2$$

Step 3: Divide both sides by \(4\pi\).

$$r^2 = \frac{196\pi}{4\pi} = 49$$

Step 4: Take the square root.

$$r = 7$$

Step 5: The question asks for diameter, not radius.

$$d = 2r = 2(7) = 14$$

Answer: The diameter is \(14\text{ cm}\).

Tips for success

• Always choose the correct formula first.
• Be careful about whether the problem gives or asks for radius or diameter.
• If \(\pi\) appears on both sides, it often cancels out.
• If you see \(r^2\), use a square root at the end.
• If you see \(r^3\), use a cube root at the end.
• Include units in your final answer.

Common mistakes

• Using diameter in place of radius without dividing by 2 first
• Forgetting the \(\frac{1}{3}\) in the cone volume formula
• Stopping after finding \(r^2\) instead of finding \(r\)
• Finding the radius when the question asks for the diameter
• Mixing up surface area formulas and volume formulas

Quick check

Ask yourself these questions when you finish:

• Did I use the formula for the correct solid?
• Did I solve for the exact measurement being asked for?
• Did I use square root or cube root if needed?
• Does my answer make sense for the size of the solid?

Summary

To find a missing dimension in a solid, start with the correct volume or surface area formula and substitute the values you know. Then use inverse operations to isolate the unknown. For cylinders and cones, you may solve for radius or height. For spheres, you often solve for radius from volume or surface area. Finally, always check whether the problem wants radius or diameter.

Volume of Pyramids, Cones, and Spheres

When we talk about volume, we mean how much space a 3D object takes up inside. Volume is measured in cubic units, such as cubic centimeters \\(cm^3\\), cubic meters \\(m^3\\), or cubic inches \\(in^3\\).

In this lesson, you will learn how to find the volume of three important 3D shapes: pyramids, cones, and spheres. These shapes have special formulas because they are not box-shaped like prisms or cylinders.

A helpful idea in this topic is that these shapes often use a fraction of the volume of a related shape. For example, a pyramid is related to a prism, and a cone is related to a cylinder.

1. Volume of a Pyramid

A pyramid has a base and triangular faces that meet at one point, called the vertex. To find its volume, we use the area of the base and the height.

The formula is:

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

In this formula:

• V = volume
• B = area of the base
• h = perpendicular height

The \\(\frac{1}{3}\\) means a pyramid has one-third the volume of a prism with the same base area and height.

Important: The height must be the straight up-and-down distance from the base to the vertex, not the slanted edge.

2. Volume of a Cone

A cone has a circular base and comes to a point. It is like a pyramid with a circle for its base.

The formula is:

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

In this formula:

• r = radius of the circular base
• h = perpendicular height
• \(\pi r^2\) = area of the circular base

A cone has one-third the volume of a cylinder with the same base radius and height.

3. Volume of a Sphere

A sphere is a perfectly round 3D shape, like a basketball or a globe.

The formula for the volume of a sphere is:

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

In this formula:

• r = radius of the sphere

Notice that the radius is cubed, so you must multiply \\(r \times r \times r\\).

Comparing a Sphere to a Bounding Cylinder

A bounding cylinder is a cylinder that fits exactly around the sphere. If a sphere has radius \\(r\\), then the bounding cylinder has:

• radius \\(r\\)
• height \\(2r\\)

The volume of that cylinder is:

$$V_{cylinder} = \pi r^2(2r) = 2\pi r^3$$

The volume of the sphere is:

$$V_{sphere} = \frac{4}{3}\pi r^3$$

Now compare them:

$$\frac{V_{sphere}}{V_{cylinder}} = \frac{\frac{4}{3}\pi r^3}{2\pi r^3} = \frac{4}{6} = \frac{2}{3}$$

So a sphere has \(\frac{2}{3}\) the volume of its bounding cylinder.

How to Solve Volume Problems

1. Identify the shape.
2. Write the correct formula.
3. Find the measurements you need, especially the radius, base area, and height.
4. Substitute the values into the formula.
5. Solve carefully using the correct order of operations.
6. Write the answer in cubic units.

Worked Example 1: Volume of a Pyramid

A rectangular pyramid has a base that is \\(6\\,cm\\) by \\(4\\,cm\\), and its height is \\(9\\,cm\\). Find the volume.

Step 1: Find the area of the base.

$$B = 6 \times 4 = 24\,cm^2$$

Step 2: Use the pyramid formula.

$$V = \frac{1}{3}Bh$$ $$V = \frac{1}{3}(24)(9)$$ $$V = \frac{1}{3}(216) = 72$$

Answer: The volume is \\(72\,cm^3\\).

Worked Example 2: Volume of a Cone

A cone has radius \\(3\\,m\\) and height \\(8\\,m\\). Find the volume.

Use the formula:

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

Substitute the values:

$$V = \frac{1}{3}\pi (3)^2(8)$$ $$V = \frac{1}{3}\pi (9)(8)$$ $$V = \frac{1}{3}\pi (72)$$ $$V = 24\pi$$

If you want a decimal approximation, use \\(\pi \approx 3.14\\):

$$V \approx 24(3.14) = 75.36$$

Answer: The exact volume is \\(24\pi\,m^3\\), or about \\(75.36\,m^3\\).

Worked Example 3: Volume of a Sphere

A sphere has radius \\(5\\,in\\). Find the volume.

Use the formula:

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

Substitute the radius:

$$V = \frac{4}{3}\pi (5)^3$$ $$V = \frac{4}{3}\pi (125)$$ $$V = \frac{500}{3}\pi$$

Approximate using \\(\pi \approx 3.14\\):

$$V \approx \frac{500}{3}(3.14) \approx 523.33$$

Answer: The exact volume is \\(\frac{500}{3}\pi\,in^3\\), or about \\(523.33\,in^3\\).

Worked Example 4: Comparing a Sphere to Its Bounding Cylinder

A sphere has radius \\(4\\,cm\\). Compare its volume to the volume of the cylinder that just fits around it.

Step 1: Find the cylinder dimensions.

• radius = \\(4\\,cm\\)
• height = diameter = \\(8\\,cm\\)

Step 2: Find the cylinder volume.

$$V_{cylinder} = \pi r^2 h$$ $$V_{cylinder} = \pi (4)^2(8) = \pi (16)(8) = 128\pi\,cm^3$$

Step 3: Find the sphere volume.

$$V_{sphere} = \frac{4}{3}\pi r^3$$ $$V_{sphere} = \frac{4}{3}\pi (4)^3 = \frac{4}{3}\pi (64) = \frac{256}{3}\pi\,cm^3$$

Step 4: Compare them.

$$\frac{V_{sphere}}{V_{cylinder}} = \frac{\frac{256}{3}\pi}{128\pi} = \frac{256}{384} = \frac{2}{3}$$

Answer: The sphere has \(\frac{2}{3}\) the volume of the bounding cylinder.

Common Mistakes to Avoid

• Forgetting the \\(\frac{1}{3}\\) in the pyramid or cone formula.
• Using diameter instead of radius in formulas.
• Confusing base area with just one side length.
• Using a slanted length instead of the vertical height.
• Forgetting to write cubic units.

Quick Formula Review

• Pyramid: \\(V = \frac{1}{3}Bh\\)
• Cone: \\(V = \frac{1}{3}\pi r^2 h\\)
• Sphere: \\(V = \frac{4}{3}\pi r^3\\)

Brief Summary

To find the volume of a pyramid or cone, first think about the related prism or cylinder, then take one-third of that amount. For a sphere, use the special formula \\(V = \frac{4}{3}\pi r^3\\). A sphere also has a useful comparison: it fills two-thirds of the volume of the cylinder that just fits around it.