Spheres and Cones

What you'll learn

How to identify the radius, diameter, height and slant height in 3D shapes.
How to calculate volumes and surface areas of cones, spheres and hemispheres.
How to give answers in terms of $π\pi$ or rounded to 3 significant figures.
How to handle compound solids and water displacement questions.

The key measurements

Before using any formula, make sure you know what each measurement means.

Definition

Radius, diameter, height and slant height

The radius is the distance from the centre of a circle to its edge.
The diameter is the full width of a circle through its centre.
The vertical height of a cone is the straight up-and-down distance from the centre of the base to the top.
The slant height of a cone is the distance along the outside edge from the base to the top.

The most common trap is using the diameter when the formula needs the radius. Always halve the diameter first.

Labelled cone showing radius, vertical height, slant height, and the right triangle relationship

Key Idea

Radius first

Most sphere and cone formulae use the radius, not the diameter. If you are given a diameter, use $\frac{d}{2}$ before doing anything else.

Cones

A cone has a circular base and a pointed top called the apex.

For a cone:

Volume: $\frac{1}{3}\pi r^2h$
Curved surface area: $\pi rl$
Total surface area of a solid cone: curved surface area + base area

So:

area=πrl+πr2\text{total surface area} = \pi rl + \pi r^2

Example: volume of a cone from a diameter

Example

Finding the volume of a cone

A cone has vertical height 15 cm and base diameter 12 cm. Find its volume, correct to 3 significant figures.

Convert the diameter to a radius.
$\frac{12}{2} = 6$
Substitute into the cone volume formula.
$\frac{1}{3}\pi \times 6^2 \times 15$
Simplify the exact value first.
$180\pi$
Convert to a decimal and round to 3 significant figures: 565 cm³.

Tip

Rounding

Keep the calculator value until the final line. Rounding too early can make your final answer slightly inaccurate.

Slant height and total surface area of a cone

For cone surface area, you often need the slant height, $l$ . If you are only given the vertical height and radius, use Pythagoras.

The radius, vertical height and slant height form a right-angled triangle:

l^2 = r^2 + h^2

Example: total surface area using Pythagoras

Example

Total surface area of a cone

A solid cone has vertical height 12 cm and base diameter 18 cm. Work out its total surface area in terms of $π\pi$ .

Convert the diameter to a radius.
$\frac{18}{2} = 9$
Use Pythagoras to find the slant height.
$l^2 = 12^2 + 9^2 = 144 + 81 = 225$
Square root to find $l$ .
$\sqrt{225} = 15$
Add the curved surface area and the circular base area.
$area=πrl+πr2=π×9×15+π×92=135π+81π=216π\begin{aligned} \text{total surface area} &= \pi rl + \pi r^2 \\ &= \pi \times 9 \times 15 + \pi \times 9^2 \\ &= 135\pi + 81\pi \\ &= 216\pi \end{aligned}$
The total surface area is $216π216\pi$ cm².

Common Mistake

Using the vertical height as the slant height

In $πrl\pi rl$ , the $l$ means slant height, not vertical height. If the surface area question gives only the vertical height, check whether you need Pythagoras first.

Spheres and hemispheres

Definition

Sphere and hemisphere

A sphere is a perfectly round 3D shape, like a ball. A hemisphere is half of a sphere.

For a sphere:

Volume: $\frac{4}{3}\pi r^3$
Surface area: $4\pi r^2$

For a hemisphere:

Volume: $\frac{2}{3}\pi r^3$
Curved surface area: $2\pi r^2$
Total surface area of a solid hemisphere: curved surface area + flat circular base

So for a solid hemisphere:

area=2πr2+πr2=3πr2\text{total surface area} = 2\pi r^2 + \pi r^2 = 3\pi r^2

Labelled hemisphere showing curved surface and flat circular base

Example: total surface area of a hemisphere

Example

Total surface area of a solid hemisphere

A solid hemisphere has radius 9 cm. Work out its total surface area in terms of $π\pi$ .

A solid hemisphere has a curved surface and a flat circular base.
Find the curved surface area.
$2πr2=2π×92=162π2\pi r^2 = 2\pi \times 9^2 = 162\pi$
Find the flat circular base area.
$πr2=π×92=81π\pi r^2 = \pi \times 9^2 = 81\pi$
Add the two areas.
$162π+81π=243π162\pi + 81\pi = 243\pi$
The total surface area is $243π243\pi$ cm².

Common Mistake

Forgetting the flat circle

If the question asks for the total surface area of a solid hemisphere, include the flat circular face. If it asks for curved surface area only, do not include it.

Answers in terms of $π\pi$

Sometimes the question asks for an answer in terms of $π\pi$ . This means you should leave $π\pi$ in your answer instead of converting it to a decimal.

For example, an exact answer might be $216π216\pi$ cm² or $2723π\frac{272}{3}\pi$ cm³.

Tip

Exact form

If the question says “in terms of $π\pi$ ”, do not round the answer. Collect the number parts together and leave $π\pi$ at the end.

Compound solids

A compound solid is made by joining two or more 3D shapes together.

For volume, you usually add the volumes of the parts.

For surface area, only count the surfaces on the outside. Any joined faces inside the solid are not visible, so they are not included.

Key Idea

Volume versus surface area

For compound solids, volumes add together. For surface area, only count the outside faces.

Example: cone on top of a hemisphere

Example

Volume of a compound solid

A cone sits on top of a hemisphere. The cone has height 9 cm and base diameter 8 cm. The hemisphere also has diameter 8 cm. Work out the total volume in terms of $π\pi$ .

Find the shared radius.
$\frac{8}{2} = 4$
Find the cone volume.
$Vcone=13πr2h=13π×42×9=48π\begin{aligned} V_{\text{cone}} &= \frac{1}{3}\pi r^2h \\ &= \frac{1}{3}\pi \times 4^2 \times 9 \\ &= 48\pi \end{aligned}$
Find the hemisphere volume.
$Vhemisphere=12×43πr3=23π×43=1283π\begin{aligned} V_{\text{hemisphere}} &= \frac{1}{2}\times \frac{4}{3}\pi r^3 \\ &= \frac{2}{3}\pi \times 4^3 \\ &= \frac{128}{3}\pi \end{aligned}$
Add the two volumes.
$48π+1283π=2723π48\pi + \frac{128}{3}\pi = \frac{272}{3}\pi$
The total volume is $2723π\frac{272}{3}\pi$ cm³.

Reverse problems: finding the radius

Sometimes you are given a volume and asked to find the radius. This means you must rearrange the formula.

For a sphere, the formula is:

\frac{4}{3}\pi r^3

Because the radius is cubed, you will need a cube root at the end.

Example: sphere with the same volume as a cube

Example

Finding the radius of a sphere

A cube has side length 6 cm. A sphere has the same volume as the cube. Work out the radius of the sphere, correct to 3 significant figures.

Find the volume of the cube.
$Vcube=63=216V_{\text{cube}} = 6^3 = 216$
Set the sphere volume equal to 216.
$43πr3=216\frac{4}{3}\pi r^3 = 216$
Rearrange to find $r^3$ .
$r3=216×34π=162πr^3 = \frac{216 \times 3}{4\pi} = \frac{162}{\pi}$
Cube root to find $r$ .
$\sqrt[3]{\frac{162}{\pi}} \approx 3.72$
The radius is 3.72 cm to 3 significant figures.

Tip

Cube root, not square root

If the formula contains $r^3$ , you need a cube root to undo it. A square root only undoes $r^2$ .

Water displacement

Definition

Displacement

When an object is fully submerged in water, it pushes up a volume of water equal to its own volume.

In a rectangular container:

container\text{rise in water level} = \frac{\text{volume displaced}}{\text{base area of container}}

The base area of a rectangular container is length times width.

Example: rise in water level

Example

Water level rise

A rectangular container is 12 cm long and 10 cm wide. It contains water to a depth of 7 cm. A metal sphere of radius 3 cm is placed in the water and sinks. Calculate the rise in the water level, correct to 3 significant figures.

Find the volume of the sphere.
$\frac{4}{3}\pi \times 3^3 = 36\pi$
Find the base area of the container.
$12 \times 10 = 120$
Divide the displaced volume by the base area.
$rise=36π120≈0.942\text{rise} = \frac{36\pi}{120} \approx 0.942$
The water level rises by 0.942 cm to 3 significant figures.

Common Mistake

Check for overflow

The displacement method assumes the object is fully submerged and the container does not overflow. If water spills out, the question needs different thinking.

Exam technique

In the exam

Underline whether the question asks for volume, curved surface area, or total surface area.
Convert any diameter to a radius before substituting into a formula.
For cones, check whether slant height is given. If only vertical height is given and surface area is needed, use Pythagoras to find $l$ .
If the answer must be in terms of $π\pi$ , leave $π\pi$ in the answer. If it asks for significant figures, round only at the end.

Self review

Check yourself

What extra circle do you include for the total surface area of a solid hemisphere?
How do you find the slant height of a cone when you know its radius and vertical height?
In a water displacement question, what volume causes the water level to rise?

Spheres and Cones

What you'll learn

How to identify the radius, diameter, height and slant height in 3D shapes.
How to calculate volumes and surface areas of cones, spheres and hemispheres.
How to give answers in terms of $π\pi$ or rounded to 3 significant figures.
How to handle compound solids and water displacement questions.

The key measurements

Before using any formula, make sure you know what each measurement means.

Definition

Radius, diameter, height and slant height

The radius is the distance from the centre of a circle to its edge.
The diameter is the full width of a circle through its centre.
The vertical height of a cone is the straight up-and-down distance from the centre of the base to the top.
The slant height of a cone is the distance along the outside edge from the base to the top.

The most common trap is using the diameter when the formula needs the radius. Always halve the diameter first.

Labelled cone showing radius, vertical height, slant height, and the right triangle relationship

Key Idea

Radius first

Most sphere and cone formulae use the radius, not the diameter. If you are given a diameter, use $\frac{d}{2}$ before doing anything else.

Cones

A cone has a circular base and a pointed top called the apex.

For a cone:

Volume: $\frac{1}{3}\pi r^2h$
Curved surface area: $\pi rl$
Total surface area of a solid cone: curved surface area + base area

So:

area=πrl+πr2\text{total surface area} = \pi rl + \pi r^2

Example: volume of a cone from a diameter

Example

Finding the volume of a cone

A cone has vertical height 15 cm and base diameter 12 cm. Find its volume, correct to 3 significant figures.

Convert the diameter to a radius.
$\frac{12}{2} = 6$
Substitute into the cone volume formula.
$\frac{1}{3}\pi \times 6^2 \times 15$
Simplify the exact value first.
$180\pi$
Convert to a decimal and round to 3 significant figures: 565 cm³.

Tip

Rounding

Keep the calculator value until the final line. Rounding too early can make your final answer slightly inaccurate.

Slant height and total surface area of a cone

For cone surface area, you often need the slant height, $l$ . If you are only given the vertical height and radius, use Pythagoras.

The radius, vertical height and slant height form a right-angled triangle:

l^2 = r^2 + h^2

Example: total surface area using Pythagoras

Example

Total surface area of a cone

A solid cone has vertical height 12 cm and base diameter 18 cm. Work out its total surface area in terms of $π\pi$ .

Convert the diameter to a radius.
$\frac{18}{2} = 9$
Use Pythagoras to find the slant height.
$l^2 = 12^2 + 9^2 = 144 + 81 = 225$
Square root to find $l$ .
$\sqrt{225} = 15$
Add the curved surface area and the circular base area.
$area=πrl+πr2=π×9×15+π×92=135π+81π=216π\begin{aligned} \text{total surface area} &= \pi rl + \pi r^2 \\ &= \pi \times 9 \times 15 + \pi \times 9^2 \\ &= 135\pi + 81\pi \\ &= 216\pi \end{aligned}$
The total surface area is $216π216\pi$ cm².

Common Mistake

Using the vertical height as the slant height

In $πrl\pi rl$ , the $l$ means slant height, not vertical height. If the surface area question gives only the vertical height, check whether you need Pythagoras first.

Spheres and hemispheres

Definition

Sphere and hemisphere

A sphere is a perfectly round 3D shape, like a ball. A hemisphere is half of a sphere.

For a sphere:

Volume: $\frac{4}{3}\pi r^3$
Surface area: $4\pi r^2$

For a hemisphere:

Volume: $\frac{2}{3}\pi r^3$
Curved surface area: $2\pi r^2$
Total surface area of a solid hemisphere: curved surface area + flat circular base

So for a solid hemisphere:

area=2πr2+πr2=3πr2\text{total surface area} = 2\pi r^2 + \pi r^2 = 3\pi r^2

Labelled hemisphere showing curved surface and flat circular base

Example: total surface area of a hemisphere

Example

Total surface area of a solid hemisphere

A solid hemisphere has radius 9 cm. Work out its total surface area in terms of $π\pi$ .

A solid hemisphere has a curved surface and a flat circular base.
Find the curved surface area.
$2πr2=2π×92=162π2\pi r^2 = 2\pi \times 9^2 = 162\pi$
Find the flat circular base area.
$πr2=π×92=81π\pi r^2 = \pi \times 9^2 = 81\pi$
Add the two areas.
$162π+81π=243π162\pi + 81\pi = 243\pi$
The total surface area is $243π243\pi$ cm².

Common Mistake

Forgetting the flat circle

If the question asks for the total surface area of a solid hemisphere, include the flat circular face. If it asks for curved surface area only, do not include it.

Answers in terms of $π\pi$

Sometimes the question asks for an answer in terms of $π\pi$ . This means you should leave $π\pi$ in your answer instead of converting it to a decimal.

For example, an exact answer might be $216π216\pi$ cm² or $2723π\frac{272}{3}\pi$ cm³.

Tip

Exact form

If the question says “in terms of $π\pi$ ”, do not round the answer. Collect the number parts together and leave $π\pi$ at the end.

Compound solids

A compound solid is made by joining two or more 3D shapes together.

For volume, you usually add the volumes of the parts.

For surface area, only count the surfaces on the outside. Any joined faces inside the solid are not visible, so they are not included.

Key Idea

Volume versus surface area

For compound solids, volumes add together. For surface area, only count the outside faces.

Example: cone on top of a hemisphere

Example

Volume of a compound solid

A cone sits on top of a hemisphere. The cone has height 9 cm and base diameter 8 cm. The hemisphere also has diameter 8 cm. Work out the total volume in terms of $π\pi$ .

Find the shared radius.
$\frac{8}{2} = 4$
Find the cone volume.
$Vcone=13πr2h=13π×42×9=48π\begin{aligned} V_{\text{cone}} &= \frac{1}{3}\pi r^2h \\ &= \frac{1}{3}\pi \times 4^2 \times 9 \\ &= 48\pi \end{aligned}$
Find the hemisphere volume.
$Vhemisphere=12×43πr3=23π×43=1283π\begin{aligned} V_{\text{hemisphere}} &= \frac{1}{2}\times \frac{4}{3}\pi r^3 \\ &= \frac{2}{3}\pi \times 4^3 \\ &= \frac{128}{3}\pi \end{aligned}$
Add the two volumes.
$48π+1283π=2723π48\pi + \frac{128}{3}\pi = \frac{272}{3}\pi$
The total volume is $2723π\frac{272}{3}\pi$ cm³.

Reverse problems: finding the radius

Sometimes you are given a volume and asked to find the radius. This means you must rearrange the formula.

For a sphere, the formula is:

\frac{4}{3}\pi r^3

Because the radius is cubed, you will need a cube root at the end.

Example: sphere with the same volume as a cube

Example

Finding the radius of a sphere

A cube has side length 6 cm. A sphere has the same volume as the cube. Work out the radius of the sphere, correct to 3 significant figures.

Find the volume of the cube.
$Vcube=63=216V_{\text{cube}} = 6^3 = 216$
Set the sphere volume equal to 216.
$43πr3=216\frac{4}{3}\pi r^3 = 216$
Rearrange to find $r^3$ .
$r3=216×34π=162πr^3 = \frac{216 \times 3}{4\pi} = \frac{162}{\pi}$
Cube root to find $r$ .
$\sqrt[3]{\frac{162}{\pi}} \approx 3.72$
The radius is 3.72 cm to 3 significant figures.

Tip

Cube root, not square root

If the formula contains $r^3$ , you need a cube root to undo it. A square root only undoes $r^2$ .

Water displacement

Definition

Displacement

When an object is fully submerged in water, it pushes up a volume of water equal to its own volume.

In a rectangular container:

container\text{rise in water level} = \frac{\text{volume displaced}}{\text{base area of container}}

The base area of a rectangular container is length times width.

Example: rise in water level

Example

Water level rise

Find the volume of the sphere.
$\frac{4}{3}\pi \times 3^3 = 36\pi$
Find the base area of the container.
$12 \times 10 = 120$
Divide the displaced volume by the base area.
$rise=36π120≈0.942\text{rise} = \frac{36\pi}{120} \approx 0.942$
The water level rises by 0.942 cm to 3 significant figures.

Common Mistake

Check for overflow

The displacement method assumes the object is fully submerged and the container does not overflow. If water spills out, the question needs different thinking.

Exam technique

In the exam

Underline whether the question asks for volume, curved surface area, or total surface area.
Convert any diameter to a radius before substituting into a formula.
For cones, check whether slant height is given. If only vertical height is given and surface area is needed, use Pythagoras to find $l$ .
If the answer must be in terms of $π\pi$ , leave $π\pi$ in the answer. If it asks for significant figures, round only at the end.

Self review

Check yourself

What extra circle do you include for the total surface area of a solid hemisphere?
How do you find the slant height of a cone when you know its radius and vertical height?
In a water displacement question, what volume causes the water level to rise?

Spheres and Cones

What you'll learn

The key measurements

Cones

Example: volume of a cone from a diameter

Slant height and total surface area of a cone

Example: total surface area using Pythagoras

Spheres and hemispheres

Example: total surface area of a hemisphere

Answers in terms of $π\pi$

Compound solids

Example: cone on top of a hemisphere

Reverse problems: finding the radius

Example: sphere with the same volume as a cube

Water displacement

Example: rise in water level

Recap questions

You've reached the end

Spheres and Cones

What you'll learn

The key measurements

Cones

Example: volume of a cone from a diameter

Slant height and total surface area of a cone

Example: total surface area using Pythagoras

Spheres and hemispheres

Example: total surface area of a hemisphere

Answers in terms of $π\pi$

Compound solids

Example: cone on top of a hemisphere

Reverse problems: finding the radius

Example: sphere with the same volume as a cube

Water displacement

Example: rise in water level

Recap questions

You've reached the end

Spheres and Cones

What you'll learn

The key measurements

Cones

Example: volume of a cone from a diameter

Slant height and total surface area of a cone

Example: total surface area using Pythagoras

Spheres and hemispheres

Example: total surface area of a hemisphere

Answers in terms of π\piπ

Compound solids

Example: cone on top of a hemisphere

Reverse problems: finding the radius

Example: sphere with the same volume as a cube

Water displacement

Example: rise in water level

Recap questions

You've reached the end

Spheres and Cones Revision Guide

Spheres and Cones

What you'll learn

The key measurements

Cones

Example: volume of a cone from a diameter

Slant height and total surface area of a cone

Example: total surface area using Pythagoras

Spheres and hemispheres

Example: total surface area of a hemisphere

Answers in terms of π\piπ

Compound solids

Example: cone on top of a hemisphere

Reverse problems: finding the radius

Example: sphere with the same volume as a cube

Water displacement

Example: rise in water level

Recap questions

You've reached the end

Spheres and Cones Revision Guide

Answers in terms of $π\pi$

Answers in terms of $π\pi$