📖 Lesson 17 ⏱ ~30 min Year 10 · Unit 1 ⚡ +50 XP

Surface Area of Cylinders, Cones and Spheres

[PATHS extension] Extend surface area to curved solids. Unroll cylinders, unwrap cones, and cover spheres.

Today's hook: Why does a soft-drink can have that exact shape? It's a quiet battle between surface area and volume.

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Core learning

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Warm-up

Think First

+5 XP each

Q1: Imagine peeling the label off a cylindrical can. What flat shape would it be? What would its dimensions be in terms of the can's radius and height?

Q2: A sphere and a cube both have volume 1000 cm$^3$. Which do you think has the larger surface area? Make a prediction and explain your reasoning.

From the lesson

Intentions

Learning Intentions

Know

The surface area formulas for cylinders, cones and spheres.
How curved surfaces unfold into flat shapes.
The surface area formula for a right pyramid.

Understand

Why the curved surface area of a cylinder is a rectangle when unrolled.
Why a sphere's surface area formula contains $r^2$.
Why a pyramid's slant height is not the same as its vertical height.

Can Do

Calculate total and curved surface area of cylinders, cones and spheres.
Solve composite solid problems involving curved surfaces.
Calculate the total surface area of a right pyramid with a square base.

From the lesson

Success Criteria

I can calculate the total surface area of a closed cylinder, open cylinder, cone and sphere.
I can identify which formula applies to the curved part and which to the base(s).
I can solve practical problems involving painting, wrapping or covering curved objects.
I can calculate the total surface area of a right pyramid by adding the base area to the area of its triangular faces.

From the lesson

Key Terms

Curved surface area (CSA)The area of the curved part only, excluding any flat bases.

Total surface area (TSA)The curved surface area plus the area of all bases.

Slant height ($l$)The distance from the apex of a cone to any point on the edge of its base, measured along the curved surface.

Great circleA circle on a sphere whose plane passes through the centre of the sphere. Its area is $\pi r^2$.

Slant height of a pyramid ($l$)The distance from the apex to the midpoint of a base edge, measured along a triangular face. Not the same as the pyramid's vertical height.

From the lesson

Misconceptions

Common Mistakes to Avoid

Wrong: Using the vertical height of a cone instead of the slant height in the curved surface area formula.

Right: CSA of cone = $\pi r l$ where $l = \sqrt{r^2 + h^2}$. Always calculate slant height first.

Wrong: Adding two circular bases to the surface area of an open cylinder or cone.

Right: A closed cylinder has two bases; an open cylinder has one or none. A cone has one base (unless open at both ends).

Wrong: Using a pyramid's vertical height instead of its slant height when finding the area of a triangular face.

Right: Each triangular face uses the slant height $l$ (apex to midpoint of a base edge, along the face). For a square base, $l = \sqrt{h^2 + (s/2)^2}$, where $h$ is the vertical height and $s$ is the base side.

Concept

Surface Area of Cylinders

+5 XP

When you unroll a cylinder, the curved surface becomes a rectangle. The rectangle's width is the cylinder's height, and its length is the circumference of the base.

Cylinder Surface Area Formulas

Curved surface area: $CSA = 2\pi r h$

Total surface area (closed): $TSA = 2\pi r h + 2\pi r^2$

Total surface area (open top): $TSA = 2\pi r h + \pi r^2$

What to write in your book

Cylinder CSA: $2\pi rh$ (unrolls to a rectangle)
Closed cylinder TSA: $2\pi rh + 2\pi r^2$
Open-top cylinder TSA: $2\pi rh + \pi r^2$
The rectangle's width = height, length = circumference

What is the curved surface area of a cylinder with radius 5 cm and height 8 cm?

Correct! $CSA = 2\pi rh = 2\pi(5)(8) = 80\pi$ cm$^2$.

Not quite. $CSA = 2\pi rh = 2\pi(5)(8) = 80\pi$ cm$^2$.

From the lesson

Worked Example 1

Worked Example 1, Closed Cylinder

Given: A closed cylindrical tin has radius 7 cm and height 12 cm. Find its total surface area.

Method: $TSA = 2\pi r h + 2\pi r^2 = 2\pi(7)(12) + 2\pi(7)^2 = 168\pi + 98\pi = 266\pi \approx 835.7$ cm$^2$.

Answer: $\mathbf{266\pi \approx 835.7}$ cm$^2$ (1 d.p.)

Concept

Surface Area of Cones and Spheres

+5 XP

A cone's curved surface unrolls into a sector of a circle. A sphere has no flat faces, every point is curved.

Cone and Sphere Surface Area Formulas

Cone curved surface area: $CSA = \pi r l$

Cone total surface area: $TSA = \pi r l + \pi r^2$

Sphere surface area: $SA = 4\pi r^2$

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

Heads up

Remember: A sphere has the smallest surface area for a given volume of any 3D shape. Nature uses spheres (bubbles, planets, cells) because they are efficient.

What to write in your book

Cone CSA: $\pi rl$ where $l = \sqrt{r^2+h^2}$
Cone TSA: $\pi rl + \pi r^2$
Sphere SA: $4\pi r^2$
Always calculate slant height first for cones

A sphere has radius 3 cm. What is its surface area?

Correct! $SA = 4\pi r^2 = 4\pi(3)^2 = 36\pi$ cm$^2$.

Not quite. $SA = 4\pi r^2 = 4\pi(3)^2 = 36\pi$ cm$^2$.

From the lesson

Worked Example 2

Worked Example 2, Cone

Given: A cone has radius 5 cm and perpendicular height 12 cm. Find its total surface area.

Method: First find slant height: $l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ cm. Then $TSA = \pi r l + \pi r^2 = \pi(5)(13) + \pi(5)^2 = 65\pi + 25\pi = 90\pi \approx 282.7$ cm$^2$.

Answer: $\mathbf{90\pi \approx 282.7}$ cm$^2$ (1 d.p.)

Concept

Surface Area of Right Pyramids

+5 XP

A right pyramid's surface area is the base area plus the area of its triangular lateral faces. Each triangular face uses the slant height, the distance from the apex to the midpoint of a base edge, not the pyramid's vertical height straight down to the centre of the base.

Right Pyramid Surface Area (square base)

Base area: $A_{base} = s^2$

Area of one triangular face: $\frac{1}{2}sl$

Total surface area: $TSA = s^2 + 4\left(\frac{1}{2}sl\right) = s^2 + 2sl$

Slant height from vertical height: $l = \sqrt{h^2 + \left(\frac{s}{2}\right)^2}$

Heads up

Slant height vs vertical height: The vertical height $h$ runs straight down from the apex to the centre of the base. The slant height $l$ runs down the middle of a triangular face to the midpoint of a base edge. They form a right-angled triangle with half the base side, so $l = \sqrt{h^2 + (s/2)^2}$.

What to write in your book

Right pyramid TSA (square base): $s^2 + 2sl$
$s^2$ is the base area, $2sl$ is the combined area of the 4 triangular faces
Slant height $l = \sqrt{h^2 + (s/2)^2}$, where $h$ is the vertical height
Slant height is measured along a face; vertical height is measured straight down inside the pyramid

A square-based pyramid has base side 10 cm and vertical height 12 cm. What is its slant height?

Correct! $l = \sqrt{h^2 + (s/2)^2} = \sqrt{12^2 + 5^2} = \sqrt{144+25} = \sqrt{169} = 13$ cm.

Not quite. $l = \sqrt{h^2 + (s/2)^2} = \sqrt{12^2 + 5^2} = \sqrt{169} = 13$ cm.

From the lesson

Worked Example 3

Worked Example 3, Square Pyramid

Given: A right pyramid has a square base of side 8 cm and a vertical height of 3 cm. Find its total surface area.

Method: First find the slant height: $l = \sqrt{h^2 + (s/2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$ cm. Base area $= s^2 = 8^2 = 64$ cm$^2$. Lateral area $= 4 \times \frac{1}{2}(8)(5) = 4 \times 20 = 80$ cm$^2$. $TSA = 64 + 80 = 144$ cm$^2$.

Answer: $\mathbf{144}$ cm$^2$

From the lesson

Interactive

Interactive: Cylinder Surface Area Calculator

From the lesson

Practice

Your Turn

Question 1: Find the total surface area of a closed cylinder with radius 4 cm and height 10 cm.

Question 2: Find the surface area of a sphere with radius 6 cm.

Question 3: A cone has radius 3 cm and height 4 cm. Find its total surface area.

From the lesson

Revisit

Revisit Your Thinking

Look back at your Think First prediction about unrolling a cylinder's label. Was the flat shape a rectangle? What were its dimensions in terms of the can's radius and height? If a can has radius 3.5 cm and height 10.5 cm, what is the exact area of the label (curved surface only)?

Reflect

Revisit your thinking

reflect

Earlier you were asked: What was your first thought on this topic?

Now that you've worked through the lesson, write a fuller answer. What changed in your thinking?

Recap

Quick Review

Cylinder CSA

$2\pi rh$

Cylinder TSA

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

Cone CSA

$\pi rl$

Cone TSA

$\pi rl + \pi r^2$

Sphere

$4\pi r^2$

Slant height (cone)

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

Square pyramid TSA

$s^2 + 2sl$

From the lesson

Real-Life Link

Hot water cylinders in Australian homes are typically insulated to reduce heat loss. The amount of insulation needed depends directly on the curved surface area. A standard 250-litre cylinder has radius about 27 cm and height 145 cm, giving a curved surface area of about 2.5 m$^2$. When designing water tanks for rural Australian properties, engineers calculate the total surface area to determine the amount of corrugated steel or concrete needed. For spherical storage tanks used in industrial settings, the spherical shape minimises surface area (and therefore heat loss and material cost) for a given volume. This is why LNG storage tanks are often spherical.

From the lesson

Game

Game Time!

Test your curved surface area skills in an interactive challenge.

Play Curved SA Challenge

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