Back to homeMathematics · 6th Year OL · Change subject
Start for free
6th Year OL · Mathematics
3D shapes · 3D shapes

Solids of revolution

Before you start, make sure you know: Pythagoras' Theorem (2nd Year, Junior Cycle)
Before you start, make sure you know: Pythagoras' Theorem (5th Year, Leaving Certificate — Ordinary Level)

A can of cola from the shop, a 99 ice-cream cone, a sliotar. None of them is a polyhedron: they are solids of revolution, which are obtained by rotating a plane shape about a line. If you rotate a rectangle about one of its sides, a cylinder "appears". Later on we will use these shapes to work out how much a 99 cone can hold.

How they are generated

Move the slider to rotate the plane shape about the axis and you will see how the solid is formed:

Shape that rotates Axis of rotation Solid
Rectangle one of its sides Cylinder
Right-angled triangle one of the shorter sides Cone
Semicircle its diameter Sphere

Elements: the radius $r$ of the base, the height $h$ and the slant height $l$ (the side that, when it rotates, "sweeps out" the curved surface). In Junior Cycle we call the area of the curved part the curved surface area (CSA) and the area of everything, including the bases, the total surface area (TSA). In the cone, $l$ is the hypotenuse: $l = \sqrt{h^2 + r^2}$.

Practice★☆☆Not started
Imagine you have a rectangular piece of card with a base of $10\text{ cm}$ and a height of $15\text{ cm}$. If you spin it quickly around one of its $15\text{ cm}$ sides…

Imagine you have a rectangular piece of card with a base of $10\text{ cm}$ and a height of $15\text{ cm}$. If you spin it quickly around one of its $15\text{ cm}$ sides (the side marked as the axis of rotation in the figure), what solid of revolution is generated? Draw the resulting solid and state its radius and its height.

Rectangle with a 10 cm base and a 15 cm height, with its left side marked as the axis of rotationAxis of rotation10 cm15 cm

You can do it on paper and send a photo with your answer.

Solve
Practice★★☆Not started
A solid of revolution is generated by rotating a rectangle with base $4\text{ cm}$ and height $7\text{ cm}$ about one of its longer sides. 1. What geometric shape is…

A solid of revolution is generated by rotating a rectangle with base $4\text{ cm}$ and height $7\text{ cm}$ about one of its longer sides.

  1. What geometric shape is obtained?

  2. Calculate its total surface area.

  3. If it were rotated about the shorter side, would it have the same volume? Justify your answer by performing the calculations.

Axis of rotation 360° 7 cm 4 cm
Solve

The cylinder

If you cut off the label of a can and flatten it out, you get a rectangle: its base is the length of the circumference, $2\pi r$, and its height is $h$.

Net of a cylinder made up of a central rectangle with sides 2πr and h next to two circles of radius r 2πrh r CSA = 2πrh TSA = 2πrh + 2πr²
Practice★☆☆Not started
A standard drink can has a cylindrical shape. If the radius of its base measures $3.25\text{ cm}$ and its height is $12\text{ cm}$, calculate the area of the paper label…

A standard drink can has a cylindrical shape. If the radius of its base measures $3.25\text{ cm}$ and its height is $12\text{ cm}$, calculate the area of the paper label that wraps around the side of the can (lateral area). Use $\pi \approx 3.14$.

r = 3.25 cm h = 12 cm Lateral area (label)
Solve
Practice★★☆Not started
A courier company uses cylindrical cardboard tubes to send blueprints. If each tube has a lateral area of $942\text{ cm}^2$ and a height of $50\text{ cm}$: 1. Calculate…

A courier company uses cylindrical cardboard tubes to send blueprints. If each tube has a lateral area of $942\text{ cm}^2$ and a height of $50\text{ cm}$:

  1. Calculate the radius of the base of the tube.

  2. Calculate the amount of cardboard needed to make the two circular lids for each tube. (Use $\pi \approx 3.14$).

50 cm r = ? Lateral area Aₗ = 942 cm² Lids A = ?
Solve
Practice★★☆Not started
A canned vegetable company in Cork designs a new cylindrical tin can with a top and bottom lid. Due to strength and stacking requirements, it is established that the…

A canned vegetable company in Cork designs a new cylindrical tin can with a top and bottom lid. Due to strength and stacking requirements, it is established that the height of the cylinder must be exactly three times the radius of its base ($h = 3r$). Furthermore, to fit the manufacturing budget, the total area of sheet metal used for each can must be exactly $128\pi\text{ cm}^2$ (about $402.12\text{ cm}^2$).

  1. a) Write the algebraic expression for the total area of the cylinder solely in terms of the radius $r$, set up the corresponding equation, and find the exact value of the radius and the height of the container.
  2. b) Calculate what percentage of the can's total area corresponds to its lateral surface and what percentage to the two flat bases.
  3. c) Reason whether these percentages would change if you decided to double both the radius and the height of the container while maintaining the proportion $h = 3r$.
Solve

The cone and the frustum of a cone

The net of the curved surface of a cone is a sector of a circle with radius equal to the slant height $l$:

Net of a cone made up of a sector of a circle of radius l and a circular base l CSA = π r l TSA = πrl + πr² l = √(h² + r²)

The sector has radius $l$ and its arc has the same length as the circumference of the base, $2\pi r$. A full circle of radius $l$ has circumference $2\pi l$, so the angle of the sector is $\theta = 360^\circ \times \frac{r}{l}$.

Practice★☆☆Not started
For a wizard costume, a student has made a cone-shaped hat. If the base radius is $10\text{ cm}$ and the height of the hat is $24\text{ cm}$, use Pythagoras' theorem to…

For a wizard costume, a student has made a cone-shaped hat. If the base radius is $10\text{ cm}$ and the height of the hat is $24\text{ cm}$, use Pythagoras' theorem to calculate the slant height ($g$) of the cone.

Solve
Practice★★☆Not started
A photography studio uses a cone-shaped reflector to focus light. If the reflector has a height of $12\text{ cm}$ and the diameter of its circular opening is…

A photography studio uses a cone-shaped reflector to focus light. If the reflector has a height of $12\text{ cm}$ and the diameter of its circular opening is $10\text{ cm}$, calculate the total surface area of reflective material required to manufacture it (excluding the base cover).

d = 10 cm h = 12 cm r = 5 cm g
Solve
Practice★☆☆Not started
For a town festival, Sara wants to cover the outside of $20$ cardboard party hats with shiny paper. Each hat has a base radius of $9\text{ cm}$ and a height of…

For a town festival, Sara wants to cover the outside of $20$ cardboard party hats with shiny paper. Each hat has a base radius of $9\text{ cm}$ and a height of $12\text{ cm}$ (the hats are hollow and have no base). Her classmate Hugo does the following calculations to find the paper needed for a single hat:

$$A = \pi \times r \times h = 3.14 \times 9 \times 12 = 339.12\text{ cm}^2$$

Therefore, Hugo concludes that for the $20$ hats they will need exactly $20 \times 339.12 = 6,782.4\text{ cm}^2$ of paper.

  1. a) Identify and explain the two conceptual mistakes Hugo has made: one in the surface area formula he used and another by not calculating a key measurement of the cone.
  2. b) Correct the calculations by first finding the slant height of the cone using the Pythagorean theorem and determine the actual minimum amount of shiny paper they need for the $20$ hats.
Solve

The sphere

A sphere cannot be flattened onto a plane without distorting it (which is why every map of Ireland on a flat page distorts something). Its surface area is exactly four times the area of its great circle:

$$A = 4\pi r^2$$ Diagram showing a sphere whose surface area equals that of four circles of the same radius r = 4 circlesof radius r

A hemisphere has half the curved surface of a sphere, $2\pi r^2$, plus a flat circular face $\pi r^2$ if it is closed. Its net is a curved piece (which cannot be flattened exactly) and a circle of radius $r$: CSA $= 2\pi r^2$, TSA $= 3\pi r^2$. The sphere has no true net: its surface is the curved part only, $4\pi r^2$.

Practice★☆☆Not started
A tennis ball has an approximate diameter of $6.7\text{ cm}$. Calculate the total surface area of felt needed to cover the ball, treating it as a perfect sphere. Express…

A tennis ball has an approximate diameter of $6.7\text{ cm}$. Calculate the total surface area of felt needed to cover the ball, treating it as a perfect sphere. Express your answer in $\text{cm}^2$.

r d = 6.7 cm
Solve
Practice★☆☆Not started
A town council wants to restore and paint a spherical steel sculpture installed on a roundabout, whose diameter measures exactly $4\text{ m}$. The painter in charge has…

A town council wants to restore and paint a spherical steel sculpture installed on a roundabout, whose diameter measures exactly $4\text{ m}$. The painter in charge has cans of rust-preventive paint where the manufacturer indicates that each can covers $12\text{ m}^2$.

  1. a) Mentally estimate, without a calculator and rounding $\pi \approx 3$, the exterior surface area of the sphere and how many cans would be needed at minimum. Explain the steps of your mental calculation.
  2. b) Perform the precise calculation using the formula for the surface area of a sphere ($A = 4\pi r^2$) with the calculator's value of $\pi$ (rounding the area to two decimal places).
  3. c) Check and compare the actual result with your estimate: would it have been enough to buy the cans estimated with $\pi \approx 3$ or would the painter have run out of paint before finishing? Justify why this difference occurs when rounding $\pi$.
Solve
Practice★★☆Not started
A spherical gas storage tank has an outer diameter of $6\text{ metres}$. Engineers want to apply a coat of anti-rust paint costing $\text{€}\ 14.50$ per square metre.…

A spherical gas storage tank has an outer diameter of $6\text{ metres}$. Engineers want to apply a coat of anti-rust paint costing $\text{€}\ 14.50$ per square metre. How much will it cost to paint the total surface area of the tank? (Use $\pi \approx 3.14$).

Solve

Volumes of solids of revolution

To see how much a 99 cone or a can of cola holds, we need volume. These formulas apply to every solid in this page.

For a composite solid, add the volumes of the parts. For the surface area, add the areas of the exposed surfaces only: do not count the faces where the parts are joined.

For similar solids with scale factor $k$, lengths scale by $k$, areas by $k^2$ and volumes by $k^3$. So a cone with half the height (and half the radius) has $\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ of the volume.

Why? A cone fits exactly three times into a cylinder with the same base and height, and a sphere takes up two thirds of the cylinder that just contains it.

Practice★☆☆Not started
Find the volume of a cylindrical water tank with a radius of $2\text{ metres}$ and a height of $5\text{ metres}$. If $1\text{ m}^3$ is equivalent to…

Find the volume of a cylindrical water tank with a radius of $2\text{ metres}$ and a height of $5\text{ metres}$. If $1\text{ m}^3$ is equivalent to $1000\text{ litres}$, how many litres of water can the tank hold?

Solve
Practice★☆☆Not started
An ice cream cone has the shape of a cone. The radius of the top opening is $3\text{ cm}$ and its depth (height) is $10\text{ cm}$. What volume of ice cream fits exactly…

An ice cream cone has the shape of a cone. The radius of the top opening is $3\text{ cm}$ and its depth (height) is $10\text{ cm}$. What volume of ice cream fits exactly inside the cone? (Assume the ice cream is level with the rim, without overflowing).

h=10 cm r=3 cm
Solve
Practice★★☆Not started
A glass tumbler has the shape of a cylinder with an internal radius of $3.5\text{ cm}$. If we pour a soft drink into it that occupies a volume of $250\text{ cm}^3$, what…

A glass tumbler has the shape of a cylinder with an internal radius of $3.5\text{ cm}$. If we pour a soft drink into it that occupies a volume of $250\text{ cm}^3$, what height will the liquid reach inside the tumbler? Round your answer to two decimal places.

Solve
Practice★★☆Not started
In an ice cream parlour, a perfectly spherical scoop of ice cream of diameter $6\text{ cm}$ is served on a cone-shaped wafer of height $10\text{ cm}$ and the same…

In an ice cream parlour, a perfectly spherical scoop of ice cream of diameter $6\text{ cm}$ is served on a cone-shaped wafer of height $10\text{ cm}$ and the same diameter as the scoop. If the ice cream melts and stays completely inside the cone, will it overflow or will there be space left? Answer by comparing the volumes of the sphere and the cone.

Solve
Practice★★★Not started
An ice cream glass has the shape of an inverted cone with height $12 \text{ cm}$ and a top diameter of $10 \text{ cm}$. It is filled with chocolate milkshake to exactly…

An ice cream glass has the shape of an inverted cone with height $12 \text{ cm}$ and a top diameter of $10 \text{ cm}$. It is filled with chocolate milkshake to exactly half its height. One student claims that the glass is "half full", but another says that there is much less than half of the total volume. Calculate what percentage of the total volume of the glass the milkshake actually occupies to check who is right.

Solve
Practice★★★Not started
A spherical lead ball with a diameter of $6 \text{ cm}$ has been completely submerged in a cylindrical container of radius $10 \text{ cm}$ containing water. 1. By how…

A spherical lead ball with a diameter of $6 \text{ cm}$ has been completely submerged in a cylindrical container of radius $10 \text{ cm}$ containing water.

  1. By how much has the water level in the container risen after introducing the ball?

  2. If after this, we melt the lead ball and make a cone with it whose base has the same radius as the original sphere ($3 \text{ cm}$), what height will the cone have?

r=3 Pb R=10 Δh=? Melt r=3 h=?
Solve

Summary

See all the exercises in this section