Solids of Revolution
A water tank on a rooftop, an ice cream cone on the Malecón, a baseball. None of them has flat faces: they are solids of revolution, which are obtained by spinning a flat shape around a line. How much material do you need to make them, and how much fits inside? That is what you will learn here.
How They Are Generated
Move the slider to spin the flat shape around the axis and watch the solid form:
| Shape that spins | Axis of rotation | Solid |
|---|---|---|
| Rectangle | one of its sides | Cylinder |
| Right triangle | one leg | Cone |
| Right trapezoid | the side perpendicular to the bases | Truncated cone |
| Semicircle | its diameter | Sphere |
Parts: the radius $r$ of the base, the height $h$, and the slant height $g$ (the side that "sweeps" the lateral surface as it spins). In the cone, $g$ is the hypotenuse of the triangle that spins: $g = \sqrt{h^2 + r^2}$ (Pythagorean theorem).
The side that lies on the axis is the height. The side perpendicular to the axis draws the circle of the base as it spins, and that is why it is the radius.
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.
You can do it on paper and send a photo with your answer.
Look at the following net of a solid of revolution. a) Which solid is it? b) If the radius of the circle is $R = 5\text{ cm}$ and the height of the rectangle is…
Look at the following net of a solid of revolution.
- a) Which solid is it?
- b) If the radius of the circle is $R = 5\text{ cm}$ and the height of the rectangle is $H = 10\text{ cm}$, calculate the total area of the figure.
The Cylinder
If you cut the label off a can and unroll it, you get a rectangle. Its base is the length of the circumference, $2\pi r$, and its height is $h$.
The volume is the area of the base times the height, as in any prism: $V = \pi r^2 h$. If the measurements are in meters, the result is in m³, and $1\text{ m}^3 = 1{,}000$ liters.
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 $\pi \approx 3.14$.
Find the volume of a cylindrical water tank with a radius of $2\text{ meters}$ and a height of $5\text{ meters}$. If $1\text{ m}^3$ is equivalent to…
Find the volume of a cylindrical water tank with a radius of $2\text{ meters}$ and a height of $5\text{ meters}$. If $1\text{ m}^3$ is equivalent to $1000\text{ liters}$, how many liters of water can the tank hold?
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.
Problems with Cylinders
Sometimes you are given the area or the volume and asked for a measurement. Solve using the formula: from $A_L = 2\pi r h$ you get $r = \dfrac{A_L}{2\pi h}$, and from $V = \pi r^2 h$ you get $h = \dfrac{V}{\pi r^2}$.
To compare two solids you don't need to remember anything new: calculate the volume and the area of each one and compare. If a cylinder has half the radius and twice the height, the radius is squared in the volume and the height is not. So the volume does not stay the same: calculate it to check.
A cylinder can also come from spinning a rectangle. Depending on which side you use as the axis, the radius and the height swap, and the volume changes.
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 $50\text{ cm}$:
-
Calculate the radius of the base of the tube.
-
Calculate the amount of cardboard needed to make the two circular lids for each tube. (Use $\pi \approx 3.14$).
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.
-
What geometric shape is obtained?
-
Calculate its total surface area.
-
If it were rotated about the shorter side, would it have the same volume? Justify your answer by performing the calculations.
A stone rolling pin has a cylindrical shape with a radius of $4 \text{ cm}$ and a length of $25 \text{ cm}$. It is to be stored in a cylindrical box such that the…
A stone rolling pin has a cylindrical shape with a radius of $4 \text{ cm}$ and a length of $25 \text{ cm}$. It is to be stored in a cylindrical box such that the rolling pin fits exactly (tangent to the walls and bases). If it is decided to replace this rolling pin with another that has half the radius but double the length:
-
Will the volume of stone used change? Justify your answer with calculations.
-
Which of the two rolling pins would require a box with a larger total surface area?
The Cone
The lateral net of a cone is a circular sector whose radius is the slant height $g$:
A cone fits three times into a cylinder with the same base and the same height, so its volume is one third: $V = \dfrac{1}{3}\pi r^2 h$.
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.
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).
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).
The Truncated Cone
If you cut a cone with a plane parallel to the base and remove the tip, what is left is a truncated cone (also called a frustum), like a paper cup or a popcorn bucket. It has two circular bases, with radii $R$ (larger) and $r$ (smaller), a height $h$, and a slant height $g$.
It is obtained by spinning a right trapezoid around the side perpendicular to the bases. In the right triangle with legs $h$ and $R - r$, the hypotenuse is $g$:
$$g = \sqrt{h^2 + (R - r)^2}$$The lateral net is a large circular sector with a small one removed. That gives the formulas:
$$A_L = \pi (R + r)\, g \qquad V = \frac{\pi h}{3}\left(R^2 + Rr + r^2\right)$$The safest strategy is to complete the cone: the truncated cone is the large cone minus the small cone that you removed. Both cones are similar (their sides keep the same proportion, as in Thales' theorem), so: $\dfrac{h_{\text{small}}}{h_{\text{large}}} = \dfrac{r}{R}$.
Calculate the lateral surface area and volume of a frustum of a cone (such as a popcorn bucket) whose bases have radii of $10\text{ cm}$ and $6\text{ cm}$, knowing that…
Calculate the lateral surface area and volume of a frustum of a cone (such as a popcorn bucket) whose bases have radii of $10\text{ cm}$ and $6\text{ cm}$, knowing that its height is $15\text{ cm}$. Note: You will need to calculate the slant height first using Pythagoras' theorem.
A lamp designer wants to create a lampshade in the shape of a frustum of a cone. The lampshade has a height of $24 \text{ cm}$, a top diameter of $18 \text{ cm}$ and a…
A lamp designer wants to create a lampshade in the shape of a frustum of a cone. The lampshade has a height of $24 \text{ cm}$, a top diameter of $18 \text{ cm}$ and a bottom diameter of $30 \text{ cm}$. To calculate the cost of the material, he needs to know the lateral surface area. However, to manufacture it from a flat piece, he must first determine the height of the original full cone from which the frustum comes. Calculate this total height and the lateral area of the lampshade (without the bases).
Calculate the volume of the solid generated by rotating $360^\circ$ the following right-angled trapezium around side $AB$. Side $AD$ measures $6 \text{ cm}$ (shorter…
Calculate the volume of the solid generated by rotating $360^\circ$ the following right-angled trapezium around side $AB$. Side $AD$ measures $6 \text{ cm}$ (shorter base), side $BC$ measures $10 \text{ cm}$ (longer base) and the height of the trapezium (side $AB$) measures $8 \text{ cm}$. Identify the resulting solid and calculate its total surface area.
The Sphere
A sphere cannot be unfolded onto a plane without distorting it (which is why every world map distorts something). Its area is exactly four times the area of its great circle, and its volume is:
$$A = 4\pi r^2 \qquad V = \frac{4}{3}\pi r^3$$A hemisphere (half a sphere) has half the volume: $\dfrac{2}{3}\pi r^3$. Its curved surface is $2\pi r^2$, and if you also count the flat circle that closes it, the total area is $3\pi r^2$.
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 $\text{cm}^2$.
If we cut an orange in half, we get two hemispheres. If the radius of the orange is $4\text{ cm}$, calculate the volume of one of these halves.
If we cut an orange in half, we get two hemispheres. If the radius of the orange is $4\text{ cm}$, calculate the volume of one of these halves.
A spherical gas storage tank has an outer diameter of $6\text{ meters}$. Engineers want to apply a coat of anti-rust paint costing $\text{RD\$}\ 870…
A spherical gas storage tank has an outer diameter of $6\text{ meters}$. Engineers want to apply a coat of anti-rust paint costing $\text{RD\$}\ 870$ per square meter. How much will it cost to paint the total surface area of the tank? (Use $\pi \approx 3.14$).
Composite Solids
Many real objects combine several solids: a rice silo (a cylinder with a hemisphere on top), a medicine capsule, an ice cream cone with a scoop on top. To calculate them:
-
Break the object into known solids.
-
Calculate the volume of each part and add.
-
For the area, add only the surfaces that are on the outside. Faces that are joined together do not count.
A brand of vitamin supplements designs its capsules in the shape of a cylinder with two hemispheres at the ends. The total length of the capsule is $22\text{ mm}$ and…
A brand of vitamin supplements designs its capsules in the shape of a cylinder with two hemispheres at the ends. The total length of the capsule is $22\text{ mm}$ and the diameter of the central cylinder is $8\text{ mm}$. Calculate the total volume of the capsule in cubic millimeters.
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.
A steel water tank is to be built, combining a cylinder of height $4 \text{ m}$ and radius $3 \text{ m}$ with a perfect hemisphere at its top. If the cost of the steel…
A steel water tank is to be built, combining a cylinder of height $4 \text{ m}$ and radius $3 \text{ m}$ with a perfect hemisphere at its top. If the cost of the steel is $\text{RD\$}\ 2\,700$ per square meter of exterior surface and sealing the joint between the hemisphere and the cylinder costs $\text{RD\$}\ 900$ per linear meter of circumference, calculate the total budget for the exterior surface materials of the tank (including the flat bottom base).
Proportions and Changes of Shape
Two cones obtained by cutting one parallel to the base are similar. If the ratio between their linear measurements is $k$:
| Quantity | Multiplied by |
|---|---|
| Lengths (radius, height) | $k$ |
| Areas | $k^2$ |
| Volumes | $k^3$ |
That is why a cone filled to half its height ($k = \tfrac12$) holds only $\left(\tfrac12\right)^3 = \tfrac18$ of the total volume, that is, $12.5\%$. Anyone who says "half full" is mistaken.
When a solid is melted down or submerged, its volume is conserved. This lets you set volumes equal:
-
A ball submerged in a cylindrical container makes the water rise by a volume equal to that of the ball: $V_{\text{ball}} = \pi R^2 \cdot \Delta h$.
-
If a ball is melted down to make a cone, $V_{\text{sphere}} = V_{\text{cone}}$.
A special case: a sphere of radius $R$ inside a cylinder of radius $R$ and height $2R$. The lateral area of the cylinder is $2\pi R \cdot 2R = 4\pi R^2$, which is exactly the area of the sphere.
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.
A perfume bottle has the shape of a sphere of radius $R = 5 \text{ cm}$. The bottle is contained inside a box in the shape of a right cylinder that has the same radius…
A perfume bottle has the shape of a sphere of radius $R = 5 \text{ cm}$. The bottle is contained inside a box in the shape of a right cylinder that has the same radius as the sphere and a height equal to the diameter of the sphere.
-
Calculate the difference between the volume of the cylindrical box and the volume of the sphere.
-
Show that the lateral area of the cylinder is equal to the surface area of the sphere.
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.
-
By how much has the water level in the container risen after introducing the ball?
-
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?
The Parallels of the Earth
The Earth is almost a sphere. A parallel is a horizontal circle on it, and it is smaller the closer it is to the pole.
A point at latitude $\varphi$ lies on a parallel whose radius is smaller than the Earth's radius: $r_{\text{parallel}} = R\cos\varphi$. The reason is the right triangle formed by the Earth's radius, the radius of the parallel, and the axis.
The length of an arc of $\alpha$ degrees on a circle of radius $r$ is:
$$L = \frac{\alpha}{360}\cdot 2\pi r$$Two cities, $P$ and $Q$, lie on the same parallel of latitude at $60^\circ \text{ N}$. City $P$ has a longitude of $15^\circ \text{ E}$ and city $Q$ of…
Two cities, $P$ and $Q$, lie on the same parallel of latitude at $60^\circ \text{ N}$. City $P$ has a longitude of $15^\circ \text{ E}$ and city $Q$ of $45^\circ \text{ E}$. Knowing that the radius of the Earth is approximately $6371 \text{ km}$:
-
Calculate the radius of the parallel $60^\circ \text{ N}$ (Hint: use trigonometry or recall that in the triangle formed, the radius of the parallel is $R \times \cos(60^\circ)$).
-
Find the minimum distance between the two cities measured along the arc of that parallel.