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This page sets out the method for spheres & cones in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Spheres & cones topic page.
When you are given a radius (or diameter) and asked for volume or surface area.
When a shape is made from a cone and a hemisphere (or a cylinder), split it up.
The volume of a sphere of radius 3 is 36π. Cube the radius, then take four thirds.
The surface area of a sphere of radius 3 is 36π. Square the radius and multiply by 4.
A solid hemisphere is half a sphere. The volume of a hemisphere of radius 3 is 18π.
The volume of a cone of radius 3 and height 4 is 12π. Square the radius, multiply by the height, and divide by 3.
Volume of a sphere cubes the radius. Surface area squares it. A cone squares the radius and uses the height once.
Leave π in the answer. Do not replace it with a rounded decimal.
Substitute the radius and cube it for volume. A sphere of radius 6 cm has r^3 = 216, so the volume is 4/3π × 216 = 288π cm^3. The calculator value is about 904.779, and to 3 significant figures that is 905 cm^3. The first three digits of 904.779 are 905 because the following digit is 7, so the 4 rounds up.
Surface area squares the radius instead of cubing it. A sphere of radius 3 cm has surface area 4π × 9 = 36π cm^2, which is 113.1 cm^2 to 1 decimal place. There is no flat face to add. If a question gives the diameter, halve it before you substitute. A diameter of 6 cm is a radius of 3 cm, and using 6 in 4π r^2 makes the area four times too big.
Method: 1. Cube the radius for volume: V=4/3π r^3. 2. Radius 6cm gives 4/3π× 216=905cm^3 to 3 significant figures. 3. Square the radius for surface area: radius 3cm gives 4π× 9=113.1cm^2 to 1 decimal place.
The volume formula uses the base radius and the perpendicular height. For radius 4 cm and height 9 cm, r^2 = 16, so the volume is 1/3π × 16 × 9 = 48π cm^3. The 1/3 is easy to forget, and without it you have found the volume of the cylinder that the cone would fit inside.
The height in this formula is perpendicular. If a diagram shows a leaning measurement along the side, that is the slant, and it does not go straight into 1/3π r^2 h. You would first need the perpendicular height from Pythagoras, rearranged. Read the labels. A height marked inside the cone, at right angles to the base, is the h you want.
Method: 1. Use the perpendicular height in V=1/3π r^2h. 2. Radius 4cm and height 9cm give 1/3π× 16× 9=48πcm^3. 3. Do not drop the 1/3, or you have the surrounding cylinder.
A cross-section through the apex of a cone shows a right-angled triangle. The two shorter sides are the radius and the perpendicular height. The hypotenuse is the slant height l. So l^2 = r^2 + h^2. With radius 5 cm and height 12 cm, l^2 = 25 + 144 = 169, and l = √(169) = 13 cm. That is a Pythagorean triple, so the slant is exact.
The curved surface is then π r l = π × 5 × 13 = 65π cm^2, which is 204.2 cm^2 to 1 decimal place. The total surface area adds the base: 65π + π × 5^2 = 65π + 25π = 90π cm^2. Read the question. Curved surface means π r l only. Total surface means π r l + π r^2.
Method: 1. Find the slant with Pythagoras: l^2=r^2+h^2. 2. Radius 5cm and height 12cm give l=√(25+144)=13cm. 3. Curved surface π rl=65π, and the total with the base is 90πcm^2.
A hemisphere has half the volume of a sphere, 1/2 × 4/3π r^3 = 2/3π r^3. For radius 4 cm that is 2/3π × 64 = 128/3π cm^3. The curved surface of a hemisphere is 2π r^2, half of 4π r^2. A solid hemisphere also has the flat circular face, area π r^2, so its total surface area is 3π r^2.
Joined solids add volumes when you want the amount of material. A cone of radius 4 cm and height 9 cm stuck to a hemisphere of radius 4 cm has volume 48π + 128/3π = 272/3π cm^3. That is 284.837ldots cm^3, which rounds to 284.8 cm^3 to 1 decimal place. The shared circular face is inside the solid, so it is not added again when you are finding volume.
Method: 1. Halve a sphere volume to get 2/3π r^3. 2. Radius 4cm gives 2/3π× 64=128/3πcm^3. 3. Add a cone of 48π to that hemisphere to get 272/3π=284.8cm^3 to 1 decimal place.
In many exam questions on spheres and cones the formula is printed above the question, for example the volume of a sphere as 4/3π r^3 or the curved surface area of a cone as π r l. Check which formula is given and which quantity the question asks for. The marks are for substituting correctly and evaluating accurately.
On a calculator, deal with the power first. Work out r^3 as the radius times itself three times, then multiply by π and by the fraction. A common slip is to type r × 3 instead of r^3. Write the substitution as a line, such as 4/3 × π × 5^3, before you press any keys, because the method mark is for that line.
Method: 1. Write down the formula and name each letter. 2. Substitute the numbers, keeping the powers, for example 4/3 × π × 5^3. 3. Work out the power first, then multiply by π and the fraction. 4. Round to the accuracy asked for and write the correct unit: cm^3 for volume, cm^2 for area.
Tip: Use the π button on your calculator, not 3.14, unless the question tells you to use an approximation.
Exam questions often join two solids, such as a cone on top of a hemisphere (an ice cream cornet) or a hemisphere on top of a cylinder (a silo). To find the volume, work out the volume of each part and add them. Only use the radius and heights that belong to each part, and check which measurement is the cone height and which is the slant height.
For surface area, count only the faces you can see from outside. When a cone and hemisphere are joined at their circular faces, those two circles are hidden, so they are not counted. A solid hemisphere has a curved surface 2π r^2 and a flat circle π r^2, so a total of 3π r^2. For a cone with a hemisphere, use π r l + 2π r^2.
Method: 1. Split the solid into cones, cylinders, hemispheres or spheres. 2. Write down the radius, height or slant height for each part. 3. For volume, add the volumes. For surface area, add only the outside faces. 4. Round at the end and give the unit with the correct power.
A solid is melted and the metal is made into smaller solids. The total volume stays the same, so the number of small solids is the large volume divided by one small volume. For a sphere of radius 6 cm melted into cones, work out 4/3π × 6^3 = 288π and divide by the volume of one cone.
When the question is in terms of π you can cancel it, which avoids rounding. Equal-volume problems with algebra, such as a sphere of radius 3x and a cone of radius 4x, are solved by setting the two volume formulas equal. The π cancels and so do some powers of x. Divide through and simplify the ratio.
Method: 1. Write the volume of the first solid in terms of π without rounding. 2. Write the volume of one new solid in the same way. 3. Divide, or set equal for an equal-volume problem, and cancel π. 4. Give the answer as a whole number of solids, or as a simplified ratio.
Capacity questions finish with units. One litre is 1000 cm^3, and one cubic metre is 1000 litres. A tank in the shape of a hemisphere of radius 1.2 m has volume in m^3, and you multiply by 1000 to change this to litres. Check the units of the radius before you start so the volume comes out in the unit you want.
For a filling or emptying rate, divide the capacity by the rate to find the time. If the time is not a whole number, read the question: it may ask for the nearest minute or for hours and minutes. Convert a decimal part of an hour by multiplying by 60. Finish with a sentence if the question asks you to decide, for example whether the container fills in a given time.
Method: 1. Find the volume in the unit of the radius. 2. Convert to litres: divide cm^3 by 1000, or multiply m^3 by 1000. 3. Divide the capacity by the rate to get the time. 4. Round as asked and compare with any time limit in the question.
Tip: A decimal of an hour is not minutes. 1.5 hours is 1 hour 30 minutes.
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