Maths · 6 Mensuration

6.4 Surface area and volume

Before you start

The area of a rectangle, a triangle and a trapezium (6.2). Of these, only the triangle formula is on the formula list.
The area and circumference of a circle, and the area and arc length of a sector as a fraction of the whole circle (6.3).
Pythagoras' theorem, to find a missing side of a right-angled triangle (7.1).
Units of volume and capacity, including $1$ ml $= 1$ cm³ and $1$ litre $= 1000$ cm³ (6.1).
Rearranging a formula to make another letter the subject.
Rounding to 3 significant figures, and finding square roots and cube roots on a calculator.

Syllabus objectives covered (1)

Carry out calculations and solve problems involving the surface area and volume of a: • cuboid • prism • cylinder • sphere • pyramid • cone.

Part 1 of 8

Volume, surface area and the formula list

Volume is the amount of space a solid takes up. It counts how many unit cubes would fill the solid, so it is measured in cubic units such as cm³ or m³.
Surface area is the total area of all the faces of a solid: everything you would have to paint to cover the outside. It is an area, so it is measured in square units such as cm².
A good way to see a surface area is to unfold the solid flat into a net. The surface area is the area of the net. A cuboid unfolds into six rectangles. A cylinder unfolds into one rectangle and two circles.
Some solids have a curved surface as well as flat faces. The curved surface area of a cylinder or a cone is the curved part only. The total surface area adds the flat faces as well. Read which one the question asks for before you start.
On Papers 2 and 4 a list of formulas is printed on page 2 of the question paper. For this subtopic it gives the curved surface area of a cylinder and of a cone, the surface area of a sphere, and the volume of a prism, a pyramid, a cylinder, a cone and a sphere. You do not need to memorise those. On Papers 5 and 6 there is no list, and any formula you need is given in the question.
The list does not tell you which faces make up a total surface area, or which length goes in for each letter. The area of a rectangle is not on it either, so you must know that one. Choosing the right faces and the right lengths is the skill this lesson teaches.
Answers may be asked for **in terms of $π$ **. That means you leave $π$ as a symbol, for example $36 π$ cm³, and do not turn it into a decimal.

Part 2 of 8

Prisms: cross-section times length

A prism is a solid with the same cross-section all the way through. Slice it anywhere at right angles to its length and you get the same shape every time.
Its volume is the area $A$ of the cross-section times the length $l$ of the prism.
$V = A l$
The reason: a slice $1$ cm thick has a volume of $A$ cm³, because it is the cross-section with a depth of $1$ . A prism $l$ cm long is $l$ of those slices stacked together, so its volume is $A \times l$ .
A cuboid is a prism with a rectangular cross-section, so its volume is length $\times$ width $\times$ height. A cuboid $3$ cm by $4$ cm by $5$ cm has volume $3 \times 4 \times 5 = 60$ cm³.
For any other prism, find the area of the cross-section first, with the area formula for that shape. Then multiply by the length. A prism $9$ cm long whose cross-section is a triangle with base $5$ cm and height $4$ cm has a cross-section of $\frac{1}{2} \times 5 \times 4 = 10$ cm², so its volume is $10 \times 9 = 90$ cm³.
graph diagram — not drawn yet
A triangular prism drawn in perspective, lying on one rectangular face. The front triangular end is shaded and labelled 'cross-section, area A', with its base labelled 5 cm and a dashed perpendicular height labelled 4 cm. The edge running back from the front end to the back end is labelled 'length l = 9 cm'.
The shaded end is the same shape at every slice along the length. Volume is that area times the length.
The trap is to multiply the three lengths you can see. That only works for a cuboid. For the triangular prism, $5 \times 4 \times 9 = 180$ is double the volume, because the triangle is only half of a $5$ by $4$ rectangle.
A cylinder is a prism too: its cross-section is a circle. The syllabus also counts a cylindrical sector as a prism. That is a solid whose cross-section is a sector of a circle, like a wedge cut from a round cheese.
Take a cylindrical sector $10$ cm long whose cross-section is a $90 °$ sector of radius $4$ cm. The cross-section is a quarter of a circle, so its area is $\frac{90}{360} \times π \times 4^{2} = 4 π$ cm². The volume is $4 π \times 10 = 40 π$ cm³, which is $126$ cm³ to 3 significant figures.
graph diagram — not drawn yet
A cylindrical sector drawn in perspective: a solid whose front face is a 90-degree sector of a circle of radius 4 cm, with the two straight radii labelled 4 cm and the angle at the centre labelled 90°, extending back 10 cm (labelled 'length 10 cm'). The curved face is shaded lightly and labelled 'curved face', and the two flat rectangular faces, each 4 cm by 10 cm, are labelled 'flat face'.
The front face is a sector, and it is the same at every slice, so this is a prism. It has two sector ends, two flat rectangles and one curved face.

Part 3 of 8

Surface area of a cuboid and of a prism

To find a surface area, list the faces, find the area of each one, then add. Write the list down, so that you can check that no face is missing.
A cuboid has six rectangular faces in three matching pairs: top and bottom, front and back, and the two ends. With length $l$ , width $w$ and height $h$ :
$surface area = 2 (l w + l h + w h)$
For the $3$ cm by $4$ cm by $5$ cm cuboid: $2 (12 + 15 + 20) = 2 \times 47 = 94$ cm².
Any prism has two identical ends, which are the cross-section. It also has one rectangle for each edge of the cross-section. Each rectangle is as long as the prism and as wide as that edge.
Those rectangles together form a strip whose width is the whole perimeter of the cross-section. So their total area is the perimeter times the length, and you then add the two ends. This gives you a quick check on your list of faces.
Every edge of the cross-section needs its own length. If the cross-section is a right-angled triangle and you are given only the two shorter sides, find the third side with Pythagoras' theorem before you work out the third rectangle. Worked example 1 does this.
A cylindrical sector has a curved face as well. That face is a rectangle bent round: one side is the length of the prism, and the other side is the arc length of the sector.
For the $90 °$ sector prism above: the two ends are $4 π$ cm² each. The arc is $\frac{90}{360} \times 2 π \times 4 = 2 π$ cm, so the curved face is $2 π \times 10 = 20 π$ cm². The two flat faces are $4 \times 10 = 40$ cm² each.
$8 π + 20 π + 80 = 28 π + 80 \approx 168$
So the total surface area is $28 π + 80$ cm², or $168$ cm² to 3 significant figures. Check that you have the two flat faces: they are the faces where the wedge was cut.

Part 4 of 8

Cylinders

A cylinder of radius $r$ and height $h$ is a prism whose cross-section is a circle of area $π r^{2}$ . So its volume is cross-section times height.
$V = π r^{2} h$
For a cylinder of radius $4$ cm and height $10$ cm: $V = π \times 4^{2} \times 10 = 160 π$ cm³, which is $503$ cm³ to 3 significant figures.
The curved surface unrolls into a rectangle. One side of the rectangle is the height $h$ . The other side went once round the circle, so its length is the circumference, $2 π r$ .
$curved surface area = 2 π r h$
graph diagram — not drawn yet
Left: a cylinder of radius r and height h, with the radius drawn on the top circle and the height marked on the side. Right: its net, a rectangle with its height labelled h and its width labelled 2πr, with a circle of radius r attached to the middle of the top edge and another to the middle of the bottom edge.
The long side of the rectangle wrapped once round the circle, so it is the circumference 2πr.
For the same cylinder, the curved surface area is $2 π \times 4 \times 10 = 80 π$ cm², which is $251$ cm² to 3 significant figures.
A closed cylinder also has two circular ends, each of area $π r^{2}$ .
$total surface area = 2 π r h + 2 π r^{2}$
For the example: $80 π + 2 \times 16 π = 112 π$ cm², which is $352$ cm² to 3 significant figures.
Read the context to decide how many ends there are. A closed can has two. A tank with an open top, or a cup, has one. A pipe, or a paper label round a can, has none: that is the curved surface only.

Part 5 of 8

Pyramids

A pyramid has a base that is a polygon and triangular faces that meet at one point, the apex.
Its volume is one third of the base area $A$ times the perpendicular height $h$ .
$V = \frac{1}{3} A h$
The perpendicular height is measured straight down from the apex to the base, at right angles to the base. It is not the length of a sloping edge.
The $\frac{1}{3}$ is there because a pyramid fills exactly one third of the prism that has the same base and the same perpendicular height. If you leave it out, your answer is three times too big.
A pyramid with a square base of side $6$ cm and perpendicular height $4$ cm has volume $\frac{1}{3} \times 36 \times 4 = 48$ cm³.
For the surface area you need the area of each triangular face. The height of a face is measured up the middle of that face, from the base edge to the apex. This slant height is longer than the perpendicular height of the pyramid, because the face leans inwards.
graph diagram — not drawn yet
A square-based pyramid with base side 6 cm and the apex directly above the centre of the base. A dashed vertical line from the apex to the centre of the base labelled 'perpendicular height 4 cm'. A dashed line from the centre of the base to the midpoint of the front base edge labelled 3 cm. A solid line up the middle of the front face from that midpoint to the apex labelled 'slant height'. The right-angled triangle formed by these three lines is shaded, with a right-angle mark at the centre of the base.
The perpendicular height, half the base side and the slant height make a right-angled triangle inside the pyramid.
When the apex is directly above the centre of a square base, the perpendicular height, half the side of the base and the slant height form a right-angled triangle inside the pyramid. The slant height is its hypotenuse.
$slant height = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5$
The surface area is the base plus four identical triangles: $36 + 4 \times \frac{1}{2} \times 6 \times 5 = 36 + 60 = 96$ cm².
If the base is a rectangle and not a square, the faces come in two pairs with different slant heights, so you need Pythagoras twice. Worked example 2 handles that case.

Part 6 of 8

Cones

A cone is like a pyramid with a circular base. Its volume is one third of the cylinder with the same base and the same perpendicular height.
$V = \frac{1}{3} π r^{2} h$
A cone has two different heights. The perpendicular height $h$ goes from the centre of the base straight up to the apex. The slant height $l$ runs down the sloping side, from the apex to the edge of the base.
$r$ , $h$ and $l$ form a right-angled triangle, with the slant height $l$ as the hypotenuse.
$l^{2} = r^{2} + h^{2}$
graph diagram — not drawn yet
A cone standing on its circular base. A dashed vertical line from the apex to the centre of the base labelled h, a dashed radius from the centre to the edge of the base labelled r, and the sloping edge from the apex to that point on the edge labelled l. The right-angled triangle with sides r, h and l is shaded, with a right-angle mark at the centre of the base.
h is inside the cone and l is on its surface. l is the hypotenuse, so it is always longer than h.
The curved surface area uses the slant height.
$curved surface area = π r l$
The reason: cut the curved surface along a straight line from the apex and flatten it. It becomes a sector of a circle of radius $l$ , and the curved edge of that sector is the circumference of the base, $2 π r$ . The area of a sector is $\frac{1}{2} \times$ radius $\times$ arc length, so the area is $\frac{1}{2} \times l \times 2 π r = π r l$ .
The trap here is to swap the two heights. The volume uses the perpendicular height $h$ . The curved surface area uses the slant height $l$ . If the question gives you one, find the other with Pythagoras.
Take a cone of radius $3$ cm and perpendicular height $4$ cm. First the slant height: $l = \sqrt{3^{2} + 4^{2}} = 5$ cm. The volume is $\frac{1}{3} \times π \times 9 \times 4 = 12 π$ cm³, or $37.7$ cm³ to 3 significant figures.
Its curved surface area is $π \times 3 \times 5 = 15 π$ cm². A solid cone also has a circular base of area $π \times 3^{2} = 9 π$ cm², so its total surface area is $15 π + 9 π = 24 π$ cm², or $75.4$ cm² to 3 significant figures.

Part 7 of 8

Spheres

A sphere is fixed by one length: its radius $r$ , the distance from the centre to the surface. If you are given the diameter, halve it before you use any formula.
The two sphere formulas are both on the formula list.
$V = \frac{4}{3} π r^{3} surface area = 4 π r^{2}$
Notice that $r$ is cubed in the volume and squared in the surface area. That matches the units: a volume is in cubic units and an area is in square units. If you get them the other way round, the units tell you.
For a sphere of radius $6$ cm: $V = \frac{4}{3} \times π \times 216 = 288 π$ cm³, which is $905$ cm³ to 3 significant figures. Surface area $= 4 \times π \times 36 = 144 π$ cm², which is $452$ cm² to 3 significant figures.
A sphere has no flat faces, so its surface area is the whole answer: there is nothing else to add.
A check on the volume: a sphere fills two thirds of the cylinder that just fits round it, of radius $r$ and height $2 r$ . That cylinder has volume $π r^{2} \times 2 r = 2 π r^{3}$ , and two thirds of $2 π r^{3}$ is $\frac{4}{3} π r^{3}$ .

Part 8 of 8

Working backwards, units and accuracy

Some questions give you the volume or the surface area and ask for a length. Write the formula, put in every value you know, then rearrange for the unknown.
A cylinder has volume $600$ cm³ and radius $5$ cm. Then $600 = π \times 5^{2} \times h = 25 π h$ . Divide both sides by $25 π$ , because that is what multiplies $h$ .
$h = \frac{600}{25 π} = 7.639 \dots \approx 7.64$
When the unknown is the radius of a sphere, you need a cube root, because $r$ is cubed. A sphere has volume $1000$ cm³:
$\frac{4}{3} π r^{3} = 1000 \Rightarrow r^{3} = \frac{3000}{4 π} = 238.73 \dots \Rightarrow r = \sqrt[3]{238.73 \dots} = 6.2035 \dots$
So $r = 6.20$ cm to 3 significant figures. Keep the unrounded $6.2035 \dots$ in your calculator for any later part. The surface area from it is $4 π r^{2} = 484$ cm² to 3 significant figures. If you round $r$ to $6.2$ first, you get $483$ cm², which is wrong.
Put every length in the same unit before you substitute. For capacity, $1$ ml $= 1$ cm³ and $1$ litre $= 1000$ cm³. Also $1$ m³ $= 1 000 000$ cm³, because $1$ m $= 100$ cm and $100^{3} = 1 000 000$ .
To give an answer in terms of $π$ , keep $π$ as a symbol and do the rest of the arithmetic exactly. Then collect the $π$ terms: $15 π + 9 π = 24 π$ . Without a calculator, this is how you give an exact answer for a curved solid.
Without a calculator, cancel the fraction before you multiply. In $\frac{1}{3} \times π \times 25 \times 12$ , divide the $12$ by $3$ first to get $4$ , then $25 \times 4 = 100$ , so the answer is $100 π$ .
On a calculator paper, give answers that are not exact to 3 significant figures, using the calculator's $π$ or $3.142$ . Do not round until the final answer.

Worked examples

Example 1. A prism is $20$ cm long. Its cross-section is a right-angled triangle whose two shorter sides are $9$ cm and $12$ cm. Calculate (a) the volume of the prism, (b) its total surface area.

Show the working

The cross-section is the triangle. Its two shorter sides meet at the right angle, so one is the base and the other is the perpendicular height.
$A = \frac{1}{2} \times 9 \times 12 = 54$
Volume is cross-section times length.
$V = 54 \times 20 = 1080$
For the surface area, list the faces: two triangular ends and three rectangles, one for each side of the triangle. You do not have the third side yet, so find it with Pythagoras' theorem.
$\sqrt{9^{2} + 12^{2}} = \sqrt{81 + 144} = \sqrt{225} = 15$
Each rectangle is $20$ cm long and as wide as one side of the triangle.
$9 \times 20 + 12 \times 20 + 15 \times 20 = 180 + 240 + 300 = 720$
Check with the perimeter: $9 + 12 + 15 = 36$ cm, and $36 \times 20 = 720$ cm². The three rectangles agree.
Add the two triangular ends.
$2 \times 54 + 720 = 108 + 720 = 828$

Answer: (a) $1080$ cm³ (b) $828$ cm²

Example 2. A pyramid has a rectangular base $8$ cm by $6$ cm. Its apex is $4$ cm vertically above the centre of the base. Calculate (a) the volume of the pyramid, (b) its total surface area.

Show the working

The base area is $8 \times 6 = 48$ cm², and $4$ cm is the perpendicular height.
$V = \frac{1}{3} \times 48 \times 4 = 64$
For the surface area, notice that the four triangles are two pairs. The two faces on the $8$ cm edges match each other, and the two faces on the $6$ cm edges match each other. The pairs have different slant heights.
For a face on an $8$ cm edge: from the centre of the base to the middle of that edge is half of $6$ cm, which is $3$ cm. That and the $4$ cm height make a right-angled triangle.
$s_{1} = \sqrt{4^{2} + 3^{2}} = \sqrt{25} = 5$
For a face on a $6$ cm edge: from the centre to the middle of that edge is half of $8$ cm, which is $4$ cm.
$s_{2} = \sqrt{4^{2} + 4^{2}} = \sqrt{32} = 5.656 \dots$
Find the area of each pair. Keep $\sqrt{32}$ unrounded in your calculator.
$2 \times \frac{1}{2} \times 8 \times 5 = 40 2 \times \frac{1}{2} \times 6 \times \sqrt{32} = 33.94 \dots$
Add the base, then round only at the end.
$48 + 40 + 33.94 \dots = 121.94 \dots \approx 122$

Answer: (a) $64$ cm³ (b) $122$ cm² (3 s.f.)

Example 3. Do not use a calculator. A solid cone has base radius $5$ cm and perpendicular height $12$ cm. Find (a) its slant height, (b) its volume, (c) its total surface area. Give (b) and (c) in terms of $π$ .

Show the working

The radius, the perpendicular height and the slant height form a right-angled triangle, with the slant height as the hypotenuse.
$l = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13$
The volume uses the perpendicular height, $12$ , not the slant height. Divide the $12$ by $3$ first to keep the arithmetic easy.
$V = \frac{1}{3} \times π \times 5^{2} \times 12 = 25 \times 4 \times π = 100 π$
The curved surface area uses the slant height, $13$ .
$π r l = π \times 5 \times 13 = 65 π$
The cone is solid, so it has a circular base as well.
$π r^{2} = π \times 5^{2} = 25 π$
Add them, collecting the $π$ terms.
$65 π + 25 π = 90 π$

Answer: (a) $13$ cm (b) $100 π$ cm³ (c) $90 π$ cm²

Example 4. A closed cylindrical can has radius $4$ cm and holds $500$ ml when it is full. Calculate (a) the height of the can, (b) the total area of metal needed to make the can, ignoring any overlaps.

Show the working

Change the capacity to a volume: $1$ ml $= 1$ cm³, so the can holds $500$ cm³.
Put the known values into the volume formula. The unknown is $h$ .
$π \times 4^{2} \times h = 500 \Rightarrow 16 π h = 500$
Divide both sides by $16 π$ .
$h = \frac{500}{16 π} = 9.947 \dots \approx 9.95$
The metal covers the whole outside of a closed can: the curved surface and two circular ends. Use the unrounded height.
$2 π \times 4 \times 9.947 \dots = 250.0$
The two ends:
$2 \times π \times 4^{2} = 32 π = 100.53 \dots$
Add, then round.
$250.0 + 100.53 \dots = 350.53 \dots \approx 351$

Answer: (a) $9.95$ cm (b) $351$ cm² (3 s.f.)

Where people go wrong

Mistake: Using the slant height in the volume of a cone, as in $\frac{1}{3} π r^{2} l$ .

Instead: Use the perpendicular height: $V = \frac{1}{3} π r^{2} h$ . If you are given $l$ , find $h$ first from $h^{2} = l^{2} - r^{2}$ .

The volume of a cone is one third of a cylinder of the same height, and that height is measured at right angles to the base. The slant height is longer, so it gives a volume that is too big.

Mistake: Using the perpendicular height of a pyramid as the height of each triangular face.

Instead: Find the slant height of each face with Pythagoras: perpendicular height and half the base edge are the two shorter sides, and the slant height is the hypotenuse.

Each face leans inwards, so its height up the middle of the face is longer than the vertical height of the pyramid.

Mistake: Giving the curved surface area when the question asks for the total surface area, or adding two ends to a container with an open top.

Instead: Before you calculate, list every face the object really has: a closed can has two ends, an open tank one, a pipe none.

The formula list gives only the curved parts of the cylinder and the cone. Which flat faces to add depends on the object, so you have to decide it from the context.

Mistake: Putting the diameter into a formula that needs the radius.

Instead: Halve the diameter first. A sphere of diameter $10$ cm has $r = 5$ .

Every formula here is written with $r$ . Using the diameter doubles $r$ , which makes an area $4$ times too big and a volume $8$ times too big.

Mistake: Rounding a length found in one part, then using the rounded value in the next part.

Instead: Keep the full calculator value, such as $r = 6.2035 \dots$ , and round only the final answer.

A rounded length can change the next answer at 3 significant figures: for the sphere of volume $1000$ cm³, $r = 6.2$ gives a surface area of $483$ cm² instead of $484$ cm².

Check yourself

A cuboid measures $4$ cm by $5$ cm by $6$ cm. Find its volume and its surface area.
Answer
Volume $4 \times 5 \times 6 = 120$ cm³. Surface area $2 (20 + 24 + 30) = 148$ cm².
A cube has volume $343$ cm³. Find its surface area.
Answer
Side $\sqrt[3]{343} = 7$ cm, so surface area $6 \times 7^{2} = 294$ cm².
A prism is $12$ cm long. Its cross-section is a trapezium with parallel sides $5$ cm and $9$ cm, which are $4$ cm apart. Find its volume.
Answer
Cross-section $\frac{1}{2} (5 + 9) \times 4 = 28$ cm², so volume $28 \times 12 = 336$ cm³.
A closed cylinder has radius $3$ cm and height $10$ cm. Find its volume and its total surface area, in terms of $π$ and to 3 significant figures.
Answer
Volume $π \times 9 \times 10 = 90 π \approx 283$ cm³. Total surface area $2 π \times 3 \times 10 + 2 π \times 9 = 60 π + 18 π = 78 π \approx 245$ cm².
A sphere has diameter $10$ cm. Find its surface area and its volume, to 3 significant figures.
Answer
$r = 5$ . Surface area $4 π \times 25 = 100 π \approx 314$ cm². Volume $\frac{4}{3} π \times 125 = \frac{500 π}{3} \approx 524$ cm³.
A cone has base radius $6$ cm and slant height $10$ cm. Find its perpendicular height, then its volume in terms of $π$ .
Answer
$h = \sqrt{10^{2} - 6^{2}} = \sqrt{64} = 8$ cm. Volume $\frac{1}{3} \times π \times 36 \times 8 = 96 π$ cm³ (about $302$ cm³).
A pyramid has a rectangular base $6$ cm by $4$ cm and a perpendicular height of $9$ cm. Find its volume.
Answer
$\frac{1}{3} \times 24 \times 9 = 72$ cm³.
A sphere has volume $900$ cm³. Calculate its radius, to 3 significant figures.
Answer
$r^{3} = \frac{3 \times 900}{4 π} = 214.85 \dots$ , so $r = \sqrt[3]{214.85 \dots} = 5.99$ cm.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A solid cone has base radius $9$ cm and perpendicular height $12$ cm.

(a) Show that the slant height of the cone is $15$ cm. 2 marks No calculator

Mark scheme

M1 $9^{2} + 12^{2}$ or $81 + 144$ seen
A1 $\sqrt{225} = 15$ , with no errors seen

(b) Find the total surface area of the cone. Give your answer in terms of $π$ . 3 marks No calculator

Mark scheme

M1 curved surface $π \times 9 \times 15$ or $135 π$
M1 base $π \times 9^{2}$ or $81 π$ , added to a curved surface area
A1 $216 π$ (cm²)

(c) Find the volume of the cone. Give your answer in terms of $π$ . 2 marks No calculator

Mark scheme

M1 $\frac{1}{3} \times π \times 9^{2} \times 12$ oe
A1 $324 π$ (cm³)

Question 2. A wedge of cheese is a prism. Its cross-section is a sector of a circle with radius $12$ cm and angle $40 °$ . The wedge is $5$ cm thick.

(a) Calculate the volume of the wedge. 3 marks

Mark scheme

M1 sector area $\frac{40}{360} \times π \times 12^{2}$ oe ( $16 π$ or $50.3$ )
M1 their sector area $\times 5$
A1 $251$ awrt or $80 π$ (cm³)

(b) Calculate the total surface area of the wedge. 4 marks

Mark scheme

B1 arc length $\frac{40}{360} \times 2 π \times 12$ oe ( $\frac{8 π}{3}$ or $8.38$ ) soi
M1 curved face: their arc length $\times 5$
M1 flat faces: $2 \times$ their sector area $+ 2 \times 12 \times 5$
A1 $262$ awrt or $120 + \frac{136 π}{3}$ oe (cm²)

Question 3. A cylindrical container has internal radius $5$ cm and stands on a horizontal table. It contains water to a depth of $8$ cm. A solid metal sphere of radius $3$ cm is placed in the container. It sinks to the bottom and is completely covered by the water.

(a) Calculate the volume of the sphere. 2 marks

Mark scheme

M1 $\frac{4}{3} \times π \times 3^{3}$
A1 $113$ awrt or $36 π$ (cm³)

(b) Calculate the new depth of the water. 3 marks

Mark scheme

M1 $π \times 5^{2} \times h =$ their (a), where $h$ is the rise in level, or $π \times 5^{2} \times H = 200 π +$ their (a), where $H$ is the new depth, oe
A1 rise $= 1.44$ (cm) soi
B1 $9.44$ (cm), FT $8 +$ their rise

How this is examined

Total surface area of a solid with a curved surface: finding every face, curved and flat: A 3- or 4-mark Calculate or Find on a cylinder, a cone or a cylindrical sector, sometimes with the answer in terms of $π$ A method mark for each kind of face, each worked out with the right length: $π r l$ with the slant height for a cone, arc length $\times$ length for the curved face of a cylindrical sector, and $π r^{2}$ for each circular end the object really has. Then the accuracy mark for the sum, for example $135 π + 81 π = 216 π$ . The curved surface area on its own is not the total.
Volume of a prism as area of cross-section times length, including cylinders and cylindrical sectors: A 2- or 3-mark Calculate, with a cross-section that is a triangle, a trapezium, a circle or a sector A method mark for the area of the cross-section, such as $\frac{40}{360} \times π \times 12^{2}$ , a method mark for multiplying it by the length, and the accuracy mark for the value to 3 significant figures or in terms of $π$ .
Finding a radius, height or side from a given volume or capacity: A 2- or 3-mark Calculate or Find: the volume or capacity is given and a length is the unknown The formula written with the given value in it, such as $16 π h = 500$ , then rearranged for the unknown. For a sphere, the cube root of $\frac{3 V}{4 π}$ . The length to 3 significant figures, with the unrounded value kept for any later part.
Using Pythagoras to get the slant height or the perpendicular height of a cone or pyramid: A 2-mark Show (that) or Find before a volume or surface area: one height is given and the other is needed $r^{2} + h^{2} = l^{2}$ with the numbers in, such as $9^{2} + 12^{2} = 225$ , then $\sqrt{225} = 15$ . For a Show (that), every step is written down and the working ends at the given value. Afterwards, $h$ goes in the volume and $l$ in the curved surface area.
Giving an exact answer in terms of $π$ without a calculator: A 2- or 3-mark Find that ends 'Give your answer in terms of $π$ ', on any of the curved solids $π$ left as a symbol, the other numbers worked out exactly ( $\frac{1}{3} \times 81 \times 12 = 324$ ) and the $π$ terms collected. A decimal such as $678.6$ is not an answer in terms of $π$ .

Words to know

Volume: The amount of space a solid takes up, measured in cubic units such as cm³.
Surface area: The total area of all the faces of a solid, curved and flat, measured in square units such as cm².
Curved surface area: The area of the curved part of a solid only, without its flat faces: $2 π r h$ for a cylinder, $π r l$ for a cone.
Net: The flat shape you get by unfolding the surface of a solid. Its area is the surface area of the solid.
Prism: A solid with the same cross-section all the way through, such as a cuboid, a triangular prism, a cylinder or a cylindrical sector. Volume $=$ area of cross-section $\times$ length.
Cross-section: The shape you see when you cut straight across a solid, at right angles to its length.
Cylindrical sector: A prism whose cross-section is a sector of a circle, like a wedge cut from a round cheese.
Apex: The point at the top of a pyramid or cone, where the sloping faces or the curved surface meet.
Perpendicular height: The height of a pyramid or cone measured from the apex straight down to the base, at right angles to the base. It is the height used in the volume.
Slant height: The distance up a sloping face or surface to the apex. For a cone it is $l$ , the length from the apex to the edge of the base, used in $π r l$ .
Capacity: How much a container holds, often in ml or litres: $1$ ml $= 1$ cm³ and $1$ litre $= 1000$ cm³.

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