Composite Solids

Find the volume and the surface area of solids made of cylinders, cones, hemispheres, pyramids and cuboids joined together, and of a solid with a part cut out of it.

Before you start

Cuboid: $V =$ length $\times$ breadth $\times$ height. Cylinder of radius $r$ and height $h$ : $V = \pi r^{2} h$ , curved surface $2\pi r h$ .
Pyramid of height $h$ : $V = \frac{1}{3} \times$ base area $\times h$ ; each triangular face $\frac{1}{2} \times$ side of the base $\times$ slant height.
Cone of radius $r$ , height $h$ and slant height $l$ : $V = \frac{1}{3}\pi r^{2} h$ , curved surface $\pi r l$ , with $l^{2} = r^{2} + h^{2}$ .
Sphere of radius $r$ : $V = \frac{4}{3}\pi r^{3}$ , surface $4\pi r^{2}$ . Hemisphere: $V = \frac{2}{3}\pi r^{3}$ , curved surface $2\pi r^{2}$ .

The Volume of a Composite Solid

A composite solid is made of two or more simple solids joined together: a cone on a cylinder (a pencil, a rocket), a hemisphere on a cylinder (a silo; a capsule has one at each end), a cone and a hemisphere (a spinning top; an ice cream, the other way up), a pyramid on a cuboid (a house with a pyramid roof).

Its volume is the sum of the volumes of its parts. Find each part's measurements from the drawing (where two parts have the same radius, it is used by both), work out each volume, and add. If a part is cut out of a solid, subtract its volume instead (A Solid with a Part Cut Out, below).

Example 1: A cone on a cylinder

A model rocket is a cone on top of a cylinder. Both have a radius of $3$ cm; the cylinder is $10$ cm tall and the cone $4$ cm tall. Find the volume of the rocket.

Solution

Cylinder: $\pi \times 3^{2} \times 10 = 90\pi \text{ cm}^{3}$

Cone: $\frac{1}{3} \times \pi \times 3^{2} \times 4 = 12\pi \text{ cm}^{3}$

Total: $90\pi + 12\pi = 102\pi = 320.44\ldots = 320$ (3 s.f.)

The volume is $320 \text{ cm}^{3}$ .

Example 2: A hemisphere on a cylinder

A grain silo is a cylinder of radius $4$ m and height $12$ m with a hemispherical roof of the same radius. Find its volume.

Solution

Cylinder: $\pi \times 4^{2} \times 12 = 192\pi \text{ m}^{3}$

Hemisphere: $\frac{2}{3} \times \pi \times 4^{3} = \frac{128}{3}\pi \text{ m}^{3}$

Total: $192\pi + \frac{128}{3}\pi = \frac{704}{3}\pi = 737.22\ldots = 737$ (3 s.f.)

The volume is $737 \text{ m}^{3}$ .

Example 3: A cone on a hemisphere

A spinning top is a hemisphere of radius $2.5$ cm with a cone of height $6$ cm on its flat face. Find its volume.

Solution

Cone: $\frac{1}{3} \times \pi \times 2.5^{2} \times 6 = 12.5\pi\text{.}$ Hemisphere: $\frac{2}{3} \times \pi \times 2.5^{3} = 10.416\ldots\pi\text{.}$

Total: $12.5\pi + 10.416\ldots\pi = 22.916\ldots\pi = 71.99\ldots = 72.0$ (3 s.f.)

The volume is $72.0 \text{ cm}^{3}$ .

Example 4: A pyramid on a cuboid

A storeroom in a community garden is a cuboid $4$ m long, $4$ m wide and $3$ m high, with a roof that is a square pyramid $1.5$ m high. Find the volume of the storeroom.

Solution

Cuboid: $4 \times 4 \times 3 = 48 \text{ m}^{3}$

Pyramid: $\frac{1}{3} \times 4 \times 4 \times 1.5 = 8 \text{ m}^{3}$

Total: $48 + 8 = 56$

The volume is $56 \text{ m}^{3}$ .

The Surface Area of a Composite Solid

The surface area of a composite solid is the area of every surface you could touch or paint. Where two parts are joined, the area where they touch is inside the solid and is not counted, for either part.

So list the surfaces you can touch. For a cone on a cylinder of the same radius, they are the cone's curved surface, the cylinder's curved surface and the cylinder's base. The top of the cylinder and the base of the cone are the same circle, the joined face, and both are left out.

If one part is narrower than the face it is joined to, it covers only part of that face. A cone of radius $2$ cm on a cylinder of radius $5$ cm hides only a circle of radius $2$ cm in the middle of the cylinder's top. The rest of the top is a ring that you can still touch, so it is counted: $\pi \times 5^{2} - \pi \times 2^{2} = 25\pi - 4\pi = 21\pi \text{ cm}^{2}\text{.}$ A small pyramid on a larger cube leaves a square frame round it in the same way. And if the wider part is on top (a dome wider than the cylinder under it), the uncovered ring is underneath it, and it is counted just the same.

Surface area of a composite solid

Count every surface you could touch. Where two parts are joined, leave out the area they share, on both parts.

If one part is narrower than the face it is joined to, leave out only the area it covers: the rest of the wider face (a ring, or a frame), above or below, is counted.

Example 5: The rocket's surface

Find the total surface area of the rocket of Example 1 (radius $3$ cm, cylinder $10$ cm tall, cone $4$ cm tall).

Solution

The cone's slant height: $l^{2} = 3^{2} + 4^{2} = 25$ , so $l = 5$ cm.

Surface	Area
Cone	$\pi \times 3 \times 5 = 15\pi$
Cylinder side	$2\pi \times 3 \times 10$ $= 60\pi$
Base	$\pi \times 3^{2} = 9\pi$
Joined face	not counted

Total: $15\pi + 60\pi + 9\pi = 84\pi = 263.89\ldots = 264$ (3 s.f.)

The total surface area is $264 \text{ cm}^{2}$ .

Example 6: The silo's surface

Find the total surface area of the silo of Example 2 (radius $4$ m, cylinder $12$ m tall), including its floor.

Solution

Hemisphere, curved: $2 \times \pi \times 4^{2} = 32\pi$ . Cylinder, curved: $2 \times \pi \times 4 \times 12 = 96\pi\text{.}$ Floor: $\pi \times 4^{2} = 16\pi$ .

The hemisphere's flat face and the cylinder's top are the joined face: not counted.

Total: $32\pi + 96\pi + 16\pi = 144\pi = 452.38\ldots = 452$ (3 s.f.)

The total surface area is $452 \text{ m}^{2}$ .

Example 7: The spinning top's surface

Find the total surface area of the spinning top of Example 3 (radius $2.5$ cm, cone $6$ cm tall).

Solution

The cone's slant height: $l^{2} = 2.5^{2} + 6^{2} = 6.25 + 36 = 42.25\text{,}$ so $l = 6.5$ cm.

Cone, curved: $\pi \times 2.5 \times 6.5 = 16.25\pi\text{.}$ Hemisphere, curved: $2 \times \pi \times 2.5^{2} = 12.5\pi\text{.}$

The cone and the hemisphere have the same radius, so the joined circle is the whole of each flat face; it is not counted, and there is no flat face at all.

Total: $16.25\pi + 12.5\pi = 28.75\pi = 90.32\ldots = 90.3$ (3 s.f.)

The total surface area is $90.3 \text{ cm}^{2}$ .

Example 8: Parts of different widths

A wooden stand is a cylinder of radius $2$ cm and height $6$ cm standing on the middle of the top of a cylinder of radius $5$ cm and height $3$ cm. Find (a) its volume, (b) its total surface area.

Solution

(a) Lower cylinder: $\pi \times 5^{2} \times 3 = 75\pi$ . Upper cylinder: $\pi \times 2^{2} \times 6 = 24\pi$ .

Total: $75\pi + 24\pi = 99\pi = 311.01\ldots = 311$ (3 s.f.)

(b) The upper cylinder covers only a circle of radius $2$ cm in the middle of the lower cylinder's top. The rest of that top is a ring you can touch, so it is counted.

Surface	Area
Base of the lower cylinder	$\pi \times 5^{2} = 25\pi$
Side of the lower cylinder	$2\pi \times 5 \times 3 = 30\pi$
Ring on its top	$25\pi - 4\pi = 21\pi$
Side of the upper cylinder	$2\pi \times 2 \times 6 = 24\pi$
Top of the upper cylinder	$\pi \times 2^{2} = 4\pi$
Joined circle	not counted

Total: $25\pi + 30\pi + 21\pi + 24\pi + 4\pi = 104\pi = 326.72\ldots = 327$ (3 s.f.)

Check: the ring and the top of the upper cylinder together make $21\pi + 4\pi = 25\pi$ , the whole top of the lower cylinder. Seen from above, all of it still shows, only at two levels.

(a) $311 \text{ cm}^{3}$ (b) $327 \text{ cm}^{2}$

A Solid with a Part Cut Out

Sometimes a part is cut out of a solid: a hole drilled into a block, a hollow scooped out of its top. The volume left is the volume of the solid less the volume of the part cut out.

The surface changes in two ways. The surface inside the hole is now part of the surface (you can touch it), so its area is added. And the face the hole is cut into loses the area of the opening: what is left of it is the face with a hole in it (a ring, if both are circles), or, if the hole fills the whole face, nothing at all. A hole drilled right through has two openings, one in each face, and no floor.

A part cut out

Volume $=$ volume of the solid $-$ volume of the part cut out

Surface area $=$ the outside surfaces, less the area of each opening, $+$ the surface inside the hole

Example 9: A kueh mould

A wooden kueh mould is a solid cylinder of radius $4$ cm and height $5$ cm. A hollow in the shape of a hemisphere of radius $3$ cm is cut into the middle of its top face. Find (a) the volume of wood, (b) the total surface area of the mould.

Solution

(a) Cylinder: $\pi \times 4^{2} \times 5 = 80\pi$ . Hollow: $\frac{2}{3} \times \pi \times 3^{3} = 18\pi\text{.}$

Wood: $80\pi - 18\pi = 62\pi = 194.77\ldots = 195$ (3 s.f.)

(b) List every surface you could touch, with the opening taken off the top and the inside of the hollow added:

Surface	Area
Base	$\pi \times 4^{2} = 16\pi$
Curved side	$2\pi \times 4 \times 5 = 40\pi$
Top, less the opening	$16\pi - 9\pi = 7\pi$
Inside the hollow	$2\pi \times 3^{2} = 18\pi$

Total: $16\pi + 40\pi + 7\pi + 18\pi = 81\pi = 254.46\ldots = 254$ (3 s.f.)

(a) $195 \text{ cm}^{3}$ of wood (b) $254 \text{ cm}^{2}$

Common Mistakes

Counting the joined face. Where a cone sits on a cylinder of the same radius, the circle between them is inside the solid: count it on neither part.
Leaving out the ring. If one part is narrower than the face it is joined to, only the area it covers is hidden: the rest of the wider face is a ring (or a frame), and it is counted.
Adding two total surface areas. $3\pi r^{2}$ for a solid hemisphere plus $2\pi r^{2} + 2\pi r h$ for a closed cylinder counts the joined circle twice.
Forgetting the inside of a hole. When a part is cut out, the surface inside the hole is added, and the opening is taken off the face it is cut into.
Rounding a part before adding. Keep $\pi$ in each part, add, and round once.

Practice Questions

Try each question before you open its solution. Use the calculator's $\pi$ , and give answers correct to 3 significant figures where they are not exact.

Question 1

A solid is a cylinder of radius $5$ cm and height $8$ cm with a cone of the same radius and height $6$ cm on top. Find its volume.

ShowHide solution

$\pi \times 25 \times 8 + \frac{1}{3} \times \pi \times 25 \times 6 = 200\pi + 50\pi = 250\pi = 785.39\ldots = 785$ (3 s.f.)

$785 \text{ cm}^{3}$

Question 2

An ice cream is a cone of radius $6$ cm and depth $10$ cm, point down, filled with ice cream and topped with a hemisphere of ice cream of the same radius. Find the volume of ice cream.

ShowHide solution

Cone: $\frac{1}{3}\pi \times 36 \times 10 = 120\pi\text{.}$ Hemisphere: $\frac{2}{3}\pi \times 216 = 144\pi\text{.}$ Total: $264\pi = 829.38\ldots = 829$ (3 s.f.)

$829 \text{ cm}^{3}$

Question 3

Find the surface area of the ice cream of Question 2: the outside of the cone and the curved top.

ShowHide solution

Cone: $l = \sqrt{6^{2} + 10^{2}} = \sqrt{136} = 11.66\ldots$ cm, curved surface $\pi \times 6 \times \sqrt{136} = 219.82\ldots$

Hemisphere, curved: $2\pi \times 36 = 72\pi = 226.19\ldots\text{.}$ The joined circle is not counted.

Total: $219.82\ldots + 226.19\ldots = 446.01\ldots = 446$ (3 s.f.)

$446 \text{ cm}^{2}$

Question 4

A paperweight is a cube of edge $6$ cm with a square pyramid of height $4$ cm on top, the pyramid's base covering the whole top face of the cube. Find (a) its volume, (b) its total surface area. (c) A second paperweight has a smaller square pyramid on the same cube: its base has sides of $4$ cm, its height is $1.5$ cm, and it stands on the middle of the top face. Find its volume and its total surface area.

ShowHide solution

(a) $6^{3} + \frac{1}{3} \times 36 \times 4 = 216 + 48 = 264$

(b) The pyramid covers the whole top, so the cube shows $5$ faces: $5 \times 36 = 180 \text{ cm}^{2}$ . The pyramid's slant height is $\sqrt{4^{2} + 3^{2}} = 5$ cm, so its faces are $4 \times \frac{1}{2} \times 6 \times 5 = 60 \text{ cm}^{2}\text{.}$ Total: $240$ .

Surface: the pyramid covers only a $4$ cm square in the middle of the top, so the rest of the top is a square frame, and it is counted: $36 - 16 = 20 \text{ cm}^{2}$ . The other $5$ faces of the cube: $180 \text{ cm}^{2}$ .

The pyramid's slant height: from the middle of its base to the middle of a side is $2$ cm, so it is $\sqrt{1.5^{2} + 2^{2}} = \sqrt{6.25} = 2.5$ cm, and its faces are $4 \times \frac{1}{2} \times 4 \times 2.5 = 20 \text{ cm}^{2}\text{.}$

Total: $180 + 20 + 20 = 220$

(a) $264 \text{ cm}^{3}$ (b) $240 \text{ cm}^{2}$ (c) $224 \text{ cm}^{3}$ and $220 \text{ cm}^{2}$

Question 5

A medicine capsule is a cylinder of radius $2$ mm and length $10$ mm with a hemisphere on each end. Find (a) its volume, (b) its surface area.

ShowHide solution

The two hemispheres make one whole sphere of radius $2$ mm.

(a) $\pi \times 4 \times 10 + \frac{4}{3}\pi \times 8 = 40\pi + \frac{32}{3}\pi = \frac{152}{3}\pi = 159.17\ldots = 159$ (3 s.f.)

(b) $2\pi \times 2 \times 10 + 4\pi \times 4 = 40\pi + 16\pi = 56\pi = 175.92\ldots = 176$ (3 s.f.)

(a) $159 \text{ mm}^{3}$ (b) $176 \text{ mm}^{2}$

Question 6

A water tank on the roof of a block of flats is a cylinder of radius $0.5$ m and height $1.2$ m standing on a hemispherical bottom of the same radius. Find its capacity in litres.

ShowHide solution

$\pi \times 0.25 \times 1.2 + \frac{2}{3}\pi \times 0.125 = 0.3\pi + \frac{1}{12}\pi = 1.2042\ldots \text{ m}^{3}$

$1 \text{ m}^{3} = 1000$ litres, so $1.2042\ldots \times 1000 = 1204.2\ldots = 1200$ litres (3 s.f.)

About $1200$ litres

Question 7

A solid is a hemisphere of radius $3$ cm on a cylinder of the same radius and height $5$ cm. Ahmad adds the total surface area of a solid hemisphere, $3\pi r^{2}$ , to that of a closed cylinder, $2\pi r^{2} + 2\pi r h$ , and gets $75\pi \text{ cm}^{2}$ . Explain his mistake and find the correct area.

ShowHide solution

He counted the joined circle twice: once as the hemisphere's flat face and once as the cylinder's top. Both are inside the solid.

Correct: hemisphere curved $2\pi \times 9 = 18\pi$ , cylinder curved $2\pi \times 3 \times 5 = 30\pi$ , base $9\pi$ .

Total: $57\pi = 179.07\ldots = 179$ (3 s.f.), which is $18\pi$ (two circles) less than his.

The joined face was counted twice; the area is $179 \text{ cm}^{2}$ .

Question 8

A trophy is a solid hemisphere of radius $4$ cm fixed to the middle of the top face of a solid cylinder of radius $6$ cm and height $4$ cm. Find (a) its volume, (b) its total surface area.

ShowHide solution

(a) Cylinder: $\pi \times 6^{2} \times 4 = 144\pi$ . Hemisphere: $\frac{2}{3} \times \pi \times 4^{3} = \frac{128}{3}\pi\text{.}$

Total: $144\pi + \frac{128}{3}\pi = \frac{560}{3}\pi = 586.43\ldots = 586$ (3 s.f.)

(b) The hemisphere covers a circle of radius $4$ cm on the cylinder's top; the rest of the top is a ring, and it is counted.

Base $\pi \times 6^{2} = 36\pi$ , curved side $2\pi \times 6 \times 4 = 48\pi$ , ring $36\pi - 16\pi = 20\pi$ , hemisphere curved $2\pi \times 4^{2} = 32\pi$ .

Total: $36\pi + 48\pi + 20\pi + 32\pi = 136\pi = 427.25\ldots = 427$ (3 s.f.)

(a) $586 \text{ cm}^{3}$ (b) $427 \text{ cm}^{2}$

Question 9

A cone of radius $5$ cm and height $12$ cm is drilled out of a solid cylinder of the same radius and height, leaving a cone-shaped hole whose circle is the whole top of the cylinder. Find (a) the volume left, (b) the total surface area of what is left.

ShowHide solution

(a) $\pi \times 25 \times 12 - \frac{1}{3}\pi \times 25 \times 12 = 300\pi - 100\pi = 200\pi = 628.31\ldots = 628$ (3 s.f.)

(b) Outside: curved $2\pi \times 5 \times 12 = 120\pi$ and base $25\pi$ . The hole's opening is the whole top, so no top is left.

Inside the hole: the cone's curved surface, with $l = \sqrt{25 + 144} = 13$ cm, so $\pi \times 5 \times 13 = 65\pi$ .

Total: $120\pi + 25\pi + 65\pi = 210\pi = 659.73\ldots = 660$ (3 s.f.)

(a) $628 \text{ cm}^{3}$ (b) $660 \text{ cm}^{2}$

Question 10

In a Design and Technology lesson, Priya makes a pen holder from a solid wooden cylinder of radius $5$ cm and height $10$ cm. She drills a cylindrical hole of radius $3$ cm and depth $8$ cm into the middle of its top face. Find (a) the volume of wood left, (b) the total surface area of the pen holder, inside and out.

ShowHide solution

(a) $\pi \times 5^{2} \times 10 - \pi \times 3^{2} \times 8 = 250\pi - 72\pi = 178\pi = 559.20\ldots = 559$ (3 s.f.)

(b) Outside: base $25\pi$ , curved side $2\pi \times 5 \times 10 = 100\pi\text{,}$ and the top less the opening, $25\pi - 9\pi = 16\pi$ .

Inside the hole: its curved side $2\pi \times 3 \times 8 = 48\pi$ and its floor $\pi \times 3^{2} = 9\pi$ .

Total: $25\pi + 100\pi + 16\pi + 48\pi + 9\pi = 198\pi = 622.03\ldots = 622$ (3 s.f.)

(a) $559 \text{ cm}^{3}$ (b) $622 \text{ cm}^{2}$

Question 11

To make a napkin ring, Wei Ming drills a cylindrical hole of radius $2$ cm right through a solid wooden cylinder of radius $5$ cm and height $10$ cm, along its axis. Find (a) the volume of wood left, (b) the total surface area of the napkin ring, inside and out.

ShowHide solution

(a) The hole is $10$ cm long, as long as the cylinder: $\pi \times 5^{2} \times 10 - \pi \times 2^{2} \times 10 = 250\pi - 40\pi = 210\pi = 659.73\ldots = 660$ (3 s.f.)

(b) The hole goes right through, so it has two openings, one in the top and one in the base, and no floor.

Outside: curved side $2\pi \times 5 \times 10 = 100\pi\text{;}$ the top and the base are each a ring, $25\pi - 4\pi = 21\pi$ , so $2 \times 21\pi = 42\pi$ .

Inside the hole: its curved side $2\pi \times 2 \times 10 = 40\pi$ .

Total: $100\pi + 42\pi + 40\pi = 182\pi = 571.76\ldots = 572$ (3 s.f.)

(a) $660 \text{ cm}^{3}$ (b) $572 \text{ cm}^{2}$

Key Points

A composite solid is made of simple solids joined together; its volume is the sum of their volumes.
Its surface area counts every surface you could touch: the area where two parts are joined is inside the solid and is not counted; the uncovered ring (or frame) of a wider face is counted.
When a part is cut out, subtract its volume; for the surface, take the opening off the face it is cut into and add the surface inside the hole.
List the surfaces one by one, in a table if it helps, before adding.
Keep each part exact (in terms of $\pi$ or unrounded), add or subtract, and round once.