Surface Area of Solids with Holes and Cuts

Find the surface area of a solid with a hole through it, of a pipe, and of the pieces of a cut solid: the new faces that drilling and cutting make.

Before you start

From Surface Area of Composite Solids: the surface area of a solid is the area of its outside only, everything you could touch or paint. There, joining solids hid faces. Here it goes the other way: a hole or a cut makes new faces on the outside.
From Surface Area of Prisms and Surface Area of Cylinders: a cuboid $l$ by $b$ by $h$ has total surface area $2(lb + bh + lh)$ and a cube of edge $l$ has $6l^{2}$ ; a cylinder of radius $r$ and height (or length) $h$ has curved surface area $2\pi rh$ and two ends of area $\pi r^{2}$ each.
From composite figures with circular parts: a ring between circles of radius $R$ and $r$ has area $\pi R^{2} - \pi r^{2}$ . Each question says which $\pi$ to use; answers that are not exact are given correct to 3 significant figures, rounding only at the end.

Holes and Pipes

Drill a round hole straight through a block, from one face to the opposite face. Two circles are taken out of the outside, one where the hole goes in and one where it comes out. But the hole has an inside wall, a curved surface you can touch with your finger, and that is new surface. A hole takes away volume but can add surface area.

A hole in a solid; a pipe

A cylindrical hole of radius $r$ drilled straight through a solid, from one face to the opposite face a distance $l$ away:

Total surface area $=$ surface area of the solid $-\ 2\pi r^{2}$ (the two circles) $+\ 2\pi rl$ (the inside of the hole).

A hole of any shape straight through: take away its two ends and add its inside walls, whose area is the perimeter of the hole $\times$ its length.

A round hole of radius $r$ and depth $d$ that stops inside the solid has one opening and a circular floor:

Total surface area $=$ surface area of the solid $-\ \pi r^{2}$ (the opening) $+\ 2\pi rd$ (the wall) $+\ \pi r^{2}$ (the floor) $=$ surface area of the solid $+\ 2\pi rd$ . The floor replaces the circle taken from the top.

A pipe (a hollow cylinder) of external radius $R$ , internal radius $r$ and length $l$ has an outer curved surface, an inner curved surface and two ring-shaped ends:

Total surface area $= 2\pi Rl + 2\pi rl + 2(\pi R^{2} - \pi r^{2})$

Example 1: A round hole

A wooden block is 10 cm wide, 6 cm high and 8 cm long. A hole of radius 2 cm is drilled straight through it, from the middle of the front face to the middle of the back face. Find the total surface area of the block with its hole, using the $\pi$ button on your calculator.

Solution

Total surface area of the block before drilling $= 2(10 \times 6 + 6 \times 8 + 10 \times 8) = 2(60 + 48 + 80) = 376\text{ cm}^{2}$

Two circles taken out $= 2\pi r^{2} = 2 \times \pi \times 2^{2} = 8\pi\text{ cm}^{2}$

The hole runs the length of the block, 8 cm. Its inside $= 2\pi rl = 2 \times \pi \times 2 \times 8 = 32\pi\text{ cm}^{2}$

Total surface area $= 376 - 8\pi + 32\pi = 376 + 24\pi = 451.39\ldots \approx 451\text{ cm}^{2}$

The total surface area of the block is $(376 + 24\pi)\text{ cm}^{2} \approx 451\text{ cm}^{2}$ , more than the $376\text{ cm}^{2}$ it had before the hole.

Example 2: A square hole

A wooden block is 10 cm by 10 cm and 6 cm high. A square hole 4 cm by 4 cm is cut straight down through the middle of it, from the top face to the bottom face. Find the total surface area of the block with its hole.

Solution

Block before the hole: $2(10 \times 10 + 10 \times 6 + 10 \times 6) = 2(100 + 60 + 60) = 440\text{ cm}^{2}$

Two squares taken out, one from the top and one from the bottom: $2 \times 4 \times 4 = 32\text{ cm}^{2}$

The inside of the hole is four rectangles, each 4 cm wide and as long as the hole, 6 cm: $4 \times 4 \times 6 = 96\text{ cm}^{2}$ . (This is the perimeter of the hole, 16 cm, times its length, 6 cm.)

Total surface area $= 440 - 32 + 96 = 504\text{ cm}^{2}$

Check, face by face: top and bottom $2(100 - 16) = 168$ ; four outside faces $4 \times 10 \times 6 = 240$ ; four inside faces 96. $168 + 240 + 96 = 504\text{ cm}^{2}$ .

The total surface area of the block is $504\text{ cm}^{2}$ .

Example 3: A hole that does not go right through

A wooden block is 12 cm long, 8 cm wide and 5 cm high. A round hole of radius 2 cm and depth 3 cm is drilled into the middle of its top; it does not go right through. Find the total surface area of the block with its hole (a) in terms of $\pi$ , (b) using the $\pi$ button on your calculator.

Solution

Block before the hole: $2(12 \times 8 + 8 \times 5 + 12 \times 5) = 2(96 + 40 + 60) = 392\text{ cm}^{2}$

The hole has only one opening, so only one circle is taken out, from the top: $\pi r^{2} = \pi \times 2^{2} = 4\pi\text{ cm}^{2}$

Its curved wall is 3 cm deep: $2\pi rd = 2 \times \pi \times 2 \times 3 = 12\pi\text{ cm}^{2}$

Its floor is a circle of the same radius: $\pi \times 2^{2} = 4\pi\text{ cm}^{2}$

(a) Total surface area $= 392 - 4\pi + 12\pi + 4\pi = (392 + 12\pi)\text{ cm}^{2}$ . The floor replaces the circle taken from the top, so the hole adds only its wall.

(b) $392 + 12\pi = 429.69\ldots \approx 430\text{ cm}^{2}$

The total surface area is $(392 + 12\pi)\text{ cm}^{2} \approx 430\text{ cm}^{2}$ .

Example 4: A concrete pipe

A concrete pipe for a drain is 2 m long. Its external diameter is 1 m and its internal diameter is 0.8 m. Find the total surface area of the pipe, inside and out, in $\text{m}^{2}$ , using the $\pi$ button on your calculator.

Solution

The external radius is $R = 1 \div 2 = 0.5$ m and the internal radius is $r = 0.8 \div 2 = 0.4$ m.

Outer curved surface $= 2\pi Rl = 2 \times \pi \times 0.5 \times 2 = 2\pi\text{ m}^{2}$

Inner curved surface $= 2\pi rl = 2 \times \pi \times 0.4 \times 2 = 1.6\pi\text{ m}^{2}$

Each end is a ring: $\pi R^{2} - \pi r^{2} = \pi \times 0.5^{2} - \pi \times 0.4^{2} = 0.25\pi - 0.16\pi = 0.09\pi\text{ m}^{2}$ . Two ends: $0.18\pi\text{ m}^{2}$ .

Total surface area $= 2\pi + 1.6\pi + 0.18\pi = 3.78\pi\text{ m}^{2} = 11.875\ldots\text{ m}^{2} \approx 11.9\text{ m}^{2}$

The total surface area of the pipe is $3.78\pi\text{ m}^{2} \approx 11.9\text{ m}^{2}$ .

Cutting a Solid into Pieces

Cut a loaf of bread in two and each half has a new face where the knife went through. The two new faces are the same shape and size as the cut. So cutting never changes the volume, but it always adds surface area.

Cutting a solid

One straight cut right through a solid makes two new faces, each the shape of the cut:

Total surface area of the two pieces $=$ surface area of the solid $+\ 2 \times$ area of the cut face.

Each further cut right across adds two more faces of its own.

Example 5: A block cut in two

A block of wood 20 cm long, 10 cm wide and 6 cm high is cut into two equal halves by a cut across its middle, parallel to its ends. Find the total surface area (a) of the block, (b) of the two halves together.

Solution

(a) Total surface area $= 2(20 \times 10 + 10 \times 6 + 20 \times 6) = 2(200 + 60 + 120) = 760\text{ cm}^{2}$

(b) The cut is the same shape as an end: a rectangle 10 cm by 6 cm, of area $60\text{ cm}^{2}$ . Each half gets one new face.

Total surface area of the halves $= 760 + 2 \times 60 = 880\text{ cm}^{2}$

Check: each half is a cuboid 10 cm by 10 cm by 6 cm, with total surface area $2(10 \times 10 + 10 \times 6 + 10 \times 6) = 2(100 + 60 + 60) = 440\text{ cm}^{2}$ . Two halves: $2 \times 440 = 880\text{ cm}^{2}$ .

(a) The block has total surface area $760\text{ cm}^{2}$ . (b) The two halves have $880\text{ cm}^{2}$ together, $120\text{ cm}^{2}$ more.

Example 6: A cube cut into small cubes

A wooden cube of edge 6 cm is cut into 27 small cubes, each of edge 2 cm. (a) Find the total surface area of the large cube and of the 27 small cubes together. (b) How many times as great is the second area as the first?

Solution

(a) Large cube: $6l^{2} = 6 \times 6^{2} = 216\text{ cm}^{2}$

One small cube: $6 \times 2^{2} = 24\text{ cm}^{2}$ . All 27: $27 \times 24 = 648\text{ cm}^{2}$

Check by the cuts: there are 2 cuts in each of the 3 directions, 6 cuts in all, and each goes right across the cube, making two new faces 6 cm by 6 cm. Added area $= 6 \times 2 \times 36 = 432\text{ cm}^{2}$ , and $216 + 432 = 648\text{ cm}^{2}$ .

(b) $648 \div 216 = 3$

(a) The large cube has $216\text{ cm}^{2}$ and the 27 small cubes have $648\text{ cm}^{2}$ together. (b) The second area is 3 times the first, though the volume is the same.

Common mistakes

Forgetting the inside of a hole. A hole takes away its two ends but adds its inside walls: $2\pi rl$ for a round hole, perimeter $\times$ length for any hole.
Taking away two circles for a hole that does not go through. A hole that stops inside the solid has one opening and a circular floor; the floor's area equals the circle taken from the top (Example 3).
Forgetting the inner curved surface of a pipe, or painting it when the question says outside only.
Using $\pi(R - r)^{2}$ for the end of a pipe. The ring is the large circle less the small one: $\pi R^{2} - \pi r^{2}$ .
Adding one cut face, not two. A cut makes a new face on each piece.
Mixing up cuts and pieces. Pieces in a row need one cut fewer than there are pieces: 5 slices, 4 cuts.
Thinking surface area and volume change together. A cut does not change the volume; a hole takes volume away. Both can add surface area.

Practice questions

Try each question before you open its solution. Answers that are not exact are correct to 3 significant figures.

Practice 1: A square hole through a cube

A wooden cube of edge 8 cm has a square hole 4 cm by 4 cm cut straight down through the middle of it, from the top face to the bottom face. Find the total surface area of the cube with its hole.

ShowHide solution

Cube: $6 \times 8^{2} = 384\text{ cm}^{2}$

Two squares taken out: $2 \times 4 \times 4 = 32\text{ cm}^{2}$

Inside of the hole: perimeter $\times$ length $= 16 \times 8 = 128\text{ cm}^{2}$

Total surface area $= 384 - 32 + 128 = 480\text{ cm}^{2}$

Check, face by face: top and bottom $2(64 - 16) = 96$ ; four outside faces $4 \times 64 = 256$ ; inside 128. $96 + 256 + 128 = 480\text{ cm}^{2}$ .

The total surface area is $480\text{ cm}^{2}$ .

Practice 2: A hollow concrete block

A concrete block is 20 cm wide, 20 cm high and 40 cm long. A hole of radius 6 cm runs straight through it along its length, from the middle of one square end to the middle of the other. Taking $\pi = 3.142$ , find the total surface area of the block.

ShowHide solution

Block before the hole: $2(20 \times 20 + 20 \times 40 + 20 \times 40) = 2(400 + 800 + 800) = 4000\text{ cm}^{2}$

Two circles taken out: $2\pi r^{2} = 2 \times \pi \times 6^{2} = 72\pi\text{ cm}^{2}$

Inside of the hole, 40 cm long: $2\pi rl = 2 \times \pi \times 6 \times 40 = 480\pi\text{ cm}^{2}$

Total surface area $= 4000 - 72\pi + 480\pi = 4000 + 408\pi = 4000 + 408 \times 3.142 = 4000 + 1281.936 = 5281.936 \approx 5280\text{ cm}^{2}$

The total surface area is about $5280\text{ cm}^{2}$ .

Practice 3: A block with two holes

A wooden block is 12 cm long, 6 cm wide and 4 cm high. Two round holes, each of radius 1.5 cm, are drilled straight down through it from the top face to the bottom face. Find the total surface area of the block, using the $\pi$ button on your calculator.

ShowHide solution

Block before the holes: $2(12 \times 6 + 6 \times 4 + 12 \times 4) = 2(72 + 24 + 48) = 288\text{ cm}^{2}$

Each hole is 4 cm long. It takes out two circles, $2 \times \pi \times 1.5^{2} = 4.5\pi\text{ cm}^{2}$ , and adds its inside, $2 \times \pi \times 1.5 \times 4 = 12\pi\text{ cm}^{2}$ : a gain of $12\pi - 4.5\pi = 7.5\pi\text{ cm}^{2}$ .

Two holes: $2 \times 7.5\pi = 15\pi\text{ cm}^{2}$

Total surface area $= 288 + 15\pi = 335.12\ldots \approx 335\text{ cm}^{2}$

The total surface area is $(288 + 15\pi)\text{ cm}^{2} \approx 335\text{ cm}^{2}$ .

Practice 4: A toy wheel

A wooden wheel for a toy car is a cylinder of diameter 12 cm and thickness 3 cm, with a round hole of diameter 3 cm straight through its centre for the axle. Find its total surface area (a) in terms of $\pi$ , (b) using the $\pi$ button on your calculator.

ShowHide solution

The radii are $12 \div 2 = 6$ cm and $3 \div 2 = 1.5$ cm. The hole is as long as the wheel is thick, 3 cm.

Wheel before the hole: $2\pi r^{2} + 2\pi rh = 2 \times \pi \times 6^{2} + 2 \times \pi \times 6 \times 3 = 72\pi + 36\pi = 108\pi\text{ cm}^{2}$

Two circles taken out: $2 \times \pi \times 1.5^{2} = 4.5\pi\text{ cm}^{2}$ . Inside of the hole: $2 \times \pi \times 1.5 \times 3 = 9\pi\text{ cm}^{2}$ .

(a) Total surface area $= 108\pi - 4.5\pi + 9\pi = 112.5\pi\text{ cm}^{2}$

Check, as a short pipe: $2\pi Rl + 2\pi rl + 2(\pi R^{2} - \pi r^{2}) = 36\pi + 9\pi + 2(36\pi - 2.25\pi) = 36\pi + 9\pi + 67.5\pi = 112.5\pi$ .

(b) $112.5\pi\text{ cm}^{2} = 353.42\ldots\text{ cm}^{2} \approx 353\text{ cm}^{2}$

(a) $112.5\pi\text{ cm}^{2}$ (b) about $353\text{ cm}^{2}$

Practice 5: A metal pipe

A hollow metal pipe is 50 cm long. Its external diameter is 10 cm and its internal diameter is 8 cm. Find its total surface area, inside and out, using the $\pi$ button on your calculator.

ShowHide solution

The radii are $R = 10 \div 2 = 5$ cm and $r = 8 \div 2 = 4$ cm.

Outer curved surface $= 2 \times \pi \times 5 \times 50 = 500\pi\text{ cm}^{2}$ ; inner curved surface $= 2 \times \pi \times 4 \times 50 = 400\pi\text{ cm}^{2}$

Two ring ends $= 2(\pi \times 5^{2} - \pi \times 4^{2}) = 2(25\pi - 16\pi) = 18\pi\text{ cm}^{2}$

Total surface area $= 500\pi + 400\pi + 18\pi = 918\pi\text{ cm}^{2} = 2883.98\ldots\text{ cm}^{2} \approx 2880\text{ cm}^{2}$

The total surface area is $918\pi\text{ cm}^{2} \approx 2880\text{ cm}^{2}$ .

Practice 6: Painting a pipe outside only

A concrete pipe is 1.5 m long, with an external diameter of 1.2 m and an internal diameter of 1 m. Its outside curved surface and its two ends are painted, but not its inside. Find the area painted, using the $\pi$ button on your calculator.

ShowHide solution

The radii are $R = 1.2 \div 2 = 0.6$ m and $r = 1 \div 2 = 0.5$ m.

Outer curved surface $= 2\pi Rl = 2 \times \pi \times 0.6 \times 1.5 = 1.8\pi\text{ m}^{2}$

Two ring ends $= 2(\pi \times 0.6^{2} - \pi \times 0.5^{2}) = 2(0.36\pi - 0.25\pi) = 0.22\pi\text{ m}^{2}$

The inner curved surface is not painted.

Area painted $= 1.8\pi + 0.22\pi = 2.02\pi\text{ m}^{2} = 6.346\ldots\text{ m}^{2} \approx 6.35\text{ m}^{2}$

The area painted is $2.02\pi\text{ m}^{2} \approx 6.35\text{ m}^{2}$ .

Practice 7: A log cut in two

A solid log is a cylinder of radius 7 cm and length 30 cm. Ravi saws it in two along its length, straight through the middle of both ends, to make two half-cylinders. Taking $\pi = \frac{22}{7}$ , find (a) the total surface area of the log, (b) the total surface area of the two pieces together.

ShowHide solution

(a) Two ends $2\pi r^{2} = 2 \times \frac{22}{7} \times 7^{2} = 308\text{ cm}^{2}$ ; curved surface $2\pi rl = 2 \times \frac{22}{7} \times 7 \times 30 = 1320\text{ cm}^{2}$ .

Total surface area $= 308 + 1320 = 1628\text{ cm}^{2}$

(b) The cut face is a rectangle as wide as the diameter, $2 \times 7 = 14$ cm, and 30 cm long: $14 \times 30 = 420\text{ cm}^{2}$ . There are two of them.

Total for the two pieces $= 1628 + 2 \times 420 = 2468\text{ cm}^{2}$

Check: each half-cylinder has curved surface $\pi rl = \frac{22}{7} \times 7 \times 30 = 660$ , ends $\pi r^{2} = 154$ and flat face 420, total 1234; two pieces $2 \times 1234 = 2468\text{ cm}^{2}$ .

(a) $1628\text{ cm}^{2}$ (b) $2468\text{ cm}^{2}$

Practice 8: Slicing a Swiss roll

A Swiss roll is a cylinder of diameter 7 cm and length 20 cm. Hui Min cuts it straight across into 5 equal slices. Taking $\pi = \frac{22}{7}$ , find the total surface area (a) of the roll, (b) of the 5 slices together.

ShowHide solution

(a) The radius is $7 \div 2 = 3.5$ cm. Two ends $2\pi r^{2} = 2 \times \frac{22}{7} \times 3.5^{2} = 77\text{ cm}^{2}$ ; curved surface $2\pi rl = 2 \times \frac{22}{7} \times 3.5 \times 20 = 440\text{ cm}^{2}$ .

Total surface area $= 77 + 440 = 517\text{ cm}^{2}$

(b) 5 slices in a row need 4 cuts. Each cut makes two new circles, each $\pi r^{2} = \frac{22}{7} \times 3.5^{2} = 38.5\text{ cm}^{2}$ , so each cut adds $77\text{ cm}^{2}$ .

Total for the slices $= 517 + 4 \times 77 = 517 + 308 = 825\text{ cm}^{2}$

Check: each slice is $20 \div 5 = 4$ cm long, with surface area $77 + 2 \times \frac{22}{7} \times 3.5 \times 4 = 77 + 88 = 165$ ; five slices $5 \times 165 = 825\text{ cm}^{2}$ .

(a) $517\text{ cm}^{2}$ (b) $825\text{ cm}^{2}$

Practice 9: Cutting a block into cubes

A block of cheese is a cuboid 12 cm long, 8 cm wide and 4 cm high. Hui Min cuts it into cubes of edge 4 cm. (a) How many cubes does she get? (b) By how much does the total surface area increase? Check your answer by counting the cuts.

ShowHide solution

(a) Along the length $12 \div 4 = 3$ , across $8 \div 4 = 2$ , up $4 \div 4 = 1$ : $3 \times 2 \times 1 = 6$ cubes.

(b) Block: $2(12 \times 8 + 8 \times 4 + 12 \times 4) = 2(96 + 32 + 48) = 352\text{ cm}^{2}$

One cube: $6 \times 4^{2} = 96\text{ cm}^{2}$ . Six cubes: $6 \times 96 = 576\text{ cm}^{2}$

Increase $= 576 - 352 = 224\text{ cm}^{2}$

Check by the cuts: 2 cuts across the length, each making two faces 8 cm by 4 cm, add $2 \times 2 \times 32 = 128$ ; 1 cut along the length, making two faces 12 cm by 4 cm, adds $2 \times 48 = 96$ . $128 + 96 = 224\text{ cm}^{2}$ .

(a) 6 cubes (b) The surface area increases by $224\text{ cm}^{2}$ .

Practice 10: How many slices?

A loaf of bread is a cuboid 24 cm long with ends 10 cm by 8 cm. Mrs Tan cuts it straight across, parallel to its ends. The slices together have a total surface area $1600\text{ cm}^{2}$ more than the loaf had. How many cuts did she make, and how many slices does she have?

ShowHide solution

Each cut is the shape of an end, $10 \times 8 = 80\text{ cm}^{2}$ , and makes two new faces: it adds $2 \times 80 = 160\text{ cm}^{2}$ .

Let the number of cuts be $n$ .

$160n = 1600$

$n = 10$

10 cuts straight across a loaf make $10 + 1 = 11$ slices.

Mrs Tan made 10 cuts and has 11 slices.

Practice 11: A pencil holder

Siti makes a pencil holder from a wooden cuboid 10 cm by 10 cm by 8 cm high. She drills a hole of radius 3 cm and depth 6 cm into the middle of its top; the hole does not go right through. Find the total surface area of the holder, using the $\pi$ button on your calculator.

ShowHide solution

Cuboid: $2(10 \times 10 + 10 \times 8 + 10 \times 8) = 2(100 + 80 + 80) = 520\text{ cm}^{2}$

The hole goes in at one end only. It takes one circle, $\pi r^{2} = 9\pi$ , from the top, and adds its curved wall, $2\pi rh = 2 \times \pi \times 3 \times 6 = 36\pi$ , and its circular floor, $9\pi$ .

Total surface area $= 520 - 9\pi + 36\pi + 9\pi = 520 + 36\pi = 633.09\ldots \approx 633\text{ cm}^{2}$

The circles cancel, as in Example 3 (and as when a cylinder stands on a cuboid in Surface Area of Composite Solids): the floor of the hole replaces the circle taken from the top.

The total surface area is $(520 + 36\pi)\text{ cm}^{2} \approx 633\text{ cm}^{2}$ .

Key points

The surface area is the outside only; holes and cuts make new faces on the outside.
A round hole right through a solid: take away the two circles, $2\pi r^{2}$ , and add the inside, $2\pi rl$ .
A hole of any shape right through: take away its two ends and add perimeter $\times$ length.
A round hole that stops inside: take away the one circle of its opening and add its wall, $2\pi rd$ , and its floor, $\pi r^{2}$ ; the floor replaces the circle, so the hole adds only its wall.
A pipe: $2\pi Rl + 2\pi rl + 2(\pi R^{2} - \pi r^{2})$ .
Each straight cut right through a solid adds $2 \times$ the area of the cut face; pieces in a row need one cut fewer than pieces; the volume does not change.
Check a second way: face by face, or as the pieces themselves.