Volume & Surface Area — Mathslive.ie

Section 1 of 5

Boxes (Cuboids)

Everything starts with the box, set by its length $\ell$, breadth $b$ and height $h$.

The box formulas

$V=\ell \times b \times h$. Closed: $A=2(\ell b+bh+\ell h)$. Open-top: drop one $\ell b$: $A=\ell b+2\ell h+2bh$.

Worked example — a solid box

A solid box has length $6$ cm, breadth $5$ cm and height $3$ cm. Find its surface area and volume.

$A = 2(\ell b + bh + \ell h) = 2(30+15+18) = 2(63)$

$A = 126\ \text{cm}^2$

$V = 6\times5\times3$

$V = 90\ \text{cm}^3$

Worked example — an open-top box

An open-top box has length $3.5$ m, width $2.8$ m and height $1.5$ m. Find its surface area and volume.

$A = \ell b + 2\ell h + 2bh = 3.5(2.8)+2(3.5)(1.5)+2(2.8)(1.5)$

$A = 9.8 + 10.5 + 8.4$

$A = 28.7\ \text{m}^2,\quad V = 14.7\ \text{m}^3$

Worked example — work back from the volume

An open-top box has length $5$ cm, width $4$ cm and volume $160\ \text{cm}^3$. Find its surface area.

$20h = 160 \Rightarrow h = 8$

$A = 20 + 2(5)(8) + 2(4)(8) = 20 + 80 + 64$

$A = 164\ \text{cm}^2$

Worked example — work back from the area

A solid box has length $5$ cm, breadth $3$ cm and surface area $120\ \text{cm}^2$. Find its volume.

$2(15 + 5h + 3h) = 120 \Rightarrow 15 + 8h = 60 \Rightarrow h = \tfrac{45}{8}$

$V = 5\times3\times\tfrac{45}{8} = 84.4\ \text{cm}^3$

Worked example — a metal block and its mass

A solid block has length $12$ cm, width $5$ cm and volume $90\ \text{cm}^3$. Each $\text{cm}^3$ has mass $8.4$ g. A stack of these blocks weighs $113.4$ kg. How many blocks are there?

Height: $12(5)h = 90 \Rightarrow 60h = 90 \Rightarrow h = 1.5$ cm

One block: $90 \times 8.4 = 756$ g

Number $= 113400 \div 756$

$150$ blocks

Worked example — the jeweller’s gold (real-life)

A jeweller buys a gold block $4\times3\times2$ cm. Gold costs €$400$ per $\text{cm}^3$. Each ring needs $250\ \text{mm}^3$ and sells for €$120$. Find the cost of the block, the number of rings, and the profit per ring.

Volume $= 24\ \text{cm}^3$, so cost $= 24\times400 = $ €$9600$

$250\ \text{mm}^3 = 0.25\ \text{cm}^3$, so rings $= 24 \div 0.25 = 96$

Cost per ring $= 9600 \div 96 = $ €$100$

€$9600$; $96$ rings; profit $=120-100=$ €$20$ per ring

Section 2 of 5

Cylinders

The cylinder formulas

$V=\pi r^2 h$. Curved $=2\pi r h$. Closed total $=2\pi r h + 2\pi r^2$.

Worked example — solid cylinder, in terms of $\pi$

A solid cylinder has radius $5$ cm and height $10$ cm. Find its volume and surface area, in terms of $\pi$.

$V = \pi(5)^2(10) = \pi(25)(10)$

$V = 250\pi\ \text{cm}^3$

$A = 2\pi(5)(10) + 2\pi(5)^2 = 100\pi + 50\pi$

$A = 150\pi\ \text{cm}^2$

Worked example — open cylinder

An open cylinder (a base but no top) has radius $8$ m and height $6$ m. Find its surface area and volume, in terms of $\pi$.

$A = 2\pi(8)(6) + \pi(8)^2 = 96\pi + 64\pi$

$A = 160\pi\ \text{m}^2,\quad V = \pi(8)^2(6) = 384\pi\ \text{m}^3$

Worked example — using $\pi=\tfrac{22}{7}$

An open cylinder has radius $7$ cm and height $5$ cm. Find its curved surface area and volume, taking $\pi=\tfrac{22}{7}$.

$A = 2\times\tfrac{22}{7}\times 7\times 5 = 220$

$V = \tfrac{22}{7}\times 49\times 5$

$A = 220\ \text{cm}^2,\quad V = 770\ \text{cm}^3$

Worked example — reverse: find the radius

A cylinder has height $5$ cm and volume $80\pi\ \text{cm}^3$. Find its radius.

$\pi r^2(5) = 80\pi \Rightarrow 5r^2 = 80 \Rightarrow r^2 = 16$

$r = 4\ \text{cm}$

Worked example — reverse: find the height

A cylinder of radius $5$ m has volume $200\pi\ \text{m}^3$. Find its height.

$25h = 200$

$h = 8\ \text{m}$

Section 3 of 5

Spheres & Hemispheres

The sphere formulas (in your tables)

$V=\dfrac{4}{3}\pi r^3$ and $A=4\pi r^2$. A hemisphere is half a sphere — its surface adds the flat circle.

Worked example — a sphere

A sphere has radius $7$ cm. Find its surface area and volume, correct to one decimal place.

$A = 4\pi(7)^2 = 196\pi$

$A = 615.8\ \text{cm}^2$

$V = \dfrac{4}{3}\pi(7)^3$

$V = 1436.8\ \text{cm}^3$

Worked example — a solid hemisphere

A solid hemisphere has radius $9$ cm. Find its volume and total surface area, in terms of $\pi$.

$V = \dfrac{2}{3}\pi(729) = 486\pi$

$A = 3\pi r^2 = 3\pi(81)$ (curved $2\pi r^2$ + flat circle $\pi r^2$)

$V = 486\pi\ \text{cm}^3,\quad A = 243\pi\ \text{cm}^2$

Section 4 of 5

Recast & Melt-down

The key idea

When a solid is melted down and recast, nothing is lost — the volume stays the same. Set the two volumes equal.

Worked example — sphere recast as a cylinder

A sphere of radius $9$ cm is melted and recast as a cylinder of radius $3$ cm. Find the height of the cylinder.

Sphere: $V = \dfrac{4}{3}\pi(9)^3 = 972\pi$

$\pi(3)^2 h = 972\pi \Rightarrow 9h = 972$

$h = 108\ \text{cm}$

Worked example — cylinder recast as a cylinder

A cylinder of radius $8$ cm and height $6$ cm is melted and recast as a cylinder of height $10$ cm. Find the new radius.

$V = \pi(8)^2(6) = 384\pi$

$\pi r^2(10) = 384\pi \Rightarrow r^2 = 38.4$

$r = \sqrt{38.4} \approx 6.2\ \text{cm}$

Worked example — how many small spheres?

A sphere of radius $6$ cm is melted and recast as smaller spheres of radius $3$ cm. How many are made?

Big: $\dfrac{4}{3}\pi(6)^3 = 288\pi$. Small: $\dfrac{4}{3}\pi(3)^3 = 36\pi$

Number $= 288\pi \div 36\pi$

$8$ spheres

Section 5 of 5

Double Shape

A double shape is two solids joined or one inside another — a sphere in a cylinder, a hemisphere on a cylinder, a capsule. Work out each piece, then add or subtract.

Worked example — space left over: sphere in a cylinder

A sphere of radius $6$ cm fits snugly in the smallest possible cylinder. Find the percentage of empty space.

Cylinder: $r=6,\ h=2r=12$, so $V=\pi(6)^2(12) = 432\pi$

Sphere: $\dfrac{4}{3}\pi(6)^3 = 288\pi$

Space $= 432\pi - 288\pi = 144\pi$, so $\dfrac{144\pi}{432\pi}\times100$

$33\tfrac{1}{3}\%$ empty

Now you try

You try

Two spheres of radius $6$ cm sit in the smallest possible cylinder. Find the percentage of empty space.

Pen and paper out — try it before you reveal.

Two spheres $= 2\times 288\pi = 576\pi$

Cylinder $r=6,\ h=24$: $V=\pi(6)^2(24) = 864\pi$

Space $= 864\pi - 576\pi = 288\pi$

$\dfrac{288\pi}{864\pi}\times100 = 33\tfrac{1}{3}\%$ empty

You try

A cylinder of radius $6$ cm holds some water. A sphere of radius $3$ cm is submerged. By how much does the water rise?

The water rises by the sphere’s volume, spread over the cylinder’s base.

Pen and paper out — try it before you reveal.

Sphere $= \dfrac{4}{3}\pi(3)^3 = 36\pi$

$\pi(6)^2 h = 36\pi \Rightarrow 36h = 36$

Water rises $1\ \text{cm}$

You try

A tube is a cylinder of radius $6$ cm and height $14$ cm with a hemispherical bottom. Find its total volume in terms of $\pi$; then, if half the total is poured in as water, how high does the water stand up the cylinder?

Pen and paper out — try it before you reveal.

Hemisphere: $\dfrac{2}{3}\pi(6)^3 = 144\pi$. Cylinder: $\pi(6)^2(14) = 504\pi$

Total $= 144\pi + 504\pi = 648\pi$. Half $= 324\pi$

Water fills the hemisphere ($144\pi$) first: $324\pi - 144\pi = 180\pi$ left in the cylinder

$\pi(6)^2 h = 180\pi \Rightarrow 36h = 180$

Total $648\pi\ \text{cm}^3$; water stands $5$ cm up the cylinder

You try

A capsule is a cylinder with hemispherical ends. Its total length is $20$ mm and its diameter is $6$ mm. Find its volume, correct to the nearest $\text{mm}^3$.

Two hemispheres make one whole sphere. Radius $=3$ mm, so the cylinder length $=20-3-3=14$ mm.

Pen and paper out — try it before you reveal.

Sphere (both ends): $\dfrac{4}{3}\pi(3)^3 = 36\pi$

Cylinder: $\pi(3)^2(14) = 126\pi$

Total $= 126\pi + 36\pi = 162\pi$

$\approx 509\ \text{mm}^3$

That’s Volume & Surface Area.

Boxes, cylinders and spheres — solid or open — plus reverse problems, recasting, and every kind of double shape. Keep area in $\text{cm}^2$ and volume in $\text{cm}^3$.