Volume & Area — Mathslive.ie

Section 1 of 6

Cube & cuboid — the box

Think of a box, or a room.

The base is a rectangle. Each side is a rectangle too.

$A$ here means surface area — the total area of all the faces.

Must learn

1.$V = \ell \times b \times h$

2.$A = 2(\ell b + \ell h + bh)$

(i) Worked example — a closed box

Length $20$ m, width $12$ m, height $15$ m. Find volume and surface area.

$\ell = 20, \quad b = 12, \quad h = 15$

$V = \ell \times b \times h$

$V = 20 \times 12 \times 15$

$V = 3600$ m³

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

$A = 2\big(20(12) + 20(15) + 12(15)\big)$

$A = 1440$ m²

(ii) Worked example — an open-top box

An open-top box has length $12$ m, height $6$ m and width $8$ m. Find area and volume.

Open top means there is no lid — so leave that face out of the area.

$\ell = 12, \quad h = 6, \quad b = 8$

$V = \ell \times b \times h = 12 \times 6 \times 8$

$V = 576$ m³

For the area: one base $+$ four walls (no lid):

$A = \ell b + 2\ell h + 2bh$

$A = 12(8) + 2(12)(6) + 2(8)(6)$

$A = 432$ m²

YOU TRY · 1

A closed box has length $10$ m, width $4$ m and height $3$ m. Find the volume and the total surface area.

Write down $\ell$, $b$, $h$ first. Then $V = \ell b h$ and $A = 2(\ell b + \ell h + bh)$.

$V = 10 \times 4 \times 3 = 120$ m³

$A = 2\big(10(4) + 10(3) + 4(3)\big) = 2(82)$

$A = 164$ m²

$V = 120$ m³, $A = 164$ m²

YOU TRY · 2

An open-top box has length $5$ m, width $4$ m and height $2$ m. Find the area of material needed to make it.

No lid — one base and four walls: $A = \ell b + 2\ell h + 2bh$.

$A = 5(4) + 2(5)(2) + 2(4)(2)$

$A = 20 + 20 + 16$

$A = 56$ m²

$56$ m²

Section 2 of 6

Cylinder

A cylinder is a tube with a circle on the top and a circle on the bottom.

The two circles each have area $A = \pi r^2$. The curved part rolls out into a rectangle.

Must learn

1.Volume: $V = \pi r^2 h$

2.Curved area: $A = 2\pi r h$

3.Each end circle: $A = \pi r^2$

Why $2\pi r h$? Unroll the curved surface — it becomes a rectangle, $h$ tall and $2\pi r$ wide (the circumference):

(i) Curved area and volume

A cylinder has radius $5.3$ cm and height $4.3$ cm. Find the volume and the curved area, to one decimal place.

$r = 5.3, \quad h = 4.3$

$A = 2\pi r h = 2\pi (5.3)(4.3)$

$A = 143.2$ cm²

$V = \pi r^2 h = \pi (5.3)^2 (4.3)$

$V = 379.46\ldots$

$V = 379.5$ cm³

(ii) Closed tank — surface area and volume

A tank has radius $8$ m and height $15$ m. Find the surface area and volume, to the nearest metre.

$r = 8, \quad h = 15$

$A = 2\pi r h = 2\pi (8)(15)$

$A = 754$ m²

$V = \pi r^2 h = \pi (8)^2 (15)$

$V = 3016$ m³

(iii) Open-top cylinder (in terms of $\pi$)

A cylinder with a base but an open top has radius $6$ cm and height $12$ cm. Find area and volume in terms of $\pi$.

Open top — so the curved surface plus one circle (the base).

$r = 6, \quad h = 12$

$A = 2\pi r h + \pi r^2$

$A = 2\pi (6)(12) + \pi (6)^2$

$A = 144\pi + 36\pi$

$A = 180\pi$ cm²

$V = \pi r^2 h = \pi (6)^2 (12)$

$V = 432\pi$ cm³

(iv) Solid cylinder (in terms of $\pi$)

Radius $8$ m, height $10$ m. Find area and volume of a solid cylinder in terms of $\pi$.

Solid — so the curved surface plus two circles.

$r = 8, \quad h = 10$

$A = 2\pi r h + 2\pi r^2$

$A = 2\pi (8)(10) + 2\pi (8)^2$

$A = 160\pi + 128\pi$

$A = 288\pi$ m²

$V = \pi r^2 h = \pi (8)^2 (10)$

$V = 640\pi$ m³

(v) Volume, curved area, then total of a solid

A cylinder has radius $5.5$ m and height $2.8$ m. To one decimal place find (i) volume, (ii) curved area, (iii) total area of the solid.

$r = 5.5, \quad h = 2.8$

$V = \pi r^2 h = \pi (5.5)^2 (2.8)$

$V = 266.1$ m³

Curved: $A = 2\pi r h = 2\pi (5.5)(2.8)$

$A = 96.8$ m²

Solid — add on the $2$ circles:

$A = 2\pi r h + 2\pi r^2 = 2\pi(5.5)(2.8) + 2\pi(5.5)^2$

$A = 286.8$ m²

(vi) Working backwards — find the height

A cylinder of radius $5$ cm has a volume of $100\pi$ cm³. Find the height.

$r = 5, \quad h = \,?\,, \quad V = 100\pi$

Backwards — keep the unknown on the left.

$\pi r^2 h = V$

$\pi (5)^2 h = 100\pi$

$25h = 100$

$h = 4$ cm

(vii) Working backwards — find the radius

A cylinder of height $6$ m has volume $24\pi$ m³. Find the radius.

$h = 6, \quad V = 24\pi, \quad r = \,?$

$\pi r^2 h = V$

$\pi r^2 (6) = 24\pi$

$6 r^2 = 24$

$r^2 = 4$

$r = \sqrt{4} = 2$ m

Same idea with decimals. Height $2.3$ m, volume $256.4$ m³:

$\pi r^2 (2.3) = 256.4$

$2.3\,\pi\, r^2 = 256.4$

$r^2 = \dfrac{256.4}{2.3\,\pi} = 35.48$

$r = \sqrt{35.48} = 5.95$

$r = 6$ m

And once more. Height $3.9$ m, volume $678.3$ m³ (to one decimal place):

$\pi r^2 (3.9) = 678.3$

$r^2 = \dfrac{678.3}{3.9\,\pi} = 55.36$

$r = 7.4$ m

YOU TRY · 3

A solid cylinder has radius $4$ cm and height $10$ cm. Find the volume and total surface area, in terms of $\pi$.

Solid — curved plus two circles: $A = 2\pi r h + 2\pi r^2$.

$V = \pi (4)^2 (10) = 160\pi$ cm³

$A = 2\pi(4)(10) + 2\pi(4)^2 = 80\pi + 32\pi$

$A = 112\pi$ cm²

$V = 160\pi$ cm³, $A = 112\pi$ cm²

YOU TRY · 4

A cylinder of radius $7$ cm has a volume of $343\pi$ cm³. Find the height.

Backwards: $\pi r^2 h = V$, keep the unknown on the left.

$\pi (7)^2 h = 343\pi$

$49h = 343$

$h = 7$ cm

YOU TRY · 5

A cylinder of height $5$ m has a volume of $180\pi$ m³. Find the radius.

$\pi r^2 h = V$, then divide and take the square root.

$\pi r^2 (5) = 180\pi$

$5 r^2 = 180 \;\Rightarrow\; r^2 = 36$

$r = \sqrt{36} = 6$ m

$r = 6$ m

Section 3 of 6

Cone

A cone has a circular base and comes to a point.

The slant height $\ell$ is the distance up the side. It is not the vertical height $h$.

Must learn

1.Slant height: $\ell^2 = h^2 + r^2$

2.Volume: $V = \dfrac{1}{3}\pi r^2 h$

3.Curved area: $A = \pi r \ell$

(i) Worked example (in terms of $\pi$)

A cone has radius $3$ cm and height $4$ cm. Find (i) volume, (ii) curved area, (iii) total area when solid.

$r = 3, \quad h = 4$

$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi (3)^2 (4)$

$V = 12\pi$ cm³

Find the slant height first:

$\ell^2 = h^2 + r^2 = 4^2 + 3^2 = 25$

$\ell = \sqrt{25} = 5$ cm

Curved: $A = \pi r \ell = \pi (3)(5)$

$A = 15\pi$ cm²

Total when solid — add the base circle:

$A = \pi r \ell + \pi r^2 = 15\pi + \pi (3)^2$

$A = 15\pi + 9\pi$

$A = 24\pi$ cm²

(ii) Worked example (in terms of $\pi$)

Radius $12$ m, height $5$ m. Find (i) volume, (ii) total area when solid.

$r = 12, \quad h = 5$

$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi (12)^2 (5)$

$V = 240\pi$ m³

$\ell^2 = h^2 + r^2 = 5^2 + 12^2 = 169$

$\ell = \sqrt{169} = 13$ m

$A = \pi r \ell + \pi r^2 = \pi (12)(13) + \pi (12)^2$

$A = 300\pi$ m²

(iii) Worked example (one decimal place)

A solid cone has radius $3.5$ m and height $2.1$ m. Find the volume and area, to one decimal place.

$r = 3.5, \quad h = 2.1$

$V = \dfrac{1}{3}\pi (3.5)^2 (2.1)$

$V = 26.9$ m³

$\ell^2 = (2.1)^2 + (3.5)^2 = 16.66$

$\ell = \sqrt{16.66} = 4.1$ m

$A = \pi r \ell + \pi r^2 = \pi (3.5)(4.1) + \pi (3.5)^2$

$A = 83.4$ m²

YOU TRY · 6

A solid cone has radius $6$ cm and height $8$ cm. Find the volume, the slant height, and the total surface area, in terms of $\pi$.

$V = \tfrac13 \pi r^2 h$, then $\ell^2 = h^2 + r^2$, then $A = \pi r \ell + \pi r^2$.

$V = \dfrac{1}{3}\pi (6)^2 (8) = 96\pi$ cm³

$\ell^2 = 8^2 + 6^2 = 100 \;\Rightarrow\; \ell = 10$ cm

$A = \pi(6)(10) + \pi(6)^2 = 60\pi + 36\pi$

$A = 96\pi$ cm²

$V = 96\pi$ cm³, $\ell = 10$ cm, $A = 96\pi$ cm²

YOU TRY · 7

A cone has radius $5$ m and height $12$ m. Find the slant height and the curved surface area, in terms of $\pi$.

Slant first: $\ell^2 = h^2 + r^2$. Curved area is $\pi r \ell$ only.

$\ell^2 = 12^2 + 5^2 = 169 \;\Rightarrow\; \ell = 13$ m

$A = \pi r \ell = \pi (5)(13)$

$A = 65\pi$ m²

$\ell = 13$ m, curved $A = 65\pi$ m²

Section 4 of 6

Sphere

A sphere is a ball. One measurement only — the radius $r$.

Must learn

1.Volume: $V = \dfrac{4}{3}\pi r^3$

2.Area: $A = 4\pi r^2$

(i) Worked example (in terms of $\pi$)

Radius $12$ cm. Find the area and volume in terms of $\pi$.

$r = 12$

$A = 4\pi r^2 = 4\pi (12)^2$

$A = 576\pi$ cm²

$V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi (12)^3$

$V = 2304\pi$ cm³

(ii) Worked example (one decimal place)

Radius $6.3$ m. Find the area and volume, to one decimal place.

$r = 6.3$

$A = 4\pi (6.3)^2$

$A = 498.8$ m²

$V = \dfrac{4}{3}\pi (6.3)^3$

$V = 1047.4$ m³

(iii) Working backwards — find the radius

The volume of a sphere is $36\pi$ m³. Find $r$.

$V = 36\pi$

$\dfrac{4}{3}\pi r^3 = 36\pi$

$r^3 = 36 \div \dfrac{4}{3}$

$r^3 = 27$

$r = \sqrt[3]{27} = 3$ m

Same method with decimals. Volume $673.8$ m³ (to one decimal place):

$\dfrac{4}{3}\pi r^3 = 673.8$

$r^3 = 673.8 \div \dfrac{4}{3}\pi = 160.85$

$r = \sqrt[3]{160.85} = 5.43$

$r = 5.4$ m

YOU TRY · 8

A sphere has radius $3$ cm. Find the area and the volume, in terms of $\pi$.

$A = 4\pi r^2$ and $V = \tfrac{4}{3}\pi r^3$.

$A = 4\pi (3)^2 = 36\pi$ cm²

$V = \dfrac{4}{3}\pi (3)^3 = 36\pi$ cm³

$A = 36\pi$ cm², $V = 36\pi$ cm³

YOU TRY · 9

A sphere has a volume of $288\pi$ cm³. Find the radius.

$\tfrac{4}{3}\pi r^3 = V$. Divide out, then cube root.

$\dfrac{4}{3}\pi r^3 = 288\pi$

$r^3 = 288 \div \dfrac{4}{3} = 216$

$r = \sqrt[3]{216} = 6$ cm

$r = 6$ cm

Section 5 of 6

Hemisphere

A hemisphere is half a sphere.

Halve the sphere formulas. The curved area is half of $4\pi r^2$:

Must learn

1.Volume: $V = \dfrac{2}{3}\pi r^3$

2.Curved area: $A = 2\pi r^2$

3.Solid total: $A = 2\pi r^2 + \pi r^2 = 3\pi r^2$

When it is solid, add the flat circle on top ($\pi r^2$) to the curved part ($2\pi r^2$) — giving $3\pi r^2$.

(i) Worked example (in terms of $\pi$)

A solid hemisphere has radius $6$ cm. Find the area and volume in terms of $\pi$.

$r = 6$

$V = \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi (6)^3$

$V = 144\pi$ cm³

Area of a solid hemisphere — curved part plus the top circle:

$A = 2\pi r^2 + \pi r^2 = 3\pi r^2$

$A = 3\pi (6)^2$

$A = 108\pi$ cm²

YOU TRY · 10

A solid hemisphere has radius $3$ cm. Find the volume and the total surface area, in terms of $\pi$.

$V = \tfrac{2}{3}\pi r^3$. Solid total area $= 3\pi r^2$.

$V = \dfrac{2}{3}\pi (3)^3 = 18\pi$ cm³

$A = 3\pi (3)^2 = 27\pi$ cm²

$V = 18\pi$ cm³, $A = 27\pi$ cm²

Section 6 of 6

Nets

A net is the solid opened out flat. It shows every face, so it is the surface area laid out where you can see it.

What each net is made of

BoxSix rectangles.

CylTwo circles (the ends) and one rectangle (the curved part). The rectangle's width is the circumference, $2\pi r$.

ConeOne circle (the base) and one sector (the curved part). The sector's arc is the base circumference, $2\pi r$.

(i) The net of a box

A solid box has length $5\,\text{cm}$, width $4\,\text{cm}$ and height $3\,\text{cm}$. Draw its net.

Open it out flat: a row of four faces around the middle, with the top and bottom folded off the front.

(ii) The net of a cylinder

Draw the net of a cylinder with radius $5\,\text{cm}$ and height $6\,\text{cm}$.

The curved part unrolls into a rectangle. Its height is the cylinder's height; its width is the circumference of the circle.

Rectangle width $= 2\pi r = 2\pi(5) = 10\pi \approx 31.4 \text{ cm}$

The two circles are the top and bottom; the rectangle wraps around to make the tube.

(iii) The net of a cone

A cone has radius $3\,\text{cm}$ and height $5\,\text{cm}$. Draw the net and find the angle of the sector.

The curved part opens into a sector. Its straight edge is the slant height $\ell$, and its curved edge (the arc) is the base circumference $2\pi r$.

$\ell^2 = h^2 + r^2 = 5^2 + 3^2 = 34 \quad\Rightarrow\quad \ell = \sqrt{34}$

arc $= 2\pi r = 2\pi(3) = 6\pi$

The sector is a slice of a circle of radius $\ell = \sqrt{34}$. Its arc is $6\pi$, so set the arc formula equal to $6\pi$ and solve for the angle.

$2\pi r\left(\dfrac{\theta}{360}\right) = 6\pi \quad\Rightarrow\quad 2\pi\sqrt{34}\left(\dfrac{\theta}{360}\right) = 6\pi$

$\theta = \dfrac{6\pi \times 360}{2\pi\sqrt{34}} = \dfrac{1080}{\sqrt{34}}$

$\theta = 185^\circ$ (to the nearest degree)

YOU TRY · 11

Draw the net of a cylinder with radius $7\,\text{cm}$ and height $10\,\text{cm}$. What are the dimensions of the rectangle?

Two circles plus a rectangle. Rectangle width $= 2\pi r$, height $=$ the cylinder height.

width $= 2\pi(7) = 14\pi \approx 44.0$

height $= 10$

Rectangle $44.0 \text{ cm} \times 10 \text{ cm}$, plus two circles of radius $7 \text{ cm}$

Rectangle $14\pi \approx 44.0 \text{ cm}$ wide by $10 \text{ cm}$ high, with two circles of radius $7 \text{ cm}$

Summary

Volume & Area toolkit

1.Cuboid: $V = \ell b h$, $A = 2(\ell b + \ell h + bh)$

2.Cylinder: $V = \pi r^2 h$, curved $2\pi r h$, each end $\pi r^2$

3.Cone: $V = \dfrac{1}{3}\pi r^2 h$, curved $\pi r \ell$, $\ell^2 = h^2 + r^2$

4.Sphere: $V = \dfrac{4}{3}\pi r^3$, $A = 4\pi r^2$

5.Hemisphere: $V = \dfrac{2}{3}\pi r^3$, solid total $3\pi r^2$

6.Open top: drop one circle. Solid: keep both.

7.Backwards: keep the unknown on the left, then divide and root.

8.Nets: box $=$ 6 rectangles; cylinder $=$ 2 circles $+$ a rectangle of width $2\pi r$; cone $=$ a circle $+$ a sector whose arc is $2\pi r$.

End of lesson

Volume & Area — HL · Mathslive.ie