MENSURATION · HL
Volume 2
Composite solids, recasting, displacement.
Section 1 of 5
Double shapes
A double shape is two solids joined together — a cylinder with a cone on the end, a tank with a rounded base, an ice-cream cone.
We want the total volume and the total area, both left in terms of $\pi$.
Volume: just add the two volumes. Area: add only the surfaces you can actually see — drop any face that's hidden where the two solids meet.
From the tables
1.Cylinder: $V = \pi r^2 h$, curved $A = 2\pi r h$
2.Cone: $V = \dfrac{1}{3}\pi r^2 h$, curved $A = \pi r l$
3.Sphere: $V = \dfrac{4}{3}\pi r^3$
4.Hemisphere: $V = \dfrac{2}{3}\pi r^3$, curved $A = 2\pi r^2$
5.Circle (disc): $A = \pi r^2$
(i) Cylinder + cone
A cylinder sits on top of a cone. The radius is $3\text{ m}$. The cylinder is $8\text{ m}$ tall, the cone $4\text{ m}$ — $12\text{ m}$ in all. Find the volume and area in terms of $\pi$.
Cylinder: $r = 3$, $h = 8$
$V = \pi r^2 h = \pi(3)^2(8) = 72\pi$
Cone: $r = 3$, $h = 4$
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(3)^2(4) = 12\pi$
Total volume $= 72\pi + 12\pi = 84\pi \text{ m}^3$
Now the area. Three pieces: the top circle, the cylinder's curved side, and the cone's curved side.
Top circle: $A = \pi r^2 = \pi(3)^2 = 9\pi$
Cylinder (curved): $A = 2\pi r h = 2\pi(3)(8) = 48\pi$
For the cone's curved area we need the slant length $l$, not the height. Use Pythagoras on $r$ and $h$.
$l^2 = 3^2 + 4^2 = 9 + 16 = 25 \;\Rightarrow\; l = 5$
Cone (curved): $A = \pi r l = \pi(3)(5) = 15\pi$
Total area $= 9\pi + 48\pi + 15\pi = 72\pi \text{ m}^2$
(ii) Cylinder + hemisphere
A tank: a cylinder with a hemisphere on the bottom. The radius is $6\text{ m}$, the total height $30\text{ m}$, so the cylinder is $24\text{ m}$ and the hemisphere $6\text{ m}$. Volume and area in terms of $\pi$.
Cylinder: $r = 6$, $h = 24$
$V = \pi r^2 h = \pi(6)^2(24) = 864\pi$
Hemisphere: $r = 6$
$V = \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi(6)^3 = 144\pi$
Total volume $= 864\pi + 144\pi = 1008\pi \text{ m}^3$
Area: the top circle (the lid), the cylinder's curved side, and the hemisphere's curved surface.
Top circle: $A = \pi r^2 = \pi(6)^2 = 36\pi$
Cylinder (curved): $A = 2\pi r h = 2\pi(6)(24) = 288\pi$
Hemisphere (curved): $A = 2\pi r^2 = 2\pi(6)^2 = 72\pi$
Total area $= 36\pi + 288\pi + 72\pi = 396\pi \text{ m}^2$
(iii) Cone + hemisphere
An ice-cream shape: a cone on top of a hemisphere. The radius is $3\text{ cm}$ and the total height is $8\text{ cm}$, so the cone height is $8 - 3 = 5\text{ cm}$. Volume in terms of $\pi$.
Hemisphere: $r = 3$
$V = \dfrac{2}{3}\pi r^3 = \dfrac{2}{3}\pi(3)^3 = 18\pi$
Cone: $r = 3$, $h = 5$
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(3)^2(5) = 15\pi$
Total volume $= 18\pi + 15\pi = 33\pi \text{ cm}^3$
YOU TRY · 1
A cylinder of radius $5\text{ cm}$ and height $10\text{ cm}$ has a cone of the same radius and height $6\text{ cm}$ on top. Find the total volume in terms of $\pi$.
Add the two volumes. Same radius for both.
Cylinder: $\pi(5)^2(10) = 250\pi$
Cone: $\dfrac{1}{3}\pi(5)^2(6) = 50\pi$
$250\pi + 50\pi = 300\pi \text{ cm}^3$
$300\pi \text{ cm}^3$
YOU TRY · 2
A solid is a cylinder of radius $4\text{ cm}$ and height $10\text{ cm}$ with a hemisphere on top. Find the total volume in terms of $\pi$.
Hemisphere volume is $\dfrac{2}{3}\pi r^3$.
Cylinder: $\pi(4)^2(10) = 160\pi$
Hemisphere: $\dfrac{2}{3}\pi(4)^3 = \dfrac{128}{3}\pi$
$160\pi + \dfrac{128}{3}\pi = \dfrac{608}{3}\pi \text{ cm}^3$
$\dfrac{608}{3}\pi \text{ cm}^3$
YOU TRY · 3
A cone has radius $6\text{ cm}$ and height $8\text{ cm}$. Find its slant length, then its curved surface area in terms of $\pi$.
$l^2 = r^2 + h^2$, then $A = \pi r l$.
$l^2 = 6^2 + 8^2 = 100 \Rightarrow l = 10$
$A = \pi r l = \pi(6)(10)$
$l = 10\text{ cm}, \quad A = 60\pi \text{ cm}^2$
$l = 10\text{ cm}, \; A = 60\pi \text{ cm}^2$
Section 2 of 5
Meltdown and recast
The whole idea
1.When a solid is melted down and recast, the volume stays the same.
2.Set $V_{\text{old}} = V_{\text{new}}$ and solve for the missing length.
3."How many can be made?" — divide, then round down to a whole shape.
(i) Sphere → cylinder
A sphere of radius $6\text{ cm}$ is melted and recast as a cylinder of radius $3\text{ cm}$. Find the height of the cylinder.
Sphere: $r = 6$
$V = \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi(6)^3 = 288\pi$
Cylinder: $r = 3$, $V = 288\pi$
$\pi r^2 h = V \;\Rightarrow\; \pi(3)^2 h = 288\pi$
$9h = 288$
$h = \dfrac{288}{9} = 32 \text{ cm}$
(ii) Cone → cylinder
A cone of radius $15\text{ cm}$ and height $12\text{ cm}$ is melted and reformed as a cylinder of radius $6\text{ cm}$. Find the height of the cylinder.
Cone: $r = 15$, $h = 12$
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(15)^2(12) = 900\pi$
Cylinder: $r = 6$, $V = 900\pi$
$\pi(6)^2 h = 900\pi \;\Rightarrow\; 36h = 900$
$h = 25 \text{ cm}$
(iii) Gold bar → cylinder
A gold bar is $30\text{ cm}$ long, $15\text{ cm}$ wide and $8\text{ cm}$ high. A cylinder of radius $4\text{ cm}$ is formed from it. Find the height of the cylinder to $1$ decimal place.
Rectangular bar: $V = l \times b \times h = 30 \times 15 \times 8 = 3600 \text{ cm}^3$
Cylinder: $r = 4$, $V = 3600$
$\pi(4)^2 h = 3600 \;\Rightarrow\; 16\pi h = 3600$
$h = \dfrac{3600}{16\pi} = 71.6 \text{ cm}$
(iv) Cylinder → cone
A cylinder of radius $9\text{ cm}$ and height $10\text{ cm}$ is melted and reformed as a cone of radius $12\text{ cm}$. Find the height of the cone.
Cylinder: $r = 9$, $h = 10$
$V = \pi r^2 h = \pi(9)^2(10) = 810\pi$
Cone: $r = 12$, $V = 810\pi$
$\dfrac{1}{3}\pi(12)^2 h = 810\pi \;\Rightarrow\; 48h = 810$
$h = \dfrac{810}{48} = 16.9 \text{ cm}$
(v) Cone → sphere
A cone of radius $9\text{ cm}$ and height $6\text{ cm}$ is reformed into a sphere. Find the radius of the sphere.
Cone: $r = 9$, $h = 6$
$V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi(9)^2(6) = 162\pi$
Sphere: $V = 162\pi$
$\dfrac{4}{3}\pi r^3 = 162\pi$
$r^3 = 162 \div \dfrac{4}{3} = 121.5$
$r = \sqrt[3]{121.5} = 4.95 \approx 5 \text{ cm}$
(vi) Sphere → how many small spheres?
A sphere of radius $15\text{ cm}$ is melted into spheres of radius $9\text{ cm}$. How many can be formed?
Big sphere: $r = 15$, $V = \dfrac{4}{3}\pi(15)^3 = 4500\pi$
Small sphere: $r = 9$, $V = \dfrac{4}{3}\pi(9)^3 = 972\pi$
Number $= \dfrac{4500\pi}{972\pi} = 4.6$
You need a whole sphere — $4.6$ rounds down.
$= 4$ spheres
(vii) Cylinder → how many cones?
A cylinder of radius $15\text{ cm}$ and height $12\text{ cm}$ is melted and recast as cones of radius $6\text{ cm}$ and height $3\text{ cm}$. How many cones can be formed?
Cylinder: $r = 15$, $h = 12$
$V = \pi r^2 h = \pi(15)^2(12) = 2700\pi$
Cone: $r = 6$, $h = 3$
$V = \dfrac{1}{3}\pi(6)^2(3) = 36\pi$
Number $= \dfrac{2700\pi}{36\pi} = 75$ cones
YOU TRY · 4
A sphere of radius $6\text{ cm}$ is melted and recast as a cylinder of radius $4\text{ cm}$. Find the height of the cylinder.
Volume stays the same. Find the sphere volume first.
Sphere: $\dfrac{4}{3}\pi(6)^3 = 288\pi$
$\pi(4)^2 h = 288\pi \Rightarrow 16h = 288$
$h = 18 \text{ cm}$
$h = 18 \text{ cm}$
YOU TRY · 5
A sphere of radius $12\text{ cm}$ is melted into spheres of radius $4\text{ cm}$. How many can be formed?
Divide the volumes. The $\pi$ cancels.
Big: $\dfrac{4}{3}\pi(12)^3 = 2304\pi$
Small: $\dfrac{4}{3}\pi(4)^3 = \dfrac{256}{3}\pi$
$\dfrac{2304\pi}{\frac{256}{3}\pi} = 27$ spheres
$27$ spheres
Section 3 of 5
Empty space
A solid sits inside a container — a sphere in a box, a sphere in a cylinder. We want the empty space around it.
Empty space $=$ container $-$ solid
Draw a front view (2-D). It tells you the container's dimensions at a glance: a ball that just fits has diameter $=$ the side of the box.
(i) Sphere in a box
A sphere of radius $8\text{ cm}$ just fits in a box. Find (a) the empty space and (b) the percentage of empty space.
The ball touches all four walls, so the box side $= 2 \times 8 = 16\text{ cm}$.
Box: $l = b = h = 16$
$V = l \times b \times h = 16 \times 16 \times 16 = 4096 \text{ cm}^3$
Sphere: $r = 8$
$V = \dfrac{4}{3}\pi(8)^3 = 2144.7 \text{ cm}^3$
Empty space $= 4096 - 2144.7 = 1951.3 \text{ cm}^3$
For the percentage, compare the empty space with the whole box.
$\% = \dfrac{\text{space}}{\text{box}} \times 100 = \dfrac{1951.3}{4096} \times 100$
$= 47.6\% \approx 48\%$
(ii) Sphere in a cylinder
A sphere of radius $9\text{ cm}$ just fits in a cylinder. Find the percentage of empty space.
A ball that just fits gives cylinder height $=$ diameter $= 18\text{ cm}$, and the cylinder radius $= 9$ too.
Sphere: $r = 9$, $V = \dfrac{4}{3}\pi(9)^3 = 972\pi$
Cylinder: $r = 9$, $h = 18$
$V = \pi(9)^2(18) = 1458\pi$
Empty space $= 1458\pi - 972\pi = 486\pi$
$\% = \dfrac{486\pi}{1458\pi} \times 100$
$= 33\%$
(iii) Cylinder in a box
A cylinder of radius $5\text{ cm}$ and height $10\text{ cm}$ is placed in a box. Find the percentage of empty space.
Cylinder: $r = 5$, $h = 10$
$V = \pi(5)^2(10) = 250\pi = 785.4 \text{ cm}^3$
The box just fits the cylinder, so $l = b = $ diameter $= 10$ and $h = 10$.
Box: $V = 10 \times 10 \times 10 = 1000 \text{ cm}^3$
Empty space $= 1000 - 785.4 = 214.6 \text{ cm}^3$
$\% = \dfrac{214.6}{1000} \times 100$
$= 21.5\%$
YOU TRY · 6
A sphere of radius $5\text{ cm}$ just fits in a box. Find the percentage of empty space.
Box side $= 10$. Space $\div$ box $\times 100$.
Box: $10^3 = 1000$
Sphere: $\dfrac{4}{3}\pi(5)^3 = 523.6$
Space $= 1000 - 523.6 = 476.4$
$\dfrac{476.4}{1000} \times 100 = 47.6\% \approx 48\%$
$\approx 48\%$
Section 4 of 5
Water displacement
Drop a solid into water and the level rises. The water rises by exactly the volume of the solid.
Carl's trick: do a front-end (2-D) sketch. The risen band of water is just a cylinder — same radius as the container, and the same volume as the object you dropped in.
The yellow band is the risen layer — its volume equals the dropped object's volume.
(i) Sphere into a cylinder
A cylinder of radius $6\text{ cm}$ is part-filled with water. A sphere of radius $3\text{ cm}$ is fully submerged. How high does the water rise?
Sphere: $r = 3$, $V = \dfrac{4}{3}\pi(3)^3 = 36\pi$
The risen band is a cylinder of radius $6$ with this same volume.
$\pi(6)^2 h = 36\pi \;\Rightarrow\; 36h = 36$
$h = 1 \text{ cm}$
(ii) Cone removed from a cylinder
A cylinder of radius $5\text{ cm}$ holds water with a submerged cone of radius $3\text{ cm}$ and height $4\text{ cm}$. How far does the water drop when the cone is taken out?
Cone: $r = 3$, $h = 4$
$V = \dfrac{1}{3}\pi(3)^2(4) = 12\pi$
The drop is a cylinder of radius $5$ with this same volume.
$\pi(5)^2 h = 12\pi \;\Rightarrow\; 25h = 12$
$h = \dfrac{12}{25} = 0.48 \text{ cm}$
(iii) Cubes into a cylinder
A cylinder of radius $6\text{ cm}$ holds water. $8$ cubes of side $2\text{ cm}$ are fully submerged. By how much does the water rise?
One cube: $V = 2 \times 2 \times 2 = 8 \text{ cm}^3$
Eight cubes: $8 \times 8 = 64 \text{ cm}^3$
Cylinder: $r = 6$, $V = 64$
$\pi(6)^2 h = 64 \;\Rightarrow\; 36\pi h = 64$
$h = \dfrac{64}{36\pi} = 0.57 \text{ cm}$
YOU TRY · 7
A cylinder of radius $4\text{ cm}$ holds water. A sphere of radius $3\text{ cm}$ is fully submerged. How high does the water rise?
Sphere volume $=$ risen cylinder volume (radius $4$).
Sphere: $\dfrac{4}{3}\pi(3)^3 = 36\pi$
$\pi(4)^2 h = 36\pi \Rightarrow 16h = 36$
$h = \dfrac{36}{16} = 2.25 \text{ cm}$
$h = 2.25 \text{ cm}$
YOU TRY · 8
A cylinder of radius $10\text{ cm}$ holds water. $5$ cubes of side $4\text{ cm}$ are submerged. By how much does the water rise, to $1$ decimal place?
Total cube volume first, then $\pi r^2 h = V$.
Cubes: $5 \times 4^3 = 320 \text{ cm}^3$
$\pi(10)^2 h = 320 \Rightarrow 100\pi h = 320$
$h = \dfrac{320}{100\pi} = 1.0 \text{ cm}$
$h = 1.0 \text{ cm}$
Section 5 of 5
Rate of flow
Water moving through a pipe makes a cylinder of water. The amount that passes any point each second is the cross-section circle pushed forward by the speed.
The idea
=Volume per second $= \pi r^2 \times \text{speed}$ (a cylinder one second long)
tTime $= \dfrac{\text{total volume}}{\text{volume per second}}$
Units you will need
L$1 \text{ litre} = 1000 \text{ cm}^3$
cm$1\,\text{m} = 100\,\text{cm}$, $1\,\text{m}^2 = 10\,000\,\text{cm}^2$, $1\,\text{m}^3 = 1\,000\,000\,\text{cm}^3$
(i) Water through a pipe
Water flows through a cylindrical pipe of radius $5\,\text{cm}$ at $8\,\text{cm/sec}$. How long will it take $12$ litres to flow through?
First change litres to $\text{cm}^3$ so the units match.
$12 \text{ litres} = 12 \times 1000 = 12\,000 \text{ cm}^3$
Now the volume that flows each second is a cylinder: radius $5$, length equal to the speed $8$.
$V \text{ per second} = \pi r^2 \times \text{speed} = \pi (5)^2 (8) = 200\pi \text{ cm}^3$
Time $= \dfrac{\text{total volume}}{\text{volume per second}} = \dfrac{12\,000}{200\pi}$
Time $= 19.1 \text{ seconds}$
YOU TRY · 9
Water flows through a cylindrical pipe of radius $4\,\text{cm}$ at $10\,\text{cm/sec}$. How long will it take $8$ litres to flow through, to one decimal place?
Change $8$ litres to $\text{cm}^3$. Volume per second $= \pi r^2 \times$ speed. Then divide.
$8 \text{ litres} = 8000 \text{ cm}^3$
$V \text{ per second} = \pi(4)^2(10) = 160\pi$
Time $= \dfrac{8000}{160\pi}$
Time $= 15.9 \text{ seconds}$
Time $= 15.9 \text{ seconds}$
SUM
The lot in one box
Volume 2 toolkit
1.Double shape: add the volumes; add only the surfaces you can see.
2.Cone area needs slant $l$, not height: $l^2 = r^2 + h^2$, then $A = \pi r l$.
3.Recast: volume stays the same — set $V_{\text{old}} = V_{\text{new}}$.
4."How many?": divide volumes, round down.
5.Empty space $=$ container $-$ solid; $\% = \dfrac{\text{space}}{\text{container}} \times 100$.
6.Displacement: risen water is a cylinder with the same volume as the object.
7.Rate of flow: volume per second $= \pi r^2 \times \text{speed}$; time $= \dfrac{\text{total}}{\text{per second}}$.
8.Units: $1 \text{ litre} = 1000 \text{ cm}^3$; $1\,\text{m}^3 = 1\,000\,000\,\text{cm}^3$.
End of lesson
Volume 2 — HL · Mathslive.ie