Chapter 12 · Surface Areas and Volumes

12.2 Volume of a Combination of Solids

How much ice cream fits in the cone, how much fuel in the tank, how much wood is left after a hollow is drilled? For volumes, joined solids simply add, and a scooped-out hollow simply subtracts.

A tanker carries fuel in a cylinder with a hemisphere at each end. A rocket model holds clay in a cylinder and a cone. An ice-cream seller wants to know how much ice cream sits in and on a cone. All three questions ask for the volume of a combination of two solids. In Section 12.1 joining two solids removed area from the outside, so we had to be careful. Volume is kinder: gluing two solids together does not change the space they fill, so the volume of the combination is just the sum of the two volumes. If one solid is scooped out of another, the volume left is the difference.

We use $\pi=\tfrac{22}{7}$ when the radius is a multiple of $7$ (or $3.5$, $10.5$) and $\pi=3.14$ otherwise, and we always say which. Answers are given to two decimal places with units (cm³, m³, litres).

Volumes of the basic solids

📘 Definition 12.3 — Volume and capacity
The volume of a solid is the amount of space it occupies, measured in cubic units (cm³, m³). The capacity of a hollow container is the volume of liquid or grain it can hold. Capacity is often given in litres, where $1\ \text{L}=1000\ \text{cm}^3$ and $1\ \text{m}^3=1000\ \text{L}$.
📐 Theorem 12.4 — Volumes of the basic solids
SolidVolume
Cuboid $l\times b\times h$$lbh$
Cube of edge $a$$a^3$
Cylinder, radius $r$, height $h$$\pi r^2h$
Cone, radius $r$, height $h$$\tfrac13\pi r^2h$
Sphere, radius $r$$\tfrac43\pi r^3$
Hemisphere, radius $r$$\tfrac23\pi r^3$

Reading the table. The volume of a cone is exactly one third of that of a cylinder with the same base and height; you can check it by filling a cone with sand and emptying it three times into a cylinder of the same size. The hemisphere is half the sphere. Use the height $h$ here, not the slant height $l$: the slant height appears only in surface areas.

Take a cylinder, a cone and a hemisphere with the same radius $r$, and let the cylinder and the cone also have height $r$. Their volumes are $\pi r^3$, $\tfrac13\pi r^3$ and $\tfrac23\pi r^3$. This gives a neat fact that is also handy for checking answers (Interactive 2).

📐 Theorem 12.5 — Cone, hemisphere and cylinder
A cone and a cylinder of radius $r$ and height $r$ and a hemisphere of radius $r$ have volumes in the ratio $$\text{cone}:\text{hemisphere}:\text{cylinder}=\tfrac13\pi r^3:\tfrac23\pi r^3:\pi r^3=1:2:3.$$ In particular, the contents of one cone and one hemisphere fill the cylinder exactly.

Joining and scooping: add and subtract

📐 Theorem 12.6 — Volume of a combination
(a) Two solids joined face to face, without overlapping: $V=V_1+V_2$.
(b) A solid scooped out of another (the hollow lies completely inside the first solid): $V=V_{\text{original}}-V_{\text{hollow}}$.

Reason. Joining two solids does not make them occupy more or less room: the space of the combination is the space of the first plus the space of the second, which share only a surface and so no volume. In (b) the hollow is a part of the space the original solid used to occupy, so we take its volume away. Nothing disappears from a volume when two solids touch: this is exactly where volume differs from surface area. Compare the capsule: its surface area is $2\pi rh+4\pi r^2$ (hidden circles removed), but its volume is $\pi r^2h+\tfrac43\pi r^3$, simply the cylinder plus two hemispheres.

🧭 Problem-solving strategy — volume of a combination
  1. Sketch the solid, name its two parts and mark every radius and height. Find a missing height by subtraction (total height $-$ radius of a hemisphere, total length $-2r$ for a capsule).
  2. Decide: do the parts add (joined) or does one subtract (scooped out)?
  3. Write each volume with its formula and the stated value of $\pi$.
  4. Add or subtract. Give the answer to two decimal places in cm³ or m³.
  5. If a capacity in litres, a mass (volume $\times$ density) or a cost is asked, convert or multiply as the last step.
⚠️ Three mistakes that cost marks
1. Using the slant height (or the total height of the toy) in $\tfrac13\pi r^2h$: the cone's own height is the total height minus the hemisphere's radius. 2. Writing $\tfrac43\pi r^3$ for a hemisphere: the sphere has $\tfrac43\pi r^3$, the hemisphere half of it. 3. Subtracting the volumes when the solids are joined, or adding them when one is scooped out. Two hemispheres on a cylinder make one sphere's worth of volume.
✏️ Example 1 — An ice-cream cone
An ice-cream cone has radius $3.5$ cm and height $12$ cm. It is filled with ice cream, and a hemisphere of ice cream of the same radius sits on top. Find the total volume of ice cream. (Use $\pi=\tfrac{22}{7}$.)
Parts A cone ($r=3.5$ cm, $h=12$ cm) and a hemisphere ($r=3.5$ cm), joined, so the volumes add.
Cone $\tfrac13\pi r^2h=\tfrac13\times\tfrac{22}{7}\times12.25\times12=154$ cm³.
Hemisphere $\tfrac23\pi r^3=\tfrac23\times\tfrac{22}{7}\times42.875=89.83$ cm³.
Total $154+89.83=243.83$ cm³.
Check Almost a quarter of a litre: $243.83$ cm³ $=0.24383$ L. The cone alone ($154$) is more than the dome ($89.83$), which is plausible because the cone's height $12$ cm is more than the dome's diameter $7$ cm (the cone is bigger than the dome exactly when $h\gt2r$).
✏️ Example 2 — Capacity of the fuel tank
The fuel tank of Section 12.1 is a cylinder of radius $0.7$ m and total length $5.4$ m with a hemisphere at each end. Find its capacity in litres. (Use $\pi=\tfrac{22}{7}$.)
Parts Cylinder height $h=5.4-2\times0.7=4$ m, and two hemispheres which together make one sphere of radius $0.7$ m.
Cylinder $\pi r^2h=\tfrac{22}{7}\times0.49\times4=6.16$ m³.
Two hemispheres $\tfrac43\pi r^3=\tfrac43\times\tfrac{22}{7}\times0.343=1.44$ m³ (more precisely $1.4373$).
Total $6.16+1.4373=7.5973$ m³ $\approx7.60$ m³. In litres: $7.5973\times1000\approx7597.33$ L.
Check The cylinder is by far the larger part, so the total is a little more than $6.16$ m³ ✓.
Interactive 1 — Add It or Scoop It: Volume Builder

Choose a solid. For the ice-cream cone and the capsule the two volumes are added (red + blue). For the cylinder with a conical hollow and the cube with a hemispherical hollow the blue hollow is subtracted from the red solid, leaving the green volume. Change $r$ and $h$ with the sliders (or press Play) and switch $\pi$ between $\tfrac{22}{7}$ and $3.14$. The bar on the right shows the parts to scale.

3.58.0
part 1—
part 2—
volume of the solid (cm³)—
capacity in litres—
✏️ Example 3 — Air inside a tent
A tent has a cylindrical wall of radius $6$ m and height $4$ m and a conical roof of the same radius and height $8$ m (slant height $10$ m). Find the volume of air inside the tent. (Use $\pi=3.14$.)
Parts A cylinder ($r=6$, $h=4$) and a cone ($r=6$, $h=8$). The volume uses the height $8$ m of the cone, not the slant height $10$ m.
Cylinder $\pi r^2h=3.14\times36\times4=452.16$ m³.
Cone $\tfrac13\pi r^2h=\tfrac13\times3.14\times36\times8=301.44$ m³.
Total $452.16+301.44=753.60$ m³.
Check Factor: $\pi r^2\left(h_1+\tfrac13h_2\right)=3.14\times36\times\left(4+\tfrac83\right)=753.6$ ✓.
✏️ Example 4 — A wooden cylinder with a conical hollow
A solid wooden cylinder of radius $7$ cm and height $24$ cm has a conical cavity of the same base radius and height drilled into it (Example 5 of Section 12.1). Find the volume of wood left and its mass if the wood has density $0.8$ g/cm³. (Use $\pi=\tfrac{22}{7}$.)
Parts The cone is scooped out of the cylinder, so subtract: $V=V_{\text{cylinder}}-V_{\text{cone}}$.
Cylinder $\pi r^2h=\tfrac{22}{7}\times49\times24=3696$ cm³.
Cone $\tfrac13\pi r^2h=\tfrac13\times3696=1232$ cm³.
Wood left $3696-1232=2464$ cm³, which is $\tfrac23$ of the cylinder. Mass $=2464\times0.8=1971.2$ g.
Check Section 12.1 gave the surface area $1760$ cm² of the same solid. The areas and the volume are different questions: the hollow adds surface but removes volume.
✏️ Example 5 — Capsule-shaped sweets in syrup
A sweet is shaped like a cylinder with a hemisphere at each end. Its diameter is $3.5$ cm and its total length is $6.5$ cm. Twelve such sweets are soaked in syrup, each absorbing syrup equal to $30\%$ of its own volume. Find the volume of syrup absorbed in all. (Use $\pi=\tfrac{22}{7}$.)
Figure $r=1.75$ cm, cylinder height $h=6.5-3.5=3$ cm.
One sweet Cylinder $\pi r^2h=\tfrac{22}{7}\times3.0625\times3=28.875$ cm³. Two hemispheres $\tfrac43\pi r^3=\tfrac43\times\tfrac{22}{7}\times5.359375=22.458$ cm³. Total $51.333$ cm³.
Twelve sweets $12\times51.333=616$ cm³.
Syrup $30\%$ of $616=0.3\times616=184.8$ cm³.
✏️ Example 6 — A cube with a hemispherical hollow
A cube of edge $14$ cm has a hemisphere of radius $7$ cm scooped out of the middle of one face. Find the volume of the remaining solid. (Use $\pi=\tfrac{22}{7}$.)
Parts Cube $a^3$, minus a hemisphere $\tfrac23\pi r^3$ with $r=7$ cm.
Cube $14^3=2744$ cm³.
Hemisphere $\tfrac23\times\tfrac{22}{7}\times343=\tfrac23\times1078=718.67$ cm³.
Remaining $2744-718.67=2025.33$ cm³.
Check The hemisphere takes about a quarter of the cube, so about $\tfrac34$ remains; $\tfrac34\times2744=2058$, close to $2025.33$ ✓.
Interactive 2 — Pour a Cone, Then a Hemisphere, Into a Cylinder

A cone, a hemisphere and a cylinder all have radius $r$, and the cone and the cylinder have height $r$. Move the pour slider (or press Play) to empty the cone and then the hemisphere into the cylinder. The liquid reaches $\tfrac13$ of the height after the cone and the top after the hemisphere: volumes $1:2:3$. Change $r$ and the ratio stays the same. Here $\pi=3.14$.

3.0
cone $\tfrac13\pi r^3$ (cm³)—
hemisphere $\tfrac23\pi r^3$ (cm³)—
cylinder $\pi r^3$ (cm³)—
liquid in cylinder (cm³)—

Exercises

I
The basic solids. State the value of $\pi$ you use ($\tfrac{22}{7}$ unless told otherwise); give units.
  • 1 Find the volume of a cone of radius $7$ cm and height $12$ cm.
  • 2 Find the volume of a cylinder of radius $3.5$ m and height $10$ m.
  • 3 Find the volume of a sphere of radius $10.5$ cm.
  • 4 Find the volume of a hemisphere of radius $21$ cm.
  • 5 Find the volume of (a) a cuboid $5$ m $\times$ $4$ m $\times$ $3$ m; (b) a cube of edge $7$ cm.
  • 6 A cone of base radius $7$ cm has volume $1232$ cm³. Find its height.
  • 7 The volume of a hemisphere is $144\pi$ cm³. Find its radius.
  • 8 A cylinder and a cone have the same radius $6$ cm and the same height $10$ cm. Find their volumes ($\pi=3.14$) and compare them.
1 $\tfrac13\times\tfrac{22}{7}\times49\times12=616$ cm³.
2 $\tfrac{22}{7}\times12.25\times10=385$ m³.
3 $\tfrac43\times\tfrac{22}{7}\times10.5^3=\tfrac43\times\tfrac{22}{7}\times1157.625=4851$ cm³.
4 $\tfrac23\times\tfrac{22}{7}\times9261=19404$ cm³.
5 (a) $5\times4\times3=60$ m³. (b) $7^3=343$ cm³.
6 $\tfrac13\times\tfrac{22}{7}\times49\times h=1232$ gives $\dfrac{154}{3}h=1232$, so $h=\dfrac{3696}{154}=24$ cm.
7 $\tfrac23\pi r^3=144\pi$ gives $r^3=216$, so $r=6$ cm.
8 Cylinder $3.14\times36\times10=1130.40$ cm³; cone $\tfrac13\times1130.40=376.80$ cm³. The cylinder holds three times as much.
II
Combinations: add when joined, subtract when scooped out.
  • 1 A toy is a cone of radius $7$ cm and height $24$ cm on a hemisphere of radius $7$ cm. Find its volume. (Use $\pi=\tfrac{22}{7}$.)
  • 2 A solid is a cylinder with a hemisphere at each end, radius $3.5$ cm and total length $20$ cm. Find its volume. (Use $\pi=\tfrac{22}{7}$.)
  • 3 A cube of edge $10$ cm has a hemisphere of radius $5$ cm scooped out of one face. Find the volume left. (Use $\pi=3.14$.)
  • 4 A solid cylinder of radius $7$ cm and height $10$ cm has a hemispherical hollow of radius $7$ cm scooped from its top. Find the volume left. (Use $\pi=\tfrac{22}{7}$.)
  • 5 A model rocket is a cylinder of radius $3$ cm and height $10$ cm with a cone of the same radius and height $4$ cm on top. Find its volume. (Use $\pi=3.14$.)
  • 6 A cuboid $10$ cm $\times$ $10$ cm $\times$ $5$ cm carries a cylinder of radius $3$ cm and height $8$ cm on its top face. Find the total volume. (Use $\pi=3.14$.)
  • 7 A hemispherical bowl of inner radius $10.5$ cm is full of soup. Find the volume of soup in litres, correct to two decimal places. (Use $\pi=\tfrac{22}{7}$.)
  • 8 A stand is a cuboid $14$ cm $\times$ $14$ cm $\times$ $4$ cm with a cylinder of radius $3.5$ cm and height $10$ cm standing on it. Find the total volume. (Use $\pi=\tfrac{22}{7}$.)
1 Cone $\tfrac13\times\tfrac{22}{7}\times49\times24=1232$; hemisphere $\tfrac23\times\tfrac{22}{7}\times343=718.67$. Total $1950.67$ cm³.
2 $h=20-7=13$ cm. Cylinder $\tfrac{22}{7}\times12.25\times13=500.5$; two hemispheres $\tfrac43\times\tfrac{22}{7}\times42.875=179.67$. Total $680.17$ cm³.
3 $1000-\tfrac23\times3.14\times125=1000-261.67=738.33$ cm³.
4 $\tfrac{22}{7}\times49\times10-\tfrac23\times\tfrac{22}{7}\times343=1540-718.67=821.33$ cm³.
5 Cylinder $3.14\times9\times10=282.6$; cone $\tfrac13\times3.14\times9\times4=37.68$. Total $320.28$ cm³.
6 Cuboid $500$; cylinder $3.14\times9\times8=226.08$. Total $726.08$ cm³.
7 $\tfrac23\times\tfrac{22}{7}\times1157.625=2425.5$ cm³ $=2.4255$ L $\approx2.43$ L.
8 Cuboid $14\times14\times4=784$; cylinder $\tfrac{22}{7}\times12.25\times10=385$. Total $1169$ cm³.
III
Applications.
  • 1 A tank is a cylinder of radius $1.4$ m with a hemisphere at each end; its total length is $6.8$ m. Find its capacity in cubic metres and in litres. (Use $\pi=\tfrac{22}{7}$.)
  • 2 A grain store is a cylinder of radius $3.5$ m and height $6$ m with a conical roof of height $12$ m on top. Find the volume of grain it can hold. (Use $\pi=\tfrac{22}{7}$.)
  • 3 An ice-cream cone has radius $3$ cm and height $10$ cm and carries a hemisphere of ice cream of radius $3$ cm. Find the cost of ice cream for $8$ cones at ₹ $0.40$ per cm³. (Use $\pi=3.14$.)
  • 4 A test tube is a cylinder of radius $1$ cm and length $12$ cm with a hemispherical bottom. Find its capacity. (Use $\pi=3.14$.)
  • 5 A paperweight is a cube of edge $8$ cm with a hemisphere of radius $4$ cm on top. Find its volume. (Use $\pi=3.14$.)
  • 6 A pencil holder is a wooden cylinder of radius $3.5$ cm and height $12$ cm with a conical hollow of the same base and height. Find the volume of wood and its mass if the density is $0.7$ g/cm³. (Use $\pi=\tfrac{22}{7}$.)
1 $h=6.8-2.8=4$ m. Cylinder $\tfrac{22}{7}\times1.96\times4=24.64$; sphere $\tfrac43\times\tfrac{22}{7}\times2.744=11.4987$. Total $36.14$ m³ $=36{,}138.67\approx36{,}139$ L.
2 Cylinder $\tfrac{22}{7}\times12.25\times6=231$; cone $\tfrac13\times\tfrac{22}{7}\times12.25\times12=154$. Total $385$ m³.
3 Cone $\tfrac13\times3.14\times9\times10=94.2$; hemisphere $\tfrac23\times3.14\times27=56.52$. One: $150.72$ cm³; eight: $1205.76$ cm³. Cost $1205.76\times0.40=$ ₹ $482.30$.
4 Cylinder $3.14\times1\times12=37.68$; hemisphere $\tfrac23\times3.14=2.0933$. Total $39.77$ cm³.
5 $512+\tfrac23\times3.14\times64=512+133.97=645.97$ cm³.
6 Cylinder $\tfrac{22}{7}\times12.25\times12=462$; cone $154$. Wood $=308$ cm³; mass $=308\times0.7=215.6$ g.
IV
Show that…
  • 1 A capsule is a cylinder with a hemisphere at each end, radius $r$, total length $L$. Show that its volume is $\pi r^2L-\tfrac23\pi r^3$, and check this for $r=3$ cm, $L=14$ cm with $\pi=3.14$.
  • 2 A cone is hollowed out of a cylinder with the same base and height. Show that the remaining solid has $\tfrac23$ of the volume of the cylinder, and find it for $r=3.5$ cm, $h=6$ cm ($\pi=\tfrac{22}{7}$).
  • 3 A hemisphere of radius $6$ cm is fixed on a cylinder of the same radius. The total height of the solid is $20$ cm. Find its volume. (Use $\pi=3.14$.)
  • 4 A cone and a hemisphere have the same radius $r$. Show that they have equal volumes when the cone's height is $2r$; find that height for $r=4.2$ cm.
1 $h=L-2r$. $V=\pi r^2(L-2r)+\tfrac43\pi r^3=\pi r^2L-2\pi r^3+\tfrac43\pi r^3=\pi r^2L-\tfrac23\pi r^3$. Check: $h=8$; $3.14\times9\times8+\tfrac43\times3.14\times27=226.08+113.04=339.12$; formula $3.14\times9\times14-\tfrac23\times3.14\times27=395.64-56.52=339.12$ ✓.
2 $V=\pi r^2h-\tfrac13\pi r^2h=\tfrac23\pi r^2h$. For $r=3.5$, $h=6$: cylinder $\tfrac{22}{7}\times12.25\times6=231$, so the remaining volume is $\tfrac23\times231=154$ cm³.
3 $h=20-6=14$ cm. Cylinder $3.14\times36\times14=1582.56$; hemisphere $\tfrac23\times3.14\times216=452.16$. Total $2034.72$ cm³.
4 $\tfrac13\pi r^2h=\tfrac23\pi r^3$ gives $h=2r$. For $r=4.2$ cm: $h=8.4$ cm. Check: $\tfrac13\times17.64\times8.4=49.392=\tfrac23\times74.088$ ✓ (each times $\pi$).
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marksBoard · Case study

How this is examined: volume questions on combinations carry $2$ to $5$ marks. The marks go to the correct split of the solid into two parts, the right height in each formula, the sum or difference, and the answer with units; a litre conversion, a mass or a cost is often the last step. MCQs test whether you use $\tfrac13$ for the cone and $\tfrac23$ for the hemisphere. State the value of $\pi$ you use.

E1. Board · 1 mark A cone of radius $r$ and height $r$ stands on a hemisphere of radius $r$. The volume of the solid is  (A) $\pi r^3$  (B) $\tfrac23\pi r^3$  (C) $\tfrac53\pi r^3$  (D) $\tfrac13\pi r^3$

Solve Joined, so add: $\tfrac13\pi r^2\cdot r+\tfrac23\pi r^3=\pi r^3$.
Trap (C) $\tfrac53\pi r^3$ uses the volume of a whole sphere $\tfrac43\pi r^3$; (B) is the hemisphere alone and (D) the cone alone.
Answer (A).

E2. Board · 1 mark Assertion (A): The volume of a toy made by joining a cone to a hemisphere is the sum of the volumes of the two solids. Reason (R): The total surface area of the toy is the sum of the total surface areas of the two solids.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check A Joining does not change the space occupied, so volumes add: true.
Check R The joined faces are hidden, so $\text{TSA}=\text{TSA}_1+\text{TSA}_2-2A_c$; the areas do not simply add. R is false.
Answer (C).

E3. Board · 2 marks A hemisphere of radius $3.5$ cm is scooped out of a cube of edge $7$ cm. Find the volume of the remaining solid. (Use $\pi=\tfrac{22}{7}$.)

Parts (1 mark) Subtract: cube $7^3=343$ cm³; hemisphere $\tfrac23\times\tfrac{22}{7}\times3.5^3=\tfrac23\times\tfrac{22}{7}\times42.875=89.83$ cm³.
Answer (1 mark) $343-89.83=253.17$ cm³.

E4. Board · 3 marks A solid is a cylinder of radius $4$ cm and height $10$ cm with a cone of the same radius and height $6$ cm on one end. Find its volume. (Use $\pi=3.14$.)

Cylinder (1 mark) $\pi r^2h=3.14\times16\times10=502.4$ cm³.
Cone (1 mark) $\tfrac13\pi r^2h=\tfrac13\times3.14\times16\times6=100.48$ cm³.
Total (1 mark) $502.4+100.48=602.88$ cm³.

E5. Board · 5 marks A vessel is a hollow cylinder of radius $10.5$ cm surmounted by a hollow hemisphere of the same radius; the total height of the vessel is $25.5$ cm. Find its capacity in litres and the cost of the milk that fills it at ₹ $60$ per litre. (Use $\pi=\tfrac{22}{7}$.)

Figure (1 mark) $r=10.5$ cm; cylinder height $h=25.5-10.5=15$ cm.
Cylinder (1 mark) $\tfrac{22}{7}\times110.25\times15=5197.5$ cm³.
Hemisphere (1 mark) $\tfrac23\times\tfrac{22}{7}\times1157.625=2425.5$ cm³.
Capacity (1 mark) $5197.5+2425.5=7623$ cm³ $=7.623$ L.
Cost (1 mark) $7.623\times60=457.38$, so ₹ $457.38$.

E6. Board · Case study A student models a rocket in clay: a solid cylinder of radius $2.1$ cm and height $10$ cm with a solid cone of the same radius and height $5$ cm on top. (Use $\pi=\tfrac{22}{7}$.)
(i) Find the volume of the cylindrical part and of the conical part. (2 marks)
(ii) Find the volume of the whole model. (1 mark)
(iii) Clay has density $1.8$ g/cm³. Find the mass of the model. (1 mark)

(i) Cylinder $\tfrac{22}{7}\times4.41\times10=138.6$ cm³; cone $\tfrac13\times\tfrac{22}{7}\times4.41\times5=23.1$ cm³. (1 + 1 marks)
(ii) $138.6+23.1=161.7$ cm³. (1 mark)
(iii) Mass $=161.7\times1.8=291.06$ g. (1 mark)
🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. How close does a volume calculated from a few measurements come to the volume you measure by pouring water — for a glass, and for a round bowl?

Why it matters. Water tanks, fuel tankers, bottles and measuring spoons are designed from a plan: an engineer works out the capacity from the dimensions before anything is built, and the factory later checks it by filling. Differences show where a real object departs from the ideal solid, for example because of rounded corners or wall thickness.

What to do.

  1. Choose a straight-sided glass or steel tumbler. Measure the inner diameter and the inner height in centimetres. Compute the volume $\pi r^2h$ with $\pi=3.14$.
  2. Fill it to the brim with water and pour the water into a measuring jug. Read the volume in millilitres ($1$ mL $=1$ cm³).
  3. Choose a round-bottomed bowl whose depth is about its radius. Measure the inner rim diameter and compute $\tfrac23\pi r^3$. Measure its capacity by pouring water as before.
  4. Calculate the percentage difference between computed and measured volume for each vessel.
  5. Calculate the total volume of the tumbler and the bowl together, and check this by filling both and pouring the water into one large jug.

Tools: A tumbler, a bowl, a ruler or tape, a measuring jug (mL), a calculator

Hand in: A table of measured sizes, computed and poured volumes, the percentage difference, and a paragraph explaining which vessel was closer to its ideal solid and why.

Going further. Find an object that really is a combination (a pencil-shaped bottle, a rounded test tube or a flask with a rounded bottom), split it into a cylinder and a hemisphere, predict its capacity and test it.

Teacher/parent note: measuring the inside of the vessel matters. Wall thickness, a slightly sloping side or a rim that is not exactly level are good conversation topics about sources of error. The bowl will rarely be an exact hemisphere, so a $5$–$10\%$ difference is normal.

🧪 Self-check quiz

0/5 correct

Q1. The volume of a cone of radius $7$ cm and height $12$ cm (with $\pi=\tfrac{22}{7}$) is

Q2. The volume of a hemisphere of radius $6$ cm is

Q3. A capsule is a cylinder of radius $7$ cm and height $10$ cm with a hemisphere at each end. Its volume ($\pi=\tfrac{22}{7}$) is

Q4. A hemispherical hollow of radius $7$ cm is scooped out of the top of a solid cylinder of radius $7$ cm and height $10$ cm. The volume left ($\pi=\tfrac{22}{7}$) is

Q5. A cone, a hemisphere and a cylinder have the same radius $r$, and the cone and the cylinder have height $r$. Their volumes are in the ratio

← 12.1 Surface Area of a Combination of SolidsChapter 13 — Statistics →