Volume of Composite Shapes

Learn how to find the volume of solids built from cubes, prisms, cylinders, cones, spheres and pyramids by adding or subtracting the volumes of their parts.

Advanced30 min18 problems
Practice Problems

Definition

A composite solid (or composite shape) is a 3D object made of two or more simple solids joined together, or a simple solid with a part taken out. A pencil is a cylinder with a cone on the end, a grain silo is a cylinder with a half sphere on top, and a pipe is a cylinder with a smaller cylinder removed.
There is no new formula to learn. You split the solid into parts you already know, find the volume of each part, then combine them:
Vtotal=V1+V2(parts joined together)V_{\text{total}} = V_1 + V_2 \quad \text{(parts joined together)}
V=Vouter−Vremoved(a hole or hollow part)V = V_{\text{outer}} - V_{\text{removed}} \quad \text{(a hole or hollow part)}

Formula recap

  • Cube: V=s3V = s^3
  • Rectangular prism: V=l×w×hV = l \times w \times h
  • Cylinder: V=πr2hV = \pi r^2 h
  • Cone: V=13πr2hV = \frac{1}{3}\pi r^2 h
  • Sphere: V=43πr3V = \frac{4}{3}\pi r^3, so a hemisphere (half sphere) has V=23πr3V = \frac{2}{3}\pi r^3
  • Pyramid: V=13BhV = \frac{1}{3}Bh, where BB is the area of the base

A method that always works

  1. 1.Describe the solid in words (or sketch it): which simple solids, how they are joined, which dimensions belong to which part.
  2. 2.Find the dimensions of each part. Parts often share a radius, but each part has its own height.
  3. 3.Find each volume. Keeping answers in terms of π\pi (like 90π90\pi) makes adding easy.
  4. 4.Add or subtract.
  5. 5.**Use π≈3.14\pi \approx 3.14 and round only at the end**, to the place the question asks for. Write cubic units (cm³, m³). Remember 11 L =1000= 1000 cm³ and 11 m³ =1000= 1000 L.

Try it now

A solid is a cylinder with a cone attached to its top, so the cone's circular base sits exactly on the top of the cylinder. How do you find the volume of the whole solid?

Worked Examples

A garden shed has a box-shaped base that is 44 m long, 33 m wide and 2.52.5 m high. On top sits a roof shaped like a pyramid with the same 44 m by 33 m rectangular base and a height of 1.51.5 m. What is the total volume of the shed?
1

Name the parts

Rectangular prism (4×3×2.54 \times 3 \times 2.5) + pyramid (base 4×34 \times 3, height 1.51.5) → Two parts, joined
2

Volume of the prism

V1=4×3×2.5=30V_1 = 4 \times 3 \times 2.5 = 30 → 3030 m³
3

Base area of the pyramid

B=4×3=12B = 4 \times 3 = 12 → 1212 m²
4

Volume of the pyramid

V2=13×12×1.5=6V_2 = \frac{1}{3} \times 12 \times 1.5 = 6 → 66 m³
5

Add the parts

V=30+6=36V = 30 + 6 = 36 → 3636 m³

Common Mistakes

Using the total height for every part (a pencil 1414 cm long treated as a cylinder of height 1414 plus a cone of height 44)

Why it's wrong: Each part has its own height. The total height belongs to the whole object, not to any single part, so using it counts the cone's part twice.

Correct: Split the height: cylinder h=10h = 10, cone h=4h = 4. Check that the heights of the parts add up to the total: 10+4=1410 + 4 = 14.

Using the full sphere formula for a dome or scoop

Why it's wrong: A dome, a scoop or the round end of a silo is only half a sphere. Using 43πr3\frac{4}{3}\pi r^3 doubles that part.

Correct: A hemisphere is 12×43πr3=23πr3\frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3. Two hemispheres (the two ends of a capsule) make one whole sphere.

Subtracting the radii of a pipe first: π(5−4)2×100=100π\pi (5 - 4)^2 \times 100 = 100\pi

Why it's wrong: Volume depends on r2r^2, so you cannot subtract the radii before squaring. The difference of the volumes is not the volume of the difference.

Correct: Subtract whole volumes: π×52×100−π×42×100=π(25−16)×100=900π\pi \times 5^2 \times 100 - \pi \times 4^2 \times 100 = \pi (25 - 16) \times 100 = 900\pi.

Rounding each part before adding

Why it's wrong: Small rounding errors in each part add up, and the final answer can be off in the last digit.

Correct: Keep each part in terms of π\pi (or keep all decimals), add or subtract, and round once at the end to the place the question asks for.

Interactive Visual

Composite solid

Volume

  • Rectangular Prism: V = lwh = 30 m³
  • + Rectangular Pyramid: V = ⅓lwh = 6 m³

Total volume: 36 m³

The volume of a composite solid is the sum of the volumes of its parts. A hole is subtracted.

Surface Area

  • Rectangular Prism: 59 − 12 (touching faces) = 47 m²
  • + Rectangular Pyramid: 27.99 − 12 (touching faces) ≈ 15.99 m²

Total surface area: 62.99 m²

Surface area counts only the outside: where two parts touch, that face is hidden and is taken away from both parts. A hole takes away its openings and adds its inner wall.

Calculated with π ≈ 3.14.

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Practice Problems

18 problems
Problem 1 of 18
Easy
A solid is a cylinder with a cone attached to its top, so the cone's circular base sits exactly on the top of the cylinder. How do you find the volume of the whole solid?

Why It Matters

Real objects are rarely a single perfect shape:
  • Engineering: Gas tanks and medicine capsules are cylinders with rounded ends (two hemispheres).
  • Farming: A silo is a cylinder with a dome or cone on top, and farmers need its volume to know how much grain fits.
  • Construction: Builders order concrete for blocks and pipes that have holes in them, so they must subtract the empty space.
  • Food: An ice cream cone with a scoop on top, a cake with a hollow center, a pencil-shaped candy.
  • Architecture: A house is a box with a pyramid or prism roof, and heating depends on the volume of air inside.
If you can split a shape into simple parts, you can find the volume of almost anything.

Real World Applications

Gas Tanks and Capsules

Tanks for gas, water or milk are often cylinders with a hemisphere on each end, because round ends handle pressure better than flat ones. The two hemispheres together make one whole sphere.

Example:

A pill capsule with radius 33 mm and a 44 mm long cylinder in the middle has volume π×32×4+43π×33=36π+36π=72π≈226.08\pi \times 3^2 \times 4 + \frac{4}{3}\pi \times 3^3 = 36\pi + 36\pi = 72\pi \approx 226.08 mm³.

1Try It Yourself

A water tank is a cylinder with radius 3030 cm and length 100100 cm, with a hemisphere of radius 3030 cm on each end. Use π≈3.14\pi \approx 3.14.

How many liters of water does the tank hold?

Step 1: Write the mathematical expression

Cylinder plus one whole sphere:

Concrete Planters

Garden planters are often cast from concrete as a block with a hole for the soil. The concrete needed is the volume of the block minus the volume of the hole.

Example:

A block 88 cm by 88 cm by 55 cm with a cylindrical hole of radius 11 cm through its height uses 320−π×12×5≈304.3320 - \pi \times 1^2 \times 5 \approx 304.3 cm³ of material.

2Try It Yourself

A planter is a concrete cube with edges of 5050 cm. A cylinder-shaped hole with radius 2020 cm and depth 4040 cm is left in the top for soil. Use π≈3.14\pi \approx 3.14.

How many cm³ of concrete does the planter need?

Step 1: Write the mathematical expression

Cube minus the cylinder hole:

Model Rockets

A model rocket body is a cylinder with a cone-shaped nose. Designers use the volume to plan how much space is inside for parts and how heavy a solid model will be.

Example:

A rocket with radius 44 cm, a 1515 cm cylinder and a 66 cm nose cone has volume 240π+32π=272π≈854.08240\pi + 32\pi = 272\pi \approx 854.08 cm³.

3Try It Yourself

A toy rocket is a cylinder with radius 22 cm and height 1212 cm, topped by a cone with radius 22 cm and height 66 cm. Use π≈3.14\pi \approx 3.14.

What is the volume of the rocket?

Step 1: Write the mathematical expression

Cylinder plus cone:

Key Takeaways

  • 1A composite solid is made of simple solids joined together, or a simple solid with a part removed
  • 2Split the solid into parts you know: prisms, cubes, cylinders, cones, spheres, hemispheres, pyramids
  • 3Joined parts: add the volumes. Holes and hollow parts: subtract the volume that is removed
  • 4Each part has its own height; parts that touch often share a radius
  • 5A hemisphere has volume 23πr3\frac{2}{3}\pi r^3, and two hemispheres make one sphere
  • 6Keep parts in terms of π\pi, use π≈3.14\pi \approx 3.14 and round only at the end; give cubic units or liters (11 L =1000= 1000 cm³)

Frequently Asked Questions

Ask what the solid is made of. If two pieces are stuck together (a cone on a cylinder), add. If something is cut out or empty (a hole, the inside of a pipe), subtract the empty part from the bigger solid.
Ask what the solid is made of. If two pieces are stuck together (a cone on a cylinder), add. If something is cut out or empty (a hole, the inside of a pipe), subtract the empty part from the bigger solid.
Read the description slowly and make a quick sketch. Write each part with its own numbers, for example: cylinder r=2r = 2, h=6h = 6; hemisphere r=2r = 2. Once the parts are listed, the problem becomes two small problems.
Use what the question says. This lesson uses 3.143.14. The π\pi button gives a slightly bigger, more exact answer (for example 102π102\pi is 320.28320.28 with 3.143.14 and 320.44320.44 with the button). If a question says leave the answer in terms of π\pi, write something like 102π102\pi.
For volume, no. The face where two parts meet has no thickness, so it adds no volume. It only matters for surface area, which is a different lesson (Surface Area of Composite Shapes).

Glossary

Composite solid
A 3D shape made of two or more simple solids, or a simple solid with a part removed
Hemisphere
Half of a sphere, cut through its center; volume 23πr3\frac{2}{3}\pi r^3
Decompose
To split a shape into simpler parts whose volumes you can find
Hollow cylinder
A cylinder with a smaller cylinder removed from its middle, like a pipe or a tube
Capacity
How much a container can hold, often given in liters or milliliters (11 L =1000= 1000 cm³)

Formula Card

Joined parts

Vtotal=V1+V2+…V_{\text{total}} = V_1 + V_2 + \dots

Add the volumes of the parts

Part removed

V=Vouter−VremovedV = V_{\text{outer}} - V_{\text{removed}}

Subtract the volume of the hole or hollow part

Cylinder and cone

V=πr2hV = \pi r^2 h, V=13πr2h\quad V = \frac{1}{3}\pi r^2 h

A cone is one third of the cylinder with the same base and height

Sphere and hemisphere

V=43πr3V = \frac{4}{3}\pi r^3, V=23πr3\quad V = \frac{2}{3}\pi r^3

A hemisphere is half a sphere

Prism and pyramid

V=BhV = Bh, V=13Bh\quad V = \frac{1}{3}Bh

$B$ is the base area; a pyramid is one third of the prism with the same base and height

Hollow cylinder (pipe)

V=π(R2−r2)hV = \pi (R^2 - r^2) h

$R$ is the outer radius, $r$ the inner radius

Today's challengeOne problem, three tries

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