Garden Shed: Prism with a Pyramid Roof

A garden shed has a box-shaped base that is $4$ m long, $3$ m wide and $2.5$ m high. On top sits a roof shaped like a pyramid with the same $4$ m by $3$ m rectangular base and a height of $1.5$ m. What is the total volume of the shed?

Name the parts: Rectangular prism ($4 \times 3 \times 2.5$) + pyramid (base $4 \times 3$, height $1.5$) = Two parts, joined

Volume of the prism: $V_1 = 4 \times 3 \times 2.5 = 30$ = $30$ m³

Base area of the pyramid: $B = 4 \times 3 = 12$ = $12$ m²

Volume of the pyramid: $V_2 = \frac{1}{3} \times 12 \times 1.5 = 6$ = $6$ m³

Add the parts: $V = 30 + 6 = 36$ = $36$ m³

Answer: The shed has a volume of $36$ m³.

Sharpened Pencil: Cylinder with a Cone

A model pencil is a cylinder with radius $3$ cm and height $10$ cm. A cone with the same radius $3$ cm and height $4$ cm is attached to one end. Find the volume. Use $\pi \approx 3.14$ and round to the nearest tenth.

Name the parts: Cylinder ($r = 3$, $h = 10$) + cone ($r = 3$, $h = 4$). The total length $14$ cm is NOT used in either formula. = Two parts, joined

Cylinder in terms of $\pi$: $V_1 = \pi \times 3^2 \times 10 = 90\pi$ = $90\pi$ cm³

Cone in terms of $\pi$: $V_2 = \frac{1}{3} \times \pi \times 3^2 \times 4 = 12\pi$ = $12\pi$ cm³

Add, then multiply by $3.14$ once: $90\pi + 12\pi = 102\pi \approx 102 \times 3.14 = 320.28$ = $\approx 320.3$ cm³

Answer: The pencil has a volume of $102\pi \approx 320.3$ cm³.

Grain Silo: Cylinder with a Hemisphere

A grain silo is a cylinder with radius $2$ m and height $6$ m. Its roof is a hemisphere (half sphere) with the same radius $2$ m. How much grain can the silo hold when full? Use $\pi \approx 3.14$ and round to the nearest tenth.

Name the parts: Cylinder ($r = 2$, $h = 6$) + hemisphere ($r = 2$) = Two parts, joined

Cylinder: $V_1 = \pi \times 2^2 \times 6 = 24\pi$ = $24\pi$ m³

Hemisphere: half of a sphere: $V_2 = \frac{2}{3} \times \pi \times 2^3 = \frac{16}{3}\pi$ = $\frac{16}{3}\pi$ m³

Add in terms of $\pi$: $24\pi + \frac{16}{3}\pi = \frac{72}{3}\pi + \frac{16}{3}\pi = \frac{88}{3}\pi$ = $\frac{88}{3}\pi$ m³

Use $\pi \approx 3.14$ and round at the end: $\frac{88 \times 3.14}{3} = \frac{276.32}{3} \approx 92.11$ = $\approx 92.1$ m³

Answer: The silo holds about $92.1$ m³ of grain.

Ice Cream Cone with a Scoop

A cone of height $10$ cm and radius $3$ cm is filled with ice cream, and a hemisphere-shaped scoop of radius $3$ cm sits on top. How many milliliters of ice cream is that? Use $\pi \approx 3.14$ and round to the nearest whole number.

Name the parts: Cone ($r = 3$, $h = 10$) + hemisphere ($r = 3$) = Two parts, joined

Cone: $V_1 = \frac{1}{3} \times \pi \times 3^2 \times 10 = 30\pi$ = $30\pi$ cm³

Hemisphere: $V_2 = \frac{2}{3} \times \pi \times 3^3 = 18\pi$ = $18\pi$ cm³

Add and use $3.14$: $30\pi + 18\pi = 48\pi \approx 48 \times 3.14 = 150.72$ = $\approx 150.72$ cm³

Convert to milliliters: $1$ cm³ $= 1$ mL, so $150.72$ cm³ $= 150.72$ mL = $\approx 151$ mL

Answer: There are about $151$ mL of ice cream.

Hollow Pipe: Subtracting a Cylinder

A concrete pipe is $100$ cm long. Its outer radius is $5$ cm and the hole through the middle has radius $4$ cm. What volume of concrete is in the pipe? Use $\pi \approx 3.14$.

Describe the solid: Big cylinder ($r = 5$, $h = 100$) with a smaller cylinder ($r = 4$, $h = 100$) removed = Subtract

Outer cylinder: $V_{\text{outer}} = \pi \times 5^2 \times 100 = 2500\pi$ = $2500\pi$ cm³

Hole (inner cylinder): $V_{\text{hole}} = \pi \times 4^2 \times 100 = 1600\pi$ = $1600\pi$ cm³

Subtract and use $3.14$: $2500\pi - 1600\pi = 900\pi \approx 900 \times 3.14 = 2826$ = $2826$ cm³

Answer: The pipe contains about $2826$ cm³ of concrete, which is about $2.8$ L.

Mistake: Using the total height for every part (a pencil $14$ cm long treated as a cylinder of height $14$ plus a cone of height $4$)

Why: 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 $h = 10$, cone $h = 4$. Check that the heights of the parts add up to the total: $10 + 4 = 14$.

Mistake: Using the full sphere formula for a dome or scoop

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

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

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

Why: Volume depends on $r^2$, so you cannot subtract the radii before squaring. The difference of the volumes is not the volume of the difference.

Correct: Subtract whole volumes: $\pi \times 5^2 \times 100 - \pi \times 4^2 \times 100 = \pi (25 - 16) \times 100 = 900\pi$.

Mistake: Rounding each part before adding

Why: 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 $\pi$ (or keep all decimals), add or subtract, and round once at the end to the place the question asks for.

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.

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

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.

A block $8$ cm by $8$ cm by $5$ cm with a cylindrical hole of radius $1$ cm through its height uses $320 - \pi \times 1^2 \times 5 \approx 304.3$ cm³ of material.

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.

A rocket with radius $4$ cm, a $15$ cm cylinder and a $6$ cm nose cone has volume $240\pi + 32\pi = 272\pi \approx 854.08$ cm³.

Q: How do I know whether to add or subtract?

A: 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.

Q: There is no picture. How do I imagine the shape?

A: Read the description slowly and make a quick sketch. Write each part with its own numbers, for example: cylinder $r = 2$, $h = 6$; hemisphere $r = 2$. Once the parts are listed, the problem becomes two small problems.

Q: Should I use $\pi \approx 3.14$ or the $\pi$ button?

A: Use what the question says. This lesson uses $3.14$. The $\pi$ button gives a slightly bigger, more exact answer (for example $102\pi$ is $320.28$ with $3.14$ and $320.44$ with the button). If a question says leave the answer in terms of $\pi$, write something like $102\pi$.

Q: Do I count the flat face where the two parts touch?

A: 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).