Volume: Cones, Cylinders, Spheres

8.H.1Cylinders01/04

Volume of cylinders.

Step01of 04

A cylinder has two circular bases connected by a curved surface. Volume = base area times height.

V = π r^{2} h

The $π r^{2}$ is the area of one circular base; $h$ is the perpendicular distance between the two bases.

8.H.1 — Volume of cylinders.

Step 01 of 04

A cylinder has two circular bases connected by a curved surface. Volume = base area times height.

V = π r^{2} h

The $π r^{2}$ is the area of one circular base; $h$ is the perpendicular distance between the two bases.

Step 02 of 04

Worked example. Cylinder with radius 4, height 10.

V = π (4)^{2} (10) = 160 π \approx 502.7 cubic units

Leave answers in terms of $π$ when possible (more exact). Use $π \approx 3.14$ only when an approximation is requested.

Step 03 of 04

Watch units carefully. Volume is in CUBIC units. If radius and height are in feet, volume is in cubic feet.

Cylinder with diameter 6, height 9 — first halve the diameter: $r = 3$ .

V = π (3)^{2} (9) = 81 π \approx 254.5

Step 04 of 04

Real-world example. A water tank is a cylinder, 4 ft radius and 10 ft tall. How many cubic feet of water does it hold?

V = π (4)^{2} (10) = 160 π \approx 502.7 ft^{3}

At about 7.48 gallons per cubic foot, that's roughly 3,760 gallons.

Key insight

Cylinder volume =

π r^{2} h

. Always use radius (halve the diameter if needed). Output in cubic units. Leave answers as exact multiples of

π

unless asked to approximate.

8.H.2 — Volume of cones.

Step 01 of 04

A cone has one circular base and tapers to a single point (the apex). Its volume is exactly ONE-THIRD that of a cylinder with the same base and height.

V = \frac{1}{3} π r^{2} h

The $h$ is the perpendicular height from base to apex (NOT slant height).

Step 02 of 04

Worked example. Cone with radius 6, height 9.

V = \frac{1}{3} π (6)^{2} (9) = \frac{1}{3} (36 π) (9) = 108 π \approx 339.3

Step 03 of 04

The 1/3 factor is geometric, not arithmetic. If you fill a cone with sand and pour it into a cylinder of the same base and height, the cylinder gets exactly 1/3 full. (This is provable but not in 8th grade — accept it as fact for now.)

Step 04 of 04

If given diameter or slant height instead of radius and height: halve the diameter for radius; for slant height, use Pythagorean to recover the perpendicular height.

Cone with radius 3 and slant height 5: perpendicular height $h = 5^{2} - 3^{2} = 4$ . Volume $= \frac{1}{3} π (3)^{2} (4) = 12 π$ .

Key insight

Cone volume =

\frac{1}{3} π r^{2} h

. Exactly 1/3 of a cylinder with the same base and height. Use perpendicular height, not slant; if only slant given, recover height via Pythagorean.

8.H.3 — Volume of spheres.

Step 01 of 04

Sphere volume.

V = \frac{4}{3} π r^{3}

The $r^{3}$ reflects that this is a 3D measurement. The coefficient $\frac{4}{3} π$ is a mathematical constant (proved with calculus).

Step 02 of 04

Worked example. Sphere with radius 3.

V = \frac{4}{3} π (3)^{3} = \frac{4}{3} π (27) = 36 π \approx 113.1

Doubling the radius gives $2^{3} = 8 \times$ the volume:

V_{r = 6} = \frac{4}{3} π (6)^{3} = \frac{4}{3} π (216) = 288 π

Step 03 of 04

Hemisphere = half a sphere.

V_{hemi} = \frac{1}{2} \cdot \frac{4}{3} π r^{3} = \frac{2}{3} π r^{3}

Common in problems that combine shapes — like a silo (cylinder with a hemisphere on top): compute each volume, add them.

Step 04 of 04

Find r given V.

Volume $= 32 π /3$ . Solve:

\frac{32 π}{3} = \frac{4}{3} π r^{3} \Rightarrow r^{3} = 8 \Rightarrow r = 2

Key insight

Sphere:

V = \frac{4}{3} π r^{3}

. Hemisphere: half that. Volume scales as

r^{3}

— doubling the radius gives 8× the volume.

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