🎯⭐ INTERACTIVE LESSON

Surface Area and Volume of Solids

Learn step-by-step with interactive practice!

← Back to Standard Lesson

Surface Area and Volume of Solids - Complete Interactive Lesson

Part 1: Prisms & Cylinders

📦 Surface Area & Volume of Solids

Part 1 of 5 — Prisms & Cylinders

Topics in This Part

Section
Surface Area vs. Volume — the big distinction
Prisms: $V = Bh$
Cylinders: $V = \pi r^2 h$
Surface area by "unfolding" the net

🔑 Key Concept: Volume measures the space inside a solid (cubic units). Surface area measures the skin around it (square units). Prisms and cylinders share one idea: a base shape stretched straight up through a height $h$ .

Surface Area vs. Volume

These two ideas are easy to mix up, so anchor them now:

	Volume	Surface Area
Measures	space inside	"skin" outside
Units	cubic ( $\text{cm}^3$ , $\text{in}^3$ )	square (, )

Concept Check 🎯

Worked Example — Rectangular Prism

A box measures $\ell = 5\text{ cm}$ , $w = 4\text{ cm}$ , $h = 6\text{ cm}$ .

Volume. The base is a rectangle, so :

Cylinders

A cylinder's base is a circle, so $B = \pi r^2$ and:

$V = \pi r^2 h$

For surface area, unroll the cylinder. You get two circles (top and bottom) plus (the curved side). The rectangle's width is the base circumference and its height is :

Compute It 🧮

Find each value. Leave $\pi$ answers as a number times $\pi$ — just enter the coefficient of $\pi$ where asked.

1) Rectangular prism $3 \times 4 \times 5$ : volume (cubic units) Cylinder , : volume (enter the coefficient) Cylinder , : surface area (enter the coefficient)

One Last Habit — Units

Every number you report should carry the right kind of unit. A quick way to self-check: count the dimensions you multiplied.

Multiplied three lengths (like $\ell \cdot w \cdot h$ )? → cubic units → it's a volume.
Multiplied two lengths (like $r \cdot \ell$ )? → square units → it's an area.

💡 If your "volume" came out in square units, you almost certainly forgot to multiply by the height. The units are a free error-detector.

Pick the Right Units 🔽

Choosing the wrong units is the #1 careless error on these problems.

Part 2: Pyramids & Cones

📦 Surface Area & Volume of Solids

Part 2 of 5 — Pyramids & Cones

🔑 The "one-third" family: A pyramid or cone that shares the same base and height as a prism or cylinder holds exactly one-third as much. That single factor of $\frac{1}{3}$ runs through this entire part.

Pyramids: $V = \frac{1}{3}Bh$

Part 3: Spheres

📦 Surface Area & Volume of Solids

Part 3 of 5 — Spheres

🔑 Two formulas, one radius: A sphere is defined entirely by its radius $r$ . Memorize the pair — $V = \frac{4}{3}\pi r^3$ and — and every sphere problem becomes plug-and-chug.

Part 4: Composite Solids & Real-World Problems

📦 Surface Area & Volume of Solids

Part 4 of 5 — Composite Solids & Real-World Problems

🔑 Divide and conquer: Real objects (a silo, an ice-cream cone, a capsule) are built from the basic solids you already know. Break the shape into pieces, compute each, then add (or subtract).

Composite Volumes — Add the Pieces

A silo is a cylinder topped by a hemisphere. Find each volume and add.

Example: cylinder $r = 3$ , $h = 10$ , capped by a hemisphere of $r = 3$

Part 5: Mixed Practice & Mastery Check

📦 Surface Area & Volume of Solids

Part 5 of 5 — Mixed Practice & Mastery Check

You can now find the volume and surface area of prisms, cylinders, pyramids, cones, and spheres, and combine them for composite real-world solids. Let's put it all together.

Quick Reference

Solid	Volume	Surface Area
Prism	$Bh$	sum of all faces (net)
Cylinder	$\pi r^2 h$

Prisms & Cylinders Volume: $V = Bh$

A prism or cylinder has two identical parallel bases joined by straight sides. Its volume is always:

where $B$ is the area of one base and $h$ is the height (the perpendicular distance between the two bases).

🔑 The base $B$ is an area, not a length. For a rectangular prism $B = \ell w$ ; for a cylinder $B = \pi r^2$ . Multiply by $h$ to "stack" that base through the height.

B = 20\text{ cm}^2

$V = Bh = 20 \cdot 6 = 120\text{ cm}^3$

Surface area. A box has 3 pairs of matching faces. Add the area of one of each pair, then double:

$SA = 2(\ell w + \ell h + w h) = 2(5\cdot 4 + 5\cdot 6 + 4\cdot 6)$

$SA = 2(20 + 30 + 24) = 2(74) = 148\text{ cm}^2$

💡 Net trick: "Unfold" the solid flat into a net. Surface area is just the total area of every face in the net — no formula memorization required if you can find each face.

$SA = \underbrace{2\pi r^2}_{\text{two circles}} + \underbrace{2\pi r h}_{\text{curved side}}$

Example: $r = 3$ , $h = 10$

$V = \pi (3)^2(10) = 90\pi \approx 282.7\text{ units}^3$

$SA = 2\pi(3)^2 + 2\pi(3)(10) = 18\pi + 60\pi = 78\pi\text{ units}^2$

💡 Leave answers in terms of $\pi$ for exactness, or multiply by $\pi \approx 3.14159$ for a decimal.

A pyramid has a polygon base and triangular faces meeting at an apex. Its volume is:

$V = \frac{1}{3} B h$

where $B$ is the base area and $h$ is the vertical height (apex straight down to the base center).

Example: square base, side $6$ , height $10$

$B = 6^2 = 36, \qquad V = \frac{1}{3}(36)(10) = 120\text{ units}^3$

⚠️ Watch the height. The volume formula needs the vertical height $h$ , not the slant height $\ell$ (the distance up a triangular face). They are different — mixing them up is the classic pyramid/cone mistake.

Cones: $V = \frac{1}{3}\pi r^2 h$

A cone is a pyramid with a circular base, so $B = \pi r^2$ :

$V = \frac{1}{3}\pi r^2 h$

For surface area we need the slant height $\ell$ , found by the Pythagorean theorem on the right triangle formed by $r$ , $h$ , and $\ell$ :

$\ell = \sqrt{r^2 + h^2}$

Then the cone's surface area is the base circle plus the curved lateral surface:

$SA = \underbrace{\pi r^2}_{\text{base}} + \underbrace{\pi r \ell}_{\text{lateral}}$

🔑 $r$ , $h$ , $\ell$ form a right triangle. The radius and vertical height are the legs; the slant height is the hypotenuse: $\ell^2 = r^2 + h^2$ .

Concept Check 🎯

Full Worked Cone — $r = 6$ , $h = 8$

Step 1 — slant height. $\ell = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Step 2 — volume. $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (36)(8) = \frac{288}{3}\pi = 96\pi\text{ units}^3$

Step 3 — surface area. $SA = \pi r^2 + \pi r \ell = \pi(36) + \pi(6)(10) = 36\pi + 60\pi = 96\pi\text{ units}^2$

💡 It's a coincidence that this cone's $V$ and $SA$ are both $96\pi$ — they measure totally different things (one in cubic units, one in square). Always track your units.

Compute It 🧮

1) Square pyramid, base side $3$ , height $9$ : volume $= \,?$ (cubic units) 2) Cone with $r = 5$ , $h = 12$ : slant height $\ell = \,?$ 3) Cone with $r = 3$ , $h = 4$ : volume $= \,? \pi$ (enter the coefficient)

Quick Recap Before You Move On

The pyramid/cone family comes down to three ideas:

Need	Use
Volume	$\frac{1}{3}Bh$ (with the vertical height $h$ )
Slant height	$\ell = \sqrt{r^2 + h^2}$
Cone surface area	$\pi r^2 + \pi r \ell$

🔑 The factor of $\frac{1}{3}$ and the right triangle $r$ - $h$ - $\ell$ are the two things that separate these solids from prisms and cylinders. Lock them in before the next check.

Match the Formula 🔽

The Two Sphere Formulas

$V = \frac{4}{3}\pi r^3 \qquad\qquad SA = 4\pi r^2$

Notice the exponents: volume uses $r^3$ (it's 3-D space), and surface area uses $r^2$ (it's a 2-D skin). That pattern holds for every solid.

Quantity	Formula	Why the exponent
Volume	$\frac{4}{3}\pi r^3$	space → cube of a length
Surface area

Example: $r = 3$

$V = \frac{4}{3}\pi (3)^3 = \frac{4}{3}\pi (27) = 36\pi\text{ units}^3$

$SA = 4\pi (3)^2 = 4\pi(9) = 36\pi\text{ units}^2$

⚠️ Order of operations: cube (or square) the radius before multiplying by the other factors. $(3)^3 = 27$ , not $3 \times 3 = 9$ .

Concept Check 🎯

Hemispheres — Don't Forget the Flat Circle

A hemisphere is half a sphere. The volume is simply half:

$V_{\text{hemi}} = \frac{1}{2}\cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3$

But the surface area is not just half. Cutting a sphere exposes a flat circular face of area $\pi r^2$ . So:

$SA_{\text{hemi}} = \underbrace{\frac{1}{2}(4\pi r^2)}_{\text{curved dome}} + \underbrace{\pi r^2}_{\text{flat base}} = 2\pi r^2 + \pi r^2 = 3\pi r^2$

⚠️ The most common hemisphere mistake is forgetting the flat circle. Half the curved skin is $2\pi r^2$ ; add the $\pi r^2$ disk to get $3\pi r^2$ .

Compute It 🧮

Enter each answer as the coefficient of $\pi$ .

1) Sphere $r = 2$ : volume $= \,? \pi$ 2) Sphere $r = 4$ : surface area $= \,? \pi$ 3) Hemisphere $r = 3$ : total surface area $= \,? \pi$

The Exponent Tells You Which Is Which

Look at the two sphere formulas side by side and the pattern jumps out:

$\underbrace{\tfrac{4}{3}\pi r^3}_{\text{volume} \,\to\, r^3} \qquad \underbrace{4\pi r^2}_{\text{surface} \,\to\, r^2}$

If you ever blank on which formula is which, ask: am I filling space (cube, $r^3$ ) or covering a skin (square, $r^2$ )? The exponent and the units always agree.

💡 This same logic flags a wrong answer instantly: a surface area should never contain $r^3$ , and a volume should never stop at $r^2$ .

Sphere Sense 🔽

Cylinder: $V = \pi r^2 h = \pi(9)(10) = 90\pi$

Hemisphere: $V = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi(27) = 18\pi$

Total: $90\pi + 18\pi = 108\pi \approx 339.3\text{ units}^3$

💡 When pieces are stacked or joined, add their volumes. When a piece is drilled out or removed (a hole), subtract it.

Composite Surface Area — Hide the Joined Faces

Surface area of a composite is trickier than volume: where two pieces meet, those faces are hidden and must not be counted.

For the silo above, the circle where the hemisphere sits on the cylinder is internal — skip both that top cylinder circle and the hemisphere's flat face. The exposed skin is:

Cylinder bottom circle: $\pi r^2 = 9\pi$
Cylinder curved side: $2\pi r h = 2\pi(3)(10) = 60\pi$
Hemisphere dome (curved only): $2\pi r^2 = 18\pi$

$SA = 9\pi + 60\pi + 18\pi = 87\pi\text{ units}^2$

⚠️ Don't count covered faces. The flat circle where two solids join is sealed inside — leaving it in is the most common composite-surface-area error.

Concept Check 🎯

Real-World Application — Filling a Tank

A cylindrical water tank has radius $2\text{ m}$ and height $5\text{ m}$ . How many cubic meters of water does it hold (full)?

$V = \pi r^2 h = \pi(2)^2(5) = 20\pi \approx 62.8\text{ m}^3$

Now suppose water costs a flat rate and the tank is filled to the brim. The volume $20\pi\text{ m}^3$ is what you'd multiply by the per-cubic-meter price. (For example, at $0.50 per $\text{m}^3$ that's about $31.40.)

💡 Read the question. "How much fills it?" → volume. "How much paint/wrap/material covers it?" → surface area. The word problem tells you which one.

Volume or Surface Area? 🔽

For each task, decide which quantity you need.

A Reliable 3-Step Strategy

For any composite solid, work it the same way every time:

Decompose — sketch the solid and name each basic piece (cylinder, hemisphere, cone, prism…).
Compute each piece with its own formula, keeping $\pi$ exact.
Combine — add joined pieces, subtract removed pieces; for surface area, drop every hidden interior face.

🔑 The math of each piece is the easy part — you already know it. The points are won or lost in step 3: adding vs. subtracting, and not counting covered faces.

Composite Practice 🧮

Enter each answer as the coefficient of $\pi$ .

1) Cylinder ( $r=2$ , $h=6$ ) topped by a hemisphere ( $r=2$ ): total volume $= \,? \pi$ 2) A cube of side $4$ has a cylindrical hole ( $r=1$ , $h=4$ ) drilled through it. The remaining volume is $64 - \,? \pi$ (enter the coefficient subtracted) 3) Cone ( $r=3$ , $h=4$ , so $\ell = 5$ ) on top of a cylinder ( $r=3$ , $h=6$ ): the joined circle that is hidden has area $\,? \pi$

🔑 Three habits that prevent errors:

Volume → cubic units; surface area → square units. Always.

For cones, slant height $\ell = \sqrt{r^2 + h^2}$ — use $h$ for volume, $\ell$ for surface area.

For composites, add joined pieces, subtract removed ones, and never count hidden (interior) faces.

Mixed Practice 🎯

Choose the Move 🔽

A grain silo is a cylinder ( $r=4$ , $h=12$ ) with a cone roof ( $r=4$ , $h=3$ ). You want the volume of grain it holds.

Final Drill 🧮

1) Triangular prism: base triangle area $B = 12$ , height (length) $h = 7$ . Volume $= \,?$ 2) Sphere $r = 6$ : volume $= \,? \pi$ (enter the coefficient) 3) Cone $r = 9$ , $h = 12$ : slant height $\ell = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.

Surface Area and Volume of Solids

Surface Area and Volume of Solids - Complete Interactive Lesson

Part 1: Prisms & Cylinders

📦 Surface Area & Volume of Solids

Topics in This Part

Surface Area vs. Volume

Worked Example — Rectangular Prism

Cylinders

One Last Habit — Units

Part 2: Pyramids & Cones

📦 Surface Area & Volume of Solids

Pyramids: V=13BhV = \frac{1}{3}BhV=

Part 3: Spheres

📦 Surface Area & Volume of Solids

Part 4: Composite Solids & Real-World Problems

📦 Surface Area & Volume of Solids

Composite Volumes — Add the Pieces

Example: cylinder r=3r = 3r=3, h=10h = 10h=10, capped by a hemisphere of r=3r = 3

Part 5: Mixed Practice & Mastery Check

📦 Surface Area & Volume of Solids

Quick Reference

Prisms & Cylinders Volume: V=BhV = BhV=Bh

Example: r=3r = 3r=3, h=10h = 10h=10

Example: square base, side 666, height 101010

Cones: V=13πr2hV = \frac{1}{3}\pi r^2 hV=31​πr2h

Full Worked Cone — r=6r = 6r=6, h=8h = 8h=8

Quick Recap Before You Move On

The Two Sphere Formulas

Example: r=3r = 3r=3

Hemispheres — Don't Forget the Flat Circle

The Exponent Tells You Which Is Which

Composite Surface Area — Hide the Joined Faces

Real-World Application — Filling a Tank

A Reliable 3-Step Strategy

Pyramids: $V = \frac{1}{3}Bh$

Example: cylinder $r = 3$ , $h = 10$ , capped by a hemisphere of $r = 3$

Prisms & Cylinders Volume: $V = Bh$

Example: $r = 3$ , $h = 10$

Example: square base, side $6$ , height $10$

Cones: $V = \frac{1}{3}\pi r^2 h$

Full Worked Cone — $r = 6$ , $h = 8$

Example: $r = 3$