LESSON PLAN

Surface Area and Volume of Prisms and Cylinders

Apothem Team

Grade 8 · Measurement

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

The soup-can label challenge: peel a real label off a can, flatten it — a rectangle appears. "What are this rectangle's dimensions, in can-language?" Height: the can's height. Width: the distance AROUND — the circumference.

That unrolled label is the whole secret of cylinder surface area, and Grade 8 SA adds cylinders to last year's prisms — same principle, one new unroll.

Explore

Unroll-everything lab: (1) cylinder → two circles + the label rectangle ( $2\pi r$ wide, $h$ tall); compute a real can's SA and compare against its label + lids by direct measurement. (2) Volume reprise alongside: same can's capacity via $\pi r^2 h$ , checked with water. (3) The combined design table: for three cans of equal volume (tall/thin, squat/wide, middling), compute both SA and V — discovering again that skin varies while filling holds.

Close with the this-or-that sort: eight scenarios (paint it, fill it, wrap it, weigh the metal, will it fit the shelf) sorted into SA questions vs V questions vs neither.

Formalize

Formalize the cylinder pair beside the prism pair:

$SA_{\text{cyl}} = 2\pi r^2 + 2\pi r h \qquad V_{\text{cyl}} = \pi r^2 h$

Read the SA formula as its net: two lids ( $2\pi r^2$ ) plus the unrolled label ( $2\pi r \times h$ — circumference times height). Formulas that can be read as pictures get remembered; formulas memorized as syllables get scrambled. Exact answers keep $\pi$ .

Practice

Practice: SA and V for two cylinders (one from diameter); one prism review item; one open-top variation (subtract a lid: $\pi r^2 + 2\pi rh$ — reading the situation edits the formula); one design question (which of two equal-volume cans uses less metal?).

Exit ticket: cylinder $r = 3$ , $h = 10$ : exact SA and V. ( $SA = 18\pi + 60\pi = 78\pi$ cm²; $V = 90\pi$ cm³.)

Exit ticket

Exit ticket: cylinder $r = 3$ , $h = 10$ : exact SA and V. ( $SA = 18\pi + 60\pi = 78\pi$ cm²; $V = 90\pi$ cm³.)

TIP The open-top / no-lid variations are where formula-chanters sink and picture-readers swim: assign one daily. Real objects (mugs, pipes, tunnels) rarely have all their lids.

WORKED EXAMPLES

Example 1 — The paint estimate: a cylindrical water tank

The tank: diameter 2 m, height 3 m, sitting on the ground — paint the outside (walls and top only).

Step 1: Radius: 1 m. Inventory the painted faces: the wall (label) + the top lid. NOT the bottom (on the ground).

Step 2: Wall: $2\pi r h = 2\pi(1)(3) = 6\pi$ m². Top: $\pi r^2 = \pi$ m².

Step 3: Total: $7\pi \approx 22.0$ m².

Step 4: Paint math: coverage 8 m²/L → $22 \div 8 = 2.75$ → buy 3 L.

Step 5: The formula-editing moment made explicit: the standard $2\pi r^2 + 2\pi rh$ would have billed the customer for painting the UNDERSIDE of a tank on concrete. The object dictated $\pi r^2 + 2\pi rh$ . Real surface area starts with an inventory, not a formula.

Example 2 — Metal vs capacity: the can-design showdown

Two 500 mL can designs: Can A — $r = 3$ cm, $h = 17.7$ cm. Can B — $r = 5$ cm, $h = 6.4$ cm. (Both ≈ 500 cm³ — verify: A: $\pi(9)(17.7) \approx 500$ ✓; B: $\pi(25)(6.4) \approx 503$ ✓.)

Step 1: SA of A: $2\pi(9) + 2\pi(3)(17.7) = 18\pi + 106.2\pi = 124.2\pi \approx 390$ cm².

Step 2: SA of B: $2\pi(25) + 2\pi(5)(6.4) = 50\pi + 64\pi = 114\pi \approx 358$ cm².

Step 3: Verdict: B (squat) uses ~8% less metal for the same soup. Neither extreme wins in general — the metal-minimizing cylinder has $h = 2r$ (a fact to state, not prove, with B nearly there: $h = 6.4$ vs $2r = 10$ … B still beats A by being closer).

Step 4: Reality check against the shelf: real soup cans are TALLER than metal-optimal — because shelf space, hand grip, and label real estate also cost money. Optimization always has more constraints than the math problem admits; naming them is the difference between a calculation and a design.

Example 3 — The composite silo: cylinder plus half-sphere… no — cylinder plus prism porch

The structure: a cylindrical grain silo ( $r = 4$ m, $h = 10$ m) with an attached rectangular loading porch ( $3 \times 2 \times 2$ m) sharing one 2×2 wall with the silo.

Step 1: Volumes add cleanly: silo $\pi(16)(10) = 160\pi \approx 502.7$ ; porch $3 \times 2 \times 2 = 12$ ; total $\approx 514.7$ m³.

Step 2: Surface areas do NOT simply add — the shared 2×2 wall is interior twice over: subtract it from BOTH bodies' skins ( $2 \times 4 = 8$ m² removed in total).

Step 3: The lesson in one line: volume is indifferent to gluing; surface area pays attention to every joint. (Compute the full SA only if the class is sturdy: silo wall $80\pi$ + lids $32\pi$ + porch faces $32$ − shared $8$ ≈ $375.9$ m².)

Step 4: Where this thinking lives professionally: HVAC sizing (volume) vs cladding quotes (SA with joints) — two trades pricing the same building read different formulas off it.

MATERIALS

•Cans with peelable labels

•Water and measuring cups

•Cylinder nets

•Scenario sort cards

•Practice set (PDF)

WATCH FOR

!Label width taken as the diameter instead of the circumference. Unroll a real one; the width is visibly the wrap-around.

!SA and V formulas cross-bred (

2\pi r^2 h

hybrids). The net-reading habit keeps each part attached to its picture.

!Open containers given full-formula SA. Count the actual faces present — geometry serves the object, not the formula sheet.