MathIsimple
Lesson 3-1

Volumes & Surface Areas of 3D Solids (Advanced)

Master composite modeling with cylinders, cones, and spheres. Compute capacity and material usage with careful shared-face handling and lateral areas.

Learning Objectives Back to Unit 3

Learning Objectives

  • Recall and apply core volume/surface formulas for cylinders, cones, and spheres.
  • Calculate lateral areas and subtract/avoid double-counting shared faces in composites.
  • Formulate multi-solid capacity problems and compute total volume/area.
  • Interpret units, round results appropriately, and justify assumptions.
  • Translate real constraints (e.g., wall thickness) into geometric adjustments.

Core Formulas

Cylinder

V=πr2hV=\pi r^2 h
Stotal=2πr2+2πrhS_{total}=2\pi r^2+2\pi r h
Slateral=2πrhS_{lateral}=2\pi r h

Cone

V=13πr2hV=\tfrac{1}{3}\pi r^2 h
Slateral=πrl,  l=r2+h2S_{lateral}=\pi r l,\; l=\sqrt{r^2+h^2}

Sphere

V=43πR3V=\tfrac{4}{3}\pi R^3
S=4πR2S=4\pi R^2

Composite Principles

Additive Volumes

Volume is additive: for disjoint components, total volume is the sum of each part. For solids formed by joining pieces, ensure overlaps are handled correctly (avoid double-counting).

Surface Adjustments

Surface area is not purely additive because shared faces are internal and not exposed. Subtract the areas of coincident faces, and add any newly exposed lateral areas (for removed parts).

Worked Example A: Cylinder + Cone

A part consists of a cylinder of radius r=2 cmr=2\,\text{cm} and height hc=4 cmh_c=4\,\text{cm}with a cone of radius r=2 cmr=2\,\text{cm} and height hk=3 cmh_k=3\,\text{cm}attached on top, sharing the circular base. Compute total volume and exposed surface area.

Volume

Sum of cylinder and cone volumes:

V=πr2hc+13πr2hk=π⋅22⋅4+13π⋅22⋅3=16π+4π=20π cm3V=\pi r^2 h_c + \tfrac{1}{3}\pi r^2 h_k=\pi\cdot 2^2\cdot 4 + \tfrac{1}{3}\pi\cdot 2^2\cdot 3=16\pi+4\pi=20\pi\,\text{cm}^3

Surface Area

Exposed surfaces include cylinder lateral + bottom, cone lateral; the shared base is internal.

S=πr2⏟bottom+2πrhc⏟cyl lateral+πrl⏟cone lateral,  l=r2+hk2=13S=\underbrace{\pi r^2}_{\text{bottom}}+\underbrace{2\pi r h_c}_{\text{cyl lateral}}+\underbrace{\pi r l}_{\text{cone lateral}},\; l=\sqrt{r^2+h_k^2}=\sqrt{13}
S=π⋅22+2π⋅2⋅4+π⋅213=4π+16π+2π13S=\pi\cdot 2^2 + 2\pi\cdot 2\cdot 4 + \pi\cdot 2\sqrt{13}=4\pi+16\pi+2\pi\sqrt{13}

Worked Example B: Cylinder + Hemisphere Tank

A tank has a cylindrical body of radius r=2 mr=2\,\text{m} and height h=5 mh=5\,\text{m}topped with a hemisphere (radius also 2 m2\,\text{m}). Compute capacity and mass of oil with density ρ=800 kg/m3\rho=800\,\text{kg/m}^3.

Volume

V=πr2h+23πr3=π⋅22⋅5+23π⋅23=20π+163π=763π m3V=\pi r^2 h + \tfrac{2}{3}\pi r^3=\pi\cdot 2^2\cdot 5 + \tfrac{2}{3}\pi\cdot 2^3=20\pi+\tfrac{16}{3}\pi=\tfrac{76}{3}\pi\,\text{m}^3

Mass

Mass = density × volume:

m=ρV=800×763π kgm=\rho V=800\times \tfrac{76}{3}\pi\,\text{kg}

Engineering Scenario: Material Cost

A cylindrical container (open top) of radius rr and height hh uses sheet metal of costCsC_s per square meter for walls and base. Find total cost as a function of r,hr,h.

Open top ⇒ exposed area = bottom + lateral:

S=πr2+2πrhS=\pi r^2 + 2\pi r h
Cost(r,h)=Cs (πr2+2πrh)\text{Cost}(r,h)=C_s\,(\pi r^2 + 2\pi r h)

Thickness Adjustment (Hollow Bodies)

For containers with wall thickness tt, capacity is the volume of the inner cavity. For a cylinder of outer radius RR and height hh, inner radius is r=R−tr=R-t (assuming bottom thickness included separately).

Capacity

Vcap=π(R−t)2hV_{cap}=\pi (R-t)^2 h

Material Volume

Vmat=πR2h−π(R−t)2h=πh (2Rt−t2)V_{mat}=\pi R^2 h - \pi (R-t)^2 h = \pi h\,(2Rt - t^2)

Practice Problems

Problem 1: Cone Lateral Area

A cone has radius 5 cm and height 12 cm. Compute lateral area.

Show Solution

Slant l=52+122=13l=\sqrt{5^2+12^2}=13 cm. Slateral=πrl=65π cm2S_{lateral}=\pi r l=65\pi\,\text{cm}^2.

Problem 2: Shared Face Subtraction

A cylinder (r=3, h=10) is capped by a cone (r=3, h=4). Compute exposed area.

Show Solution

Area = cylinder bottom + cylinder lateral + cone lateral. S=pir2+2pirh+pirlS=pi r^2 + 2pi r h + pi r l, l=sqrtr2+h2=5l=sqrt{r^2+h^2}=5 ⇒ S=9pi+60pi+15pi=84piS=9pi+60pi+15pi=84pi.

Problem 3: Hemisphere Volume

Find volume of a hemisphere with radius 6 cm.

Show Solution

V=23πR3=23π⋅216=144π cm3V=\tfrac{2}{3}\pi R^3=\tfrac{2}{3}\pi \cdot 216=144\pi\,\text{cm}^3.

Problem 4: Open-Top Container Cost

Given r=0.4 mr=0.4\,\text{m}, h=0.8 mh=0.8\,\text{m}, cost Cs=12 $/m2C_s=12\,\$/\text{m}^2. Find total cost.

Show Solution

Area S=πr2+2πrh=π⋅0.16+2π⋅0.4⋅0.8S=\pi r^2 + 2\pi r h=\pi\cdot 0.16 + 2\pi\cdot 0.4\cdot 0.8 = 0.16π+0.64π=0.8π0.16\pi + 0.64\pi=0.8\pi m2\text{m}^2. Cost ≈ 12×0.8π12\times 0.8\pi ≈ $30.16.

Problem 5: Wall Thickness

Outer R=10 cm, thickness t=0.5 cm, h=20 cm. Find capacity and material volume.

Show Solution

Inner r=9.5 cm. Capacity Vcap=π⋅9.52⋅20V_{cap}=\pi\cdot 9.5^2\cdot 20; Material Vmat=π⋅20 (2⋅10⋅0.5−0.52)V_{mat}=\pi\cdot 20\,(2\cdot 10\cdot 0.5-0.5^2) cm3\text{cm}^3.

Key Takeaways

  • Volumes add directly; surface areas require shared-face adjustments.
  • Know lateral vs. total area to match real exposure in applications.
  • Units matter: report m3\text{m}^3 for volume, m2\text{m}^2 for area, and justify rounding.
  • Thickness changes inner capacity and material usage; model carefully.

Advanced Modeling Extensions

Optimization: Max Volume with Fixed Surface

For an open-top cylinder with fixed surface area S0S_0, find r,hr,h maximizing volume.

S=πr2+2πrh=S0S=\pi r^2+2\pi r h=S_0
V=πr2h=πr2 S0−πr22πr=S0r2−πr32V=\pi r^2 h=\pi r^2\,\frac{S_0-\pi r^2}{2\pi r}=\tfrac{S_0 r}{2}-\tfrac{\pi r^3}{2}
dVdr=S02−3πr22=0  ⇒  r=S03π,  h=S0−πr22πr=S0−S032πr=S03πr\frac{dV}{dr}=\tfrac{S_0}{2}-\tfrac{3\pi r^2}{2}=0\;\Rightarrow\; r=\sqrt{\tfrac{S_0}{3\pi}},\; h=\tfrac{S_0-\pi r^2}{2\pi r}=\tfrac{S_0-\tfrac{S_0}{3}}{2\pi r}=\tfrac{S_0}{3\pi r}

Tolerance & Manufacturing Margins

If radius tolerance is pmdeltapmdelta, estimate worst-case error in volume for a cylinder.

V=πr2h  ⇒  dV≈2πrh drV=\pi r^2 h\;\Rightarrow\; dV\approx 2\pi r h\,dr

Worst-case ∣DeltaV∣approx2pirh,delta|Delta V|approx 2pi r h,delta.

Extended Examples Library

Example 5: Cylinder + Hemisphere

Join cylinder (r=a,h=b) with hemisphere (R=a). Find exposed surface area.

Shared circular face is internal.

S=2piab+pia2+2pia2S=2pi a b+pi a^2+2pi a^2

Example 6: Cylinder with Conical Frustum

Attach a frustum (r1=ar_1=a, r2=a/2r_2=a/2, height=b/2) on a cylinder (r=a,h=b). Compute total surface.

S=2piahcyl+pia2+pi(r1+r2)sS=2pi a h_{cyl}+pi a^2 + pi (r_1+r_2)s

s is slant of frustum; subtract shared circle.

Example 7: Sphere Hole (Spherical Segment)

A cylindrical hole (radius a) drilled through sphere (R). Find removed volume.

Napkin-ring result depends only on height h.

V=πh36V=\tfrac{\pi h^3}{6}

Example 8: Composite with Shared Rectangle

Two rectangular prisms share a face A×B. Compute exposed area.

Subtract 2AB from naive sum.

Example 9: Truncated Cone Capacity

Compute volume of a frustum vase; compare to full cone difference.

V=πh3(r12+r1r2+r22)V=\tfrac{\pi h}{3}(r_1^2+r_1 r_2+r_2^2)

Example 10: Spherical Cap Area

Find the area of a spherical cap of height h on radius R.

S=2piRhS=2pi R h

Example 11: Cone Inside Cylinder

A cone carved from a cylinder (same r,h). Compute remaining volume.

Vremain=πr2h−13πr2h=23πr2hV_{remain}=\pi r^2 h-\tfrac{1}{3}\pi r^2 h=\tfrac{2}{3}\pi r^2 h

Example 12: Hemisphere + Cone Joint

Glue a cone to a hemisphere (same radius). Find exposed area.

Shared circle is internal; include cone lateral + hemisphere area.

Deep Practice Sets

Set 1: Frustum by Difference

  1. Model as big cone − small cone.
  2. State all radii/heights and units.
  3. Compute exact and decimal volume.
  4. Comment on sensitivity to r2r_2.

Set 2: Open-Top Constraint

  1. Target capacity V0V_0, minimize surface.
  2. Derive KKT or use calculus.
  3. Report optimal r,h and cost.

Set 3: Tolerance Bounds

  1. Given r±δ, h±δ, bound ΔV via differentials.
  2. Compare with brute-force endpoints.
  3. Explain which dimension dominates.

Set 4: Composite Exposed Area

  1. Join cylinder+cone; subtract shared disk.
  2. Express S(r,hc,hk)S(r,h_c,h_k).
  3. Check units and limiting cases.

Set 5: Hemisphere Tank

  1. Capacity with cylinder + hemisphere.
  2. Mass from density ρ; report units.
  3. What if headspace reduces fill height?

Set 6: Spherical Cap

  1. Given R,h, find cap area and volume.
  2. Relate to drilling problem.
  3. Check h→0 and h→2R limits.

Set 7: Material Optimization

  1. Minimize weight given thickness t(r).
  2. Analyze tradeoffs in r vs h.
  3. Provide design recommendation.

Set 8: Heat Exchange Area

  1. Maximize lateral area for fixed volume.
  2. Solve for r,h and interpret.
  3. Discuss manufacturability.

Set 9: Cost with Waste

  1. Add waste factor w% to surface area.
  2. Recompute cost function.
  3. Sensitivity to w.

Set 10: Reverse Engineering

  1. Given S and V, solve for r,h.
  2. Check multiple solutions.
  3. Pick feasible design by constraint.
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