Sphere, Cone & Cylinder – definitions, equations, volumes & surface areas

1. Introduction to Quadric Solids

A quadric solid is a three‑dimensional surface defined by a second‑degree polynomial in $x,y,z$. The three most common in the syllabus are:

These equations arise directly from the distance formula in three dimensions.

2. Sphere

2.1 Definition & Geometry

A sphere is the set of all points at a fixed distance $r$ from a centre $C(a,b,c)$.

• Radius $r$: distance from centre to any point on the surface.
• Diameter $d=2r$.

2.2 Volume

Derived by integrating circular cross‑sections perpendicular to any axis:

$$V_{\text{sphere}}={\int }_{-r}^r\pi (r^2-z^2)dz=\frac{4}{3}\pi r^3.$$

2.3 Surface Area

A sphere’s surface area is the limit of the areas of infinitesimal patches:

$$A_{\text{sphere}}=4\pi r^2.$$

2.4 Worked Example

Problem: A satellite dish is modelled as a spherical cap of radius $r=12$ cm and height $h=3$ cm. Find the volume of the cap.

Solution steps

1. The cap volume formula (derived from integration) is

   $$V_{\text{cap}}=\frac{\pi h^2}{3}(3r-h).$$

2. Substitute $r=12$ cm, $h=3$ cm:

   $$V_{\text{cap}}=\frac{\pi (3{)}^2}{3}(36-3)=\frac{9\pi }{3}\times 33=3\pi \times 33=99\pi {\text{cm}}^3.$$

3. Numerically, $V_{\text{cap}}\approx 311{\text{cm}}^3$.

3. Right Circular Cylinder

3.1 Definition & Geometry

A right circular cylinder consists of all points at a distance $r$ from a fixed line (the axis) and bounded between two parallel planes a distance $h$ apart.

• Base: a circle of radius $r$.
• Height $h$: distance between the two circular bases.

3.2 Volume

$$V_{\text{cyl}}=\text{Base area}\times \text{height}=\pi r^{2}h.$$

3.3 Surface Area

• Lateral (curved) area: $A_{\text{lat}}=2\pi rh$.

• Total area (including two bases):

  $$A_{\text{cyl}}=2\pi rh+2\pi r^2=2\pi r(h+r).$$

3.4 Worked Example (Real Situation)

Problem: NTC installs a water‑storage cylinder at a rural tower site. The cylinder has radius $r=1.5$ m and height $h=4$ m. Compute the amount of water (in litres) it can hold, assuming it is filled to the brim.

Solution

1. Volume in cubic metres:

   $$V=\pi (1.5{)}^2\times 4=\pi \times 2.25\times 4=9\pi {\text{m}}^3.$$

2. Convert to litres (1 m³ = 1000 L):

   $$V=9\pi \times 1000\approx 28274\text{L}.$$

3. The tank can store roughly 28 kL of water, sufficient for a month’s consumption of the tower staff.

4. Right Circular Cone

4.1 Definition & Geometry

A right circular cone is generated by rotating a right‑angled triangle about one of its legs (the axis).

• Vertex $V$: tip of the cone.
• Base: a circle of radius $r$.
• Height $h$: perpendicular distance from vertex to base plane.
• Slant height $l=\sqrt{r^2+h^2}$.

4.2 Volume

$$V_{\text{cone}}=\frac{1}{3}\pi r^{2}h.$$

4.3 Surface Area

• Lateral area: $A_{\text{lat}}=\pi rl$.

• Total area (including base):

  $$A_{\text{cone}}=\pi rl+\pi r^2=\pi r(l+r).$$

4.4 Worked Example (Daraz Order)

Problem: Daraz ships an ice‑cream cone set. The cone has radius $r=2$ cm and height $h=8$ cm. Find the amount of ice‑cream (in cm³) that fits if the cone is filled to the brim.

Solution

1. Volume:

   $$V=\frac{1}{3}\pi (2{)}^2\times 8=\frac{1}{3}\pi \times 4\times 8=\frac{32}{3}\pi \approx 33.51{\text{cm}}^3.$$

2. The cone holds about 33.5 cm³ of ice‑cream.

5. Cross‑Sections & Generating Curves

Understanding how a solid is sliced helps in both integration and practical design.

6. Comparison of Formulas

flowchart LR
    Q["Quadric Solids"]
    Q --> S["Sphere"]
    Q --> C["Right Circular Cone"]
    Q --> Y["Right Circular Cylinder"]
    S --> Vs["V = 4/3 π r³"]
    S --> As["A = 4 π r²"]
    C --> Vc["V = 1/3 π r² h"]
    C --> Ac["A = π r (l + r)"]
    Y --> Vy["V = π r² h"]
    Y --> Ay["A = 2π r (h + r)"]

Advantages & Disadvantages

7. Real‑World Applications

7.1 Water‑Storage Cylinders (NTC)

NTC’s remote tower sites often use vertical steel cylinders to store rainwater. The volume formula directly determines how many days of operation the tower can sustain without external supply.

7.2 Traffic Cones (Daraz)

Daraz sells orange traffic cones used by road‑maintenance crews. The cone’s slant height determines the material needed for the outer shell, while the volume helps estimate the amount of foam filler for stability.

7.3 Earth as a Sphere (Google Earth)

Google Earth renders the planet as a sphere of radius $R\approx 6.371\times 10^6$ m. Surface‑area calculations are essential for satellite‑coverage planning and for converting raster image data (pixels per square kilometre).

Orange traffic cone used on Nepali roads (Image: Ser Amantio di Nicolao, CC BY-SA 4.0, via Wikimedia Commons) Steel water‑storage cylinder at a telecom tower (Image: Evelyn Simak, CC BY-SA 2.0, via Wikimedia Commons)

8. Worked Example Integrated with Real Situation

Scenario: A new e‑commerce fulfillment centre (Daraz) plans to install a cylindrical storage silo for bulk rice. The silo must hold 150 m³ of rice. The design constraints are: radius cannot exceed 2 m due to floor‑space limits. Determine the minimum height required and compute the total surface area (including top and bottom) for painting.

Solution

1. Required volume $V=150{\text{m}}^3$.

2. Choose the maximum allowable radius $r=2\text{m}$.

3. Solve for height $h$ from $V=\pi r^{2}h$:

   $$h=\frac{V}{\pi r^2}=\frac{150}{\pi \times 4}=\frac{150}{4\pi }\approx \frac{150}{12.566}\approx 11.94\text{m}.$$

   So a height of ≈ 12 m satisfies the volume requirement.

4. Total surface area:

   $$A=2\pi r(h+r)=2\pi \times 2(11.94+2)=4\pi \times 13.94\approx 4\times 3.1416\times 13.94\approx 175.2{\text{m}}^2.$$

5. The centre will need roughly 175 m² of paint.

9. Common Mistakes & How to Avoid Them

10. Quick Reference Sheet

• Sphere: $V=\frac{4}{3}\pi r^3$, $A=4\pi r^2$.
• Cylinder: $V=\pi r^{2}h$, $A_{\text{lat}}=2\pi rh$, $A_{\text{tot}}=2\pi r(h+r)$.
• Cone: $V=\frac{1}{3}\pi r^{2}h$, $l=\sqrt{r^2+h^2}$, $A_{\text{lat}}=\pi rl$, $A_{\text{tot}}=\pi r(l+r)$.

Exam tip

• Memorise the three core formulas (volume & total surface area) as a single block; they differ only by a constant factor (1, 1/3, 4/3).
• Always draw a quick sketch with labelled $r,h,l$ before substituting numbers – the sketch prevents unit‑mix‑ups and reveals which radius (base vs slant) the problem needs.
• For cross‑section questions, write the generic circle radius expression first (e.g., $\sqrt{r^2-z^2}$ for a sphere) then integrate; many marks are awarded for the correct set‑up even if integration is left incomplete.
• Check the units: convert litres ↔ m³, cm ↔ m, etc., before finalising the answer.
• In multi‑part questions, answer the easiest part first (usually volume) to secure marks, then proceed to surface‑area or slant‑height calculations.