Educational Solutions • Real-World Case Studies

Cone Volume Step-by-Step Worked Examples

Explore fully worked mathematical solutions for right circular cones, slant height substitutions, reverse height and radius calculations, ice cream waffle cones, gravel stockpiles, and industrial frustum hoppers.

Example 1

Basic Volume Calculation from Radius and Height

Problem: A cone has a circular base radius of $r = 6\text{ cm}$ and a vertical height of $h = 10\text{ cm}$. Find its volume in cubic centimeters.

Given: $r = 6\text{ cm},\quad h = 10\text{ cm}$

Step-by-Step Solution:

  1. Write down the standard cone volume formula: $$V = \frac{1}{3}\pi r^2 h$$
  2. Square the radius: $$r^2 = 6^2 = 36\text{ cm}^2$$
  3. Multiply by the vertical height and $\pi$: $$36 \times 10 \times \pi = 360\pi\text{ cm}^3$$
  4. Divide by 3: $$V = \frac{360\pi}{3} = 120\pi \approx \mathbf{376.99\text{ cm}^3}$$
Final Answer: $$120\pi \approx 376.99\text{ cm}^3$$
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Example 2

Finding Volume When Diameter is Given

Problem: A conical tent has a base diameter of $d = 14\text{ meters}$ and a center pole height of $h = 9\text{ meters}$. What is the interior volume?

Given: $d = 14\text{ m},\quad h = 9\text{ m}$

Step-by-Step Solution:

  1. Find the radius by dividing diameter by 2: $$r = \frac{d}{2} = \frac{14}{2} = 7\text{ m}$$
  2. Apply the volume formula: $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (7)^2 (9)$$
  3. Simplify: $$V = \frac{1}{3}\pi (49)(9) = 49 \times 3 \times \pi = 147\pi\text{ m}^3$$
  4. Calculate decimal value: $$V \approx 147 \times 3.14159265 = \mathbf{461.81\text{ m}^3}$$
Final Answer: $$147\pi \approx 461.81\text{ m}^3$$
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Example 3

Finding Volume from Slant Height (Pythagorean Theorem)

Problem: A party hat has a base radius of $r = 5\text{ inches}$ and a slant height along its slope of $l = 13\text{ inches}$. Find its enclosed volume.

Given: $r = 5\text{ in},\quad l = 13\text{ in}$

Step-by-Step Solution:

  1. Use the Pythagorean theorem to calculate vertical height $h$: $$h = \sqrt{l^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ in}$$
  2. Substitute $r = 5$ and $h = 12$ into the volume formula: $$V = \frac{1}{3}\pi (5)^2 (12)$$
  3. Simplify the terms: $$V = \frac{1}{3}\pi (25)(12) = 25 \times 4 \times \pi = 100\pi\text{ in}^3$$
  4. Final decimal answer: $$V \approx \mathbf{314.16\text{ in}^3}$$
Final Answer: $$100\pi \approx 314.16\text{ in}^3$$
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Example 4

Reverse Calculation: Finding Height Given Volume and Radius

Problem: A conical vessel holds exactly $V = 300\text{ cm}^3$ of liquid and has a base radius of $r = 4\text{ cm}$. Determine the depth (vertical height) of the vessel.

Given: $V = 300\text{ cm}^3,\quad r = 4\text{ cm}$

Step-by-Step Solution:

  1. Rearrange the volume formula $V = \frac{1}{3}\pi r^2 h$ to solve for $h$: $$h = \frac{3V}{\pi r^2}$$
  2. Calculate the denominator (Base Area $B$): $$B = \pi r^2 = \pi (4^2) = 16\pi \approx 50.2655\text{ cm}^2$$
  3. Multiply the volume by 3: $$3V = 3 \times 300 = 900$$
  4. Divide to find $h$: $$h = \frac{900}{16\pi} = \frac{900}{50.2655} \approx \mathbf{17.90\text{ cm}}$$
Final Answer: $$h \approx 17.90\text{ cm}$$
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Example 5

Reverse Calculation: Finding Radius Given Volume and Height

Problem: An engineered conical hopper has a height of $h = 15\text{ cm}$ and must achieve a capacity of $V = 1000\text{ cm}^3$. What must its base radius be?

Given: $V = 1000\text{ cm}^3,\quad h = 15\text{ cm}$

Step-by-Step Solution:

  1. Rearrange the volume formula to solve for $r$: $$r = \sqrt{\frac{3V}{\pi h}}$$
  2. Multiply volume by 3: $$3V = 3 \times 1000 = 3000$$
  3. Calculate denominator: $$\pi h = \pi \times 15 \approx 47.1239$$
  4. Compute radius: $$r = \sqrt{\frac{3000}{47.1239}} = \sqrt{63.662} \approx \mathbf{7.98\text{ cm}}$$
Final Answer: $$r \approx 7.98\text{ cm}$$
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Example 6

Real-World: Waffle Ice Cream Cone Capacity in Milliliters

Problem: A waffle cone has an interior diameter of $d = 6\text{ cm}$ and an interior depth of $h = 14\text{ cm}$. How many milliliters of melted ice cream can it contain?

Given: $r = 3\text{ cm},\quad h = 14\text{ cm}$

Step-by-Step Solution:

  1. Calculate radius: $$r = 6 / 2 = 3\text{ cm}$$
  2. Compute volume in cubic centimeters: $$V = \frac{1}{3}\pi (3)^2 (14) = 3 \times 14 \times \pi = 42\pi \approx 131.95\text{ cm}^3$$
  3. Since $1\text{ cm}^3 = 1\text{ mL}$, the capacity is exactly: $$\mathbf{131.95\text{ mL}}$$
Final Answer: $$131.95\text{ mL} \text{ (approx. } 4.46\text{ fl oz)}$$
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Example 7

Civil Engineering: Gravel Stockpile Volume & Angle of Repose

Problem: A gravel conveyor dumps crushed stone into a conical pile with an angle of repose of $35^\circ$ and base diameter of $d = 30\text{ feet}$. Find the total volume of gravel.

Given: $r = 15\text{ ft},\quad \theta = 35^\circ$

Step-by-Step Solution:

  1. Find the vertical height using trigonometry: $$h = r \tan(35^\circ) = 15 \times 0.7002 \approx 10.50\text{ ft}$$
  2. Calculate pile volume: $$V = \frac{1}{3}\pi (15)^2 (10.50) = \frac{1}{3}\pi (225)(10.50) = 787.5\pi \approx \mathbf{2474.0\text{ cu ft}}$$
  3. Convert to cubic yards ($27\text{ ft}^3 = 1\text{ yd}^3$): $$2474.0 / 27 \approx \mathbf{91.6\text{ cubic yards}}$$
Final Answer: $$2474.0\text{ ft}^3 \approx 91.6\text{ yd}^3$$
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Example 8

Industrial Frustum: Hopper Volume with Top & Bottom Openings

Problem: A grain feed hopper is a truncated cone with top radius $R = 2.5\text{ m}$, bottom discharge radius $r = 0.5\text{ m}$, and height $h = 3\text{ m}$. Find its total capacity.

Given: $R = 2.5\text{ m},\quad r = 0.5\text{ m},\quad h = 3\text{ m}$

Step-by-Step Solution:

  1. Apply the frustum volume formula: $$V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$$
  2. Substitute values: $$V = \frac{1}{3}\pi (3) \left(2.5^2 + 0.5^2 + (2.5)(0.5)\right)$$
  3. Simplify bracketed terms: $$2.5^2 = 6.25,\quad 0.5^2 = 0.25,\quad 2.5 \times 0.5 = 1.25$$
  4. Sum terms: $$6.25 + 0.25 + 1.25 = 7.75$$
  5. Final volume: $$V = \pi \times 7.75 \approx \mathbf{24.35\text{ m}^3} \text{ (24,347 Liters)}$$
Final Answer: $$24.35\text{ m}^3 \text{ (approx. } 6432\text{ US gal)}$$
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