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:
- Write down the standard cone volume formula: $$V = \frac{1}{3}\pi r^2 h$$
- Square the radius: $$r^2 = 6^2 = 36\text{ cm}^2$$
- Multiply by the vertical height and $\pi$: $$36 \times 10 \times \pi = 360\pi\text{ cm}^3$$
- Divide by 3: $$V = \frac{360\pi}{3} = 120\pi \approx \mathbf{376.99\text{ cm}^3}$$
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:
- Find the radius by dividing diameter by 2: $$r = \frac{d}{2} = \frac{14}{2} = 7\text{ m}$$
- Apply the volume formula: $$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (7)^2 (9)$$
- Simplify: $$V = \frac{1}{3}\pi (49)(9) = 49 \times 3 \times \pi = 147\pi\text{ m}^3$$
- Calculate decimal value: $$V \approx 147 \times 3.14159265 = \mathbf{461.81\text{ m}^3}$$
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:
- Use the Pythagorean theorem to calculate vertical height $h$: $$h = \sqrt{l^2 - r^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ in}$$
- Substitute $r = 5$ and $h = 12$ into the volume formula: $$V = \frac{1}{3}\pi (5)^2 (12)$$
- Simplify the terms: $$V = \frac{1}{3}\pi (25)(12) = 25 \times 4 \times \pi = 100\pi\text{ in}^3$$
- Final decimal answer: $$V \approx \mathbf{314.16\text{ in}^3}$$
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:
- Rearrange the volume formula $V = \frac{1}{3}\pi r^2 h$ to solve for $h$: $$h = \frac{3V}{\pi r^2}$$
- Calculate the denominator (Base Area $B$): $$B = \pi r^2 = \pi (4^2) = 16\pi \approx 50.2655\text{ cm}^2$$
- Multiply the volume by 3: $$3V = 3 \times 300 = 900$$
- Divide to find $h$: $$h = \frac{900}{16\pi} = \frac{900}{50.2655} \approx \mathbf{17.90\text{ cm}}$$
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:
- Rearrange the volume formula to solve for $r$: $$r = \sqrt{\frac{3V}{\pi h}}$$
- Multiply volume by 3: $$3V = 3 \times 1000 = 3000$$
- Calculate denominator: $$\pi h = \pi \times 15 \approx 47.1239$$
- Compute radius: $$r = \sqrt{\frac{3000}{47.1239}} = \sqrt{63.662} \approx \mathbf{7.98\text{ cm}}$$
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:
- Calculate radius: $$r = 6 / 2 = 3\text{ cm}$$
- Compute volume in cubic centimeters: $$V = \frac{1}{3}\pi (3)^2 (14) = 3 \times 14 \times \pi = 42\pi \approx 131.95\text{ cm}^3$$
- Since $1\text{ cm}^3 = 1\text{ mL}$, the capacity is exactly: $$\mathbf{131.95\text{ mL}}$$
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:
- Find the vertical height using trigonometry: $$h = r \tan(35^\circ) = 15 \times 0.7002 \approx 10.50\text{ ft}$$
- Calculate pile volume: $$V = \frac{1}{3}\pi (15)^2 (10.50) = \frac{1}{3}\pi (225)(10.50) = 787.5\pi \approx \mathbf{2474.0\text{ cu ft}}$$
- Convert to cubic yards ($27\text{ ft}^3 = 1\text{ yd}^3$): $$2474.0 / 27 \approx \mathbf{91.6\text{ cubic yards}}$$
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:
- Apply the frustum volume formula: $$V = \frac{1}{3}\pi h (R^2 + r^2 + Rr)$$
- Substitute values: $$V = \frac{1}{3}\pi (3) \left(2.5^2 + 0.5^2 + (2.5)(0.5)\right)$$
- Simplify bracketed terms: $$2.5^2 = 6.25,\quad 0.5^2 = 0.25,\quad 2.5 \times 0.5 = 1.25$$
- Sum terms: $$6.25 + 0.25 + 1.25 = 7.75$$
- Final volume: $$V = \pi \times 7.75 \approx \mathbf{24.35\text{ m}^3} \text{ (24,347 Liters)}$$