Wentworth's Plane and Solid Geometry · Book VII, Exs. 678–691 · Book 8: Book VII: Polyhedrons, Cylinders, and Cones
110. Problems of computation: cylinders
Goal: Your student solves practical problems about wells, tunnels, funnels and pails, and writes general formulas for cylinders.
You'll need
- The book, open to these pages
- A ruler (straightedge) and a pair of compasses
- Sharp pencil, eraser and plain paper (or a geometry notebook)
- A calculator (optional)
What's in the book
Page 341: Problems of Computation, Ex. 678–691.
The lesson, step by step
Warm up
Ask: Lateral area, total area and volume of a cylinder of revolution?
Practical problems
Ex. 678. The diameter of a well is 6 feet, and the water is 7 feet deep. How many gallons of water are there in the well, reckoning 7½ gallons to the cubic foot?
Ex. 679. When a body is placed under water in a right circular cylinder 60 centimeters in diameter, the level of the water rises 40 centimeters. Find the volume of the body.
Ex. 680. How many cubic yards of earth must be removed in constructing a tunnel 100 yards long, whose section is a semicircle with a radius of 18 feet?
Finding a radius
Ex. 682. Find the radius of a cylindrical pail 14 inches high that will hold exactly 2 cubic feet.
Remember: a cubic foot is 1728 cubic inches.
Formulas
Your student does two of Ex. 684–691, answering with formulas. Ex. 684 and 686 are a good start.
Wrap up
Check the answers together.
Answers to the book's problems
Ex. 678: about 197.9 cu ft, about 1484 gallons. Ex. 679: 36,000π, about 113,100 cubic centimeters (about 113 liters). Ex. 680: about 5655 cu yd. Ex. 681: 60π, about 188.5 sq ft. Ex. 682: about 8.86 in. Ex. 683: 20 liters = 20,000 cu cm; R ≈ 14.71 cm, H ≈ 29.42 cm. Ex. 684: H = (T − 2πR²)/(2πR). Ex. 685: R = 2V/S, H = S²/(4πV). Ex. 686: V = C²H/(4π). Ex. 687: T = 6πR², so V = 2πR³ with R = √(T/6π). Ex. 689: R = √(V/πH), T = 2V/H + 2√(πVH). Ex. 690: H = ∛(4V/π), T = 1.5πH².
Worked out for this site from the scan. If a number in your copy differs, trust the book.