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[ S = 4\pi^2 R r ]
[ V = 2\pi^2 R r^2 ]
It sounds like you’re referring to the — a common playful name for a math problem involving torus volume or surface area , often found in calculus or geometry classes.
[ V = (\pi r^2) \times (2\pi R) = 2\pi^2 R r^2 ] For surface area ( S ):
This comes from Pappus’s theorem: Volume = (cross-sectional area of tube) × (distance traveled by its centroid) Cross-sectional area of tube = ( \pi r^2 ) Distance traveled by centroid = ( 2\pi R ) So:
Derivation: Perimeter of tube cross-section = ( 2\pi r ) Distance traveled by its centroid = ( 2\pi R ) Product = ( (2\pi r)(2\pi R) = 4\pi^2 R r ). Example 1: ( R = 5 ) cm, ( r = 1 ) cm [ V = 2\pi^2 (5)(1^2) = 10\pi^2 \ \text{cm}^3 ] [ S = 4\pi^2 (5)(1) = 20\pi^2 \ \text{cm}^2 ]
[ S = 4\pi^2 R r ]
[ V = 2\pi^2 R r^2 ]
It sounds like you’re referring to the — a common playful name for a math problem involving torus volume or surface area , often found in calculus or geometry classes. Law Of The Donut Math Answer Key
[ V = (\pi r^2) \times (2\pi R) = 2\pi^2 R r^2 ] For surface area ( S ): [ S = 4\pi^2 R r ] [
This comes from Pappus’s theorem: Volume = (cross-sectional area of tube) × (distance traveled by its centroid) Cross-sectional area of tube = ( \pi r^2 ) Distance traveled by centroid = ( 2\pi R ) So: Law Of The Donut Math Answer Key
Derivation: Perimeter of tube cross-section = ( 2\pi r ) Distance traveled by its centroid = ( 2\pi R ) Product = ( (2\pi r)(2\pi R) = 4\pi^2 R r ). Example 1: ( R = 5 ) cm, ( r = 1 ) cm [ V = 2\pi^2 (5)(1^2) = 10\pi^2 \ \text{cm}^3 ] [ S = 4\pi^2 (5)(1) = 20\pi^2 \ \text{cm}^2 ]
