But if the original square had an initial surface of S2 =1 sq units, that is, side of 1 times side of 1 = 1 sq unit, how can the resulting shape, that is, distorting a square to form a hyperbolic paraboloid, have a surface area of 5.708 sq units??
But if the original square had an initial surface of S2 =1 sq units, that is, side of 1 times side of 1 = 1 sq unit, how can the resulting shape, that is, distorting a square to form a hyperbolic paraboloid, have a surface area of 5.708 sq units??
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Thanks for prompt response. But if the original square had an initial surface of S2 =1 sq units, that is, side of 1 times side of 1 = 1 sq unit, how can the resulting shape, that is, distorting a square to form a hyperbolic paraboloid, have a surface area of 5.708 sq units??
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Is it pi/8 times the square root of 2 to the 3rd power, that is 8, minus one in that order??
Using that sequence i don't get 1.166
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