5 5. Use an integral to find the area of the surface parameterized byr(u, v) = (u cos v, u sin v, u²),u [1,1], v [0, π].

Answered: Use an integral to find the area of the…

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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Question

5. Use an integral to find the area of the surface parameterized by
r(u, v) = (u cos v, u sin v, u²),
u [1,1], v [0, π].

Expert Solution

Step 1: Set up the integral

The surface parameterized by

$r open parentheses u comma v close parentheses space equals space left parenthesis space u space cos space v comma space u space sin space v comma space u squared space right parenthesis$ , $u space element of space open square brackets negative 1 comma space 1 close square brackets space comma space space v space element of space open square brackets 0 comma space straight pi close square brackets$

$r subscript u space equals space fraction numerator partial differential r over denominator partial differential u end fraction space equals space open parentheses cos space v comma space sin space v space comma space 2 u close parentheses$

$r subscript v space equals space fraction numerator partial differential r over denominator partial differential v end fraction space equals space open parentheses space minus u space sin space v comma space u space space cos space space v space comma space space 0 close parentheses$

$r subscript u space cross times space r subscript v space equals space open vertical bar table row i j k row cell cos space v end cell cell sin space v end cell cell 2 u end cell row cell negative u space sin space v end cell cell u space cos space v end cell 0 end table close vertical bar$

$equals space i space open square brackets space 0 minus 2 u squared cos space v space close square brackets space minus space j open square brackets space 0 plus 2 u to the power of 2 space end exponent sin space v space close square brackets space plus k open square brackets space u space cos squared v space plus u sin squared space v space close square brackets$

$equals space left parenthesis space minus 2 u squared space cos space v space comma space minus 2 u squared space sin space v comma space u space right parenthesis$

$open vertical bar r subscript u space cross times space r subscript v close vertical bar space equals space square root of open parentheses negative 2 u squared space cos space v close parentheses squared plus open parentheses negative 2 u squared space sin space v close parentheses squared plus open parentheses u close parentheses squared space end root$

$equals space square root of 4 u to the power of 4 plus u squared end root$

$equals space u square root of 1 plus open parentheses 2 u close parentheses squared end root$