For this problem why did we integrated from -90degre to 90degree, instead of integrating from 90degree to 270 degree? Is not the half of the cadroid fall into first and fourth quadrant? Example: Find the length of the cardioid r= 1 + sin 0, 002T.Fund length fromandmultiply yfwo2Cr-TrI+2sin+siiptca- 2+ 2sneL =deT(22+2sine dD2T(2lenth I+S\ns 2L 2( 8S2+25in2-2N2-2sinDde2-25inJ4-4SinS2- 2sin&dtlet72h= 2-2sinDdu -2Co9d2 Cos8de2-25inD-212-2sin2 vS-du411FIN

To discuss the suggested solution for the length of the given curve

Answered: Example: Find the length 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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For this problem why did we integrated from -90degre to 90degree, instead of integrating from 90degree to 270 degree? Is not the half of the cadroid fall into first and fourth quadrant?

Example: Find the length of the cardioid r
= 1 + sin 0, 0
02T.
Fund length from
and
multi
ply yfwo
2
Cr
-Tr
I+2sin+siiptca
- 2+ 2sne
L =
de
T(2
2+2sine dD
2
T(2
lenth I+S\n
s 2L 2( 8
S2+25in2-2
N2-2sinD
de
2-25in
J4-4Sin
S
2- 2sin&
dt
let
72
h= 2-2sinD
du -2Co9d
2 Cos8
de
2-25inD
-212-2sin
2 v
S-du
4
11
FIN

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