Question
Asked Oct 31, 2019
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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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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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