Answers

2015-12-15T22:05:10+05:30
 r=a(1+cos x) 
r^2=a^2(1+cos x)^2 = a^2 + 2[a^2][cos(x)] + [a^2][cos^2(x)] 
dr/dx = r' = -a sin(x) 
(r')^2 = [a^2][sin^2 (x)] 
Therefore perimeter (s) of curve r=a(1+cos x) in polar coordinate with x vary from 0 to Pi (due to curve is symmetry on axis x=0) is 
s = 2 Int { sqrt[r^2 + (r')^2] } dx where x vary from 0 to Pi. 
Thus 
sqrt[r^2 + (r')^2] 
= sqrt { a^2 + 2[a^2][cos(x)] + [a^2][cos^2 (x)] + [a^2][sin^2 (x)] } 
= sqrt { (2a^2)[1+cos(x)] } 
= [sqrt(2)]a {sqrt [1+cos(x)]} 
Then 
s = 2[sqrt(2)]a . Int {sqrt [1+cos(x)]} dx 
Let 1+cos(x) = 1+2cos^2 (x/2) - 1 = 2cos^2 (x/2) 
s = 2[sqrt(2)]a . Int {sqrt [2cos^2 (x/2)]} dx 
s = 4a . Int [cos(x/2)] dx where x vary from 0 to Pi 
s = 4a [sin(x/2)]/(1/2) 
s = 8a [sin(Pi/2) - sin(0)] 
s = 8a 
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