What is the area enclosed by #r=-cos(theta-(7pi)/4) # between #theta in [4pi/3,(5pi)/3]#?

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Answer 1 · 2017-02-17T16:57:42

#=pi/12-1/8sqrt1.5=0.1087#, nearly.

The graph is a circle of radius 1/2, through pole, with center on

#theta=5/4pi#.

Despite that the non-negative r = sqrt(x^2+y^2) is #-cos15^o < 0#, at

the higher limit #theta=5/3pi#, I work out for the answer.

graph{(sqrt2(x^2+y^2)+x+y)(y+sqrt3x)(y-sqrt3x)=0 [-2.5, 2.5, -1.25, 1.25]}

Area= #1/2 int r^2 d theta#, with# r =-cos(theta-7/4pi) and theta# f

from #4/3pi# to #5/3pi#

#=1/2 int cos^2(theta-7/4pi) d theta#, for the limis

#=1/4int(1+cos(2theta-7/4pi) d theta#, for the limite

#=1/4[theta+1/2sin(2theta-7/4pi)]#, between #theta= 4/3pi and 5/3pi#

#=1/4[(5/3pi-4/3pi)+1/2(sin(19/12pi)-sin(11/12pi)]#

#=1/4(pi/3+cos(5/4pi)sin(pi/3))#

#=pi/12+1/4((-1/sqrt2)(sqrt3/2))#

#=pi/12-1/8sqrt1.5#