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

Question

William Abdalla · Accepted Answer

#=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#

Aurora Alvarado · Answer

undefined To find the area enclosed by $ r = -\cos\left(\theta - \frac{7\pi}{4}\right) $ between $ \theta = \frac{4\pi}{3} $ and $ \theta = \frac{5\pi}{3} $, you integrate $ \frac{1}{2} r^2 $ with respect to $ \theta $ from the lower bound to the upper bound. The integral will yield the desired area.