What is the area enclosed by #r=2sin(theta+(7pi)/4) # between #theta in [pi/8,(pi)/4]#?

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Answer 1 · 2018-01-04T18:01:46

#= pi/8 -1/(2sqrt2)#

Area of a polar curve is #int_a^b 1/2 r^2 d theta#

Accordingly, in the present case it would be #int_(pi/8) ^ (pi/4) 1/2 *4sin^2 (theta + (7pi)/4) d theta#

#=int_(pi/8)^(pi/4) (1- cos(2 theta + (7pi)/2) d theta#

#[theta -1/2 sin(2theta +(7pi)/2)]_(pi/8)^(pi/4)#

#= (pi/4 -1/2 sin(pi/2 +(7pi)/2)-(pi/8 -1/2sin(pi/4+(7pi)/2)#

#(pi/4 -pi/8 +1/2 sin(4pi -pi/4))#

#=pi/8 +1/2 sin (- pi/4)#

#= pi/8 -1/(2sqrt2)#