What is the area enclosed by #r=2sin(4theta+(11pi)/12) # between #theta in [pi/8,(pi)/4]#?

1 Answer
Jun 21, 2016

# (pi + 1)/8#

Explanation:

It's the red segment.
Desmos

Plug and play

#A = 1/2 \int_{\theta = pi/8}^{pi/4} \ r^2(theta) \ d theta #
# = \int_{\theta = pi/8}^{pi/4} \ 2sin^2 (4theta+(11pi)/12) \ d theta#

using # cos 2A = 1 - 2 sin^2 A#

# = \int_{\theta = pi/8}^{pi/4} \ 1 - cos (8theta+(11pi)/6) \ d theta#

# = [ \ theta - 1/8 sin (8theta+(11pi)/6) ]_{\theta = pi/8}^{pi/4}#

# = [ \ pi/4 - 1/8 sin (2 pi+(11pi)/6) ] - [ \ pi/8 - 1/8 sin ( pi+(11pi)/6) ]#

# = pi/8 + 1/8 [ sin ( pi+(11pi)/6)- sin (2 pi+(11pi)/6) ]#

# = pi/8 + 1/8 [ sin ( pi) cos ((11pi)/6) + cos ( pi) sin ((11pi)/6) - sin (2 pi) cos ((11pi)/6) - cos (2 pi) sin ((11pi)/6) ]#

# = pi/8 + 1/8 [ (-1) (- 1/2) - (1)(-1/2) ]#

# = (pi + 1)/8#