Given: #r=cos((2theta)/3-(pi)/8)+sin((7theta)/8+(pi)/4)#
Required: The area enclosed by #r(theta), theta in [0,pi]#
Solution Strategy: Use the Area Integral in polar coordinated give by:
#A_o. = int_(theta_1)^(theta_2)1/2r^2 d"theta#
#=1/2int [cos((2theta)/3-(pi)/8)+sin((7theta)/8+(pi)/4)]^2dx#
Expand the square:
#=1/2int [cos^2((2theta)/3-(pi)/8)+sin^2((7theta)/8+(pi)/4)+2cos()sin()dx=1/2(I_1+I_2+I_3)# Apply linearity separate each component into, #I_1+I_2+I_3)# respectively. I am going to work with x it is easy to type instead of theta, OK:
#I_1= 1/2int cos^2((2x)/3-(pi)/8)dx#
from integral tables we see that:
#I_1=[x/4+3/16sin(4/3x-pi/4)]_0^pi#
#I_1=pi/4+3/16sin(13/12pi)-(3/16sin(-pi/4))#
#I_1=pi/4+3/16[sin(13/12pi)+sinpi/4]~~0.86945#
#I_2= 1/2int sin^2((7x)/8+(pi)/4)dx# This similar to the above
#I_2=[x/8-1/14sin(7/4x-pi/4) ]_0^pi#
#I_2= pi/8-1/14sin(9/4pi )+1/14 ~~0.8272#
#I_3=intcos((2x)/3-(pi)/8)sin((7x)/8+(pi)/4)dx#
For this one we will us the trigonometric identity:
#sin(ax+b)cos(cx-d)=1/2{sin[(ax+b)+(cx-d)] + sin[(ax+b)-(cx-d)]}#
We have reduce #I_3# into two simpler integrals,
#I_(3.1)= 1/2int sin(37/24x+pi/8)dx#
# = -24/37 [cos(37/24x+pi/8)]_0^pi #
#I_(3.2)=1/2int sin(5/24x+3/8pi) dx#
# =-24/5 [sin(5/24x+3/8pi)]_0^pi#
#I_3= [-24/37 cos(37/24x+pi/8 ) -24/5 sin(5/24x+3/8pi)]_0^pi = 1.677#
And #A_(o.)=.86945+.8272+1.677=3.37365 #
