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
