Given: r=cos((2theta)/3-(pi)/8)+sin((7theta)/8+(pi)/4)r=cos(2θ3−π8)+sin(7θ8+π4)
Required: The area enclosed by r(theta), theta in [0,pi]r(θ),θ∈[0,π]
Solution Strategy: Use the Area Integral in polar coordinated give by:
A_o. = int_(theta_1)^(theta_2)1/2r^2 d"thetaA⊙=∫θ2θ112r2dθ
=1/2int [cos((2theta)/3-(pi)/8)+sin((7theta)/8+(pi)/4)]^2dx=12∫[cos(2θ3−π8)+sin(7θ8+π4)]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)=12∫[cos2(2θ3−π8)+sin2(7θ8+π4)+2cos()sin()dx=12(I1+I2+I3) Apply linearity separate each component into, I_1+I_2+I_3)I1+I2+I3) 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)dxI1=12∫cos2(2x3−π8)dx
from integral tables we see that:
I_1=[x/4+3/16sin(4/3x-pi/4)]_0^piI1=[x4+316sin(43x−π4)]π0
I_1=pi/4+3/16sin(13/12pi)-(3/16sin(-pi/4))I1=π4+316sin(1312π)−(316sin(−π4))
I_1=pi/4+3/16[sin(13/12pi)+sinpi/4]~~0.86945I1=π4+316[sin(1312π)+sinπ4]≈0.86945
I_2= 1/2int sin^2((7x)/8+(pi)/4)dxI2=12∫sin2(7x8+π4)dx This similar to the above
I_2=[x/8-1/14sin(7/4x-pi/4) ]_0^piI2=[x8−114sin(74x−π4)]π0
I_2= pi/8-1/14sin(9/4pi )+1/14 ~~0.8272I2=π8−114sin(94π)+114≈0.8272
I_3=intcos((2x)/3-(pi)/8)sin((7x)/8+(pi)/4)dxI3=∫cos(2x3−π8)sin(7x8+π4)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)]}sin(ax+b)cos(cx−d)=12{sin[(ax+b)+(cx−d)]+sin[(ax+b)−(cx−d)]}
We have reduce I_3I3 into two simpler integrals,
I_(3.1)= 1/2int sin(37/24x+pi/8)dxI3.1=12∫sin(3724x+π8)dx
= -24/37 [cos(37/24x+pi/8)]_0^pi =−2437[cos(3724x+π8)]π0
I_(3.2)=1/2int sin(5/24x+3/8pi) dxI3.2=12∫sin(524x+38π)dx
=-24/5 [sin(5/24x+3/8pi)]_0^pi=−245[sin(524x+38π)]π0
I_3= [-24/37 cos(37/24x+pi/8 ) -24/5 sin(5/24x+3/8pi)]_0^pi = 1.677I3=[−2437cos(3724x+π8)−245sin(524x+38π)]π0=1.677
And A_(o.)=.86945+.8272+1.677=3.37365 A⊙=.86945+.8272+1.677=3.37365
