Find the area inside one leaf of the rose r3sin3theta The ar

Solution

dA = (r*dT)dr = (r*dr)dT We can integrate first in r and this is done form r=0 to r=3*sin(3*T) Integrate{(r*dr)dT} = (r^2/2)dT Evaluate this to get: (1/2)[3*sin(3*T)]^2 dT = (9/2)*sin^2(3*T) dT Now use the trig identity: sin^2(u) = [1 - cos(2*u)]/2 (9/2)*sin^2(3*T) dT = (9/4)*[1 - cos(6*T)] dT And integrate this to get: (9/4)*[T - sin(6*T)/6] Evaluate this from T = 0 to T = pi/3 This gives: Area =.1744 The area is 7pi/12