Find the area inside one loop of the lemniscate r2 8 sin 2

Find the area inside one loop of the lemniscate r2 = 8 sin 2 theta. The area inside one loop of the lemniscate is . (Simplify your answer.)

Solution

r^2= 8 sin(2t) sin(2t) =0 ==> 2t=Pi ==> t= pi/2 Area inside the one loop = A/2 = (1/2) \\int_{0}^{pi/2} (8 sin(2t))^2 dt A/2 =(1/2) \\int_{0}^{pi/2} 64 ( sin(2t))^2 dt A/2 =(32) \\int_{0}^{pi/2} ( (1- cos(4t) ) /2 ) dt A/2 = 256 pi Area inside the one loop