Find the area that lies inside the polar curve r6sint and ou

Solution

they intersect at 6sint = 3sqrt(2) => sin t = 1/sqrt(2) t = pi/4 , 3pi/4 area between the curves = integ (6sint)^2 - (3sqrt(2))^2 dt from pi/4 to 3pi/4 = integ 36 sin^2 t - 18 dt from pi/4 to 3pi/4 = 18 integ 2sin^2 t - 1 dt from pi/4 to 3pi/4 = 18 integ -cos 2t dt from pi/4 to 3pi/4 = - 18 sin 2t /2 from pi/4 to 3pi/4 = -9 sin 2t from pi/4 to 3pi/4 = -9 [sin 3pi/2 - sin pi/2] = -9[ -1 - 1] = 18 units area outside curve 2 and in curve 1 = area of curve 1 - 18 units radius of curve one is 3 units pi(3)^2 - 18 = 9pi - 18 units