I'm trying to plot the graph r = sin 3t and find its area. This is how far I've gotten: The graph looks like a plane propellor with one propellor pointing downward, and two pointing up-left / up-right, with the length of each equal to 1. Now to get the area... I have to figure out where the graph reaches 0... which I'm not sure how to do (I solve sin3t = 0 and get t = 0, pi, 2pi ?? )... or... I can just get the area from 0 to 2pi... so i'll try that: A = intergral of (1/2 (sin 3t)^2 dt) from 0 to 2pi the solution is: 1/2( t/2 - 1/12 sin(6t) ) If I plug in 0 or 2pi into this and solve, the last term becomes 0 and I get a weird answer... what do I do?!?