Someone else correct me if I'm wrong, but if I remember correctly, I believe this is what is happening in your problem:
As theta goes from 0 to 2pi, the polar graph makes two separate loops. The graph actually completes each loop in pi radians, so integrating from 0 to pi gives you the area of one loop (the outer loop, I believe). Integrating from pi to 2pi gives you the area of the second loop, but notice what is happening between pi and 2pi: the graph seems to loop back on its self (instead of the point at pi radians being located on the negative x axis, it appears on the positive x axis (note that I have not graphed this function so I may be mistaken on its orientation)). This represents a negative radius, which means the integrating from pi to 2pi gives a negative area. If you integrate from 0 to 2pi, you actually are getting the area of the first loop minus the magnitude of the area of the second loop. In other words, to get the area you said you need, just integrate from 0 to 2pi.