I'm trying to find the region inside r=sinθ and outside r=1+cosθ. My issue is my limits of integration. I get an intersection at $\frac π2$ and one at the pole. What are my limits for the integral? sinθ will intersect the pole at 0 or π but 1+cosθ only intersects the pole at π. From looking at the graph this is integral I came up with:
$$\frac12\int_{\frac π2}^π(sinθ)^2-(1+cosθ)^2dθ$$ Is this correct?
This is the picture:
How do I deal with intersections occurring at different angles?
Edit: New picture, the red region is what I'm looking for.
For $\pi /2 \leq \theta \leq \pi$, which is the region you want, the function $r = F(\theta) = \sin \theta$ is your “outer” function, and $r = g(\theta) = 1 + \cos \theta$ is your “inner” function. Here, you have two options: you can integrate $F(\theta)-g(\theta)$ in one motion, as you have done, or you can find the areas bounded by $F(\theta)$ and $g(\theta)$ and subtract them. Either way, they perfectly equivalent.
You will want to use the polar area formula $A = \frac{1}{2} \int_\alpha^\beta r^2 \, d\theta$ for each curve:
$$\begin{align} A &= \frac{1}{2} \int_\alpha^\beta \left( F^2- g^2 \right) \, d\theta \\ &= \frac{1}{2}\int_{\pi/2}^{\pi} \left[ (\sin\theta)^2 - (1 + \cos\theta)^2 \right] \, d\theta \end{align}$$
A quick check with the calculator tells me this would reduce to $$\ \left[ -\theta - 2\sin(\theta) - \sin(\theta)\cos(\theta)\right]_{\pi/2}^\pi$$ so I am going to make an educated guess and say that this was a with-calculator problem.
So, to conclude, your integral was spot on!