I've been trying to find a way to do this: given a point $P(\alpha,\beta)$, a function $f(x)$, and an angle $\theta$, find the area of the sector determined by extending a horizontal line from $P$ to $f$, and then another one from $P$ to $f$ at the angle $\theta$ from the first line. If you look at this picture:
it's a little more clear what I'm talking about. I want to find the area of that sector $S$, which is determined by that triangle $R_1$ and the other region $R_2$. Let the value of $x$ where the horizontal line from $P$ hits $f$ be $q$ (that is, $q=f^{-1}(\beta)$). Now, the line $\overline{PA}$ has the equation $y=\tan\theta(x-\alpha)+\beta$. The intersection of that and $f$, the point $A$ (the first intersection point), is the beginning of the interval determining $R_2$; I'll call the root $p$ of $f(x)=\tan\theta(x-\alpha)+\beta$ that value of $x$, so $p=f^{-1}(\tan\theta(x-\alpha)+\beta)$. The area of $R_1$ is simply the area of the triangle: $R_1=\frac{1}{2}(p-\alpha)(f(p)-\beta)$. The area of $R_2$ is the area under the curve from $p$ to $q$, minus the initial height $\beta$: $R_2=\displaystyle\int_p^q (f(x)-\beta) dx$. Therefore, the total area is $$S=\frac{1}{2}\tan\theta(p-\alpha)^2-\beta(q-p)+\displaystyle\int_p^q f(x) dx$$
This is the way I did it, but there has to be a simpler way; is there one involving polar equations? Is there a general formula when $f$ is non-invertible?
Edit: If the function $f$ is translated so that $P$ is now the origin, the area is just $\displaystyle\frac{1}{2}\int_0^\theta r^2 d\phi$ where $r(\phi)$ is $f(x)$ in polar form. However, translating $f$ over to where $P$ is on the origin turns it into $f(x+\alpha)-\beta$. Is there a nice way of putting that into polar form?