Consider the curve $C':=(t, t^2)$ on the interval $t\in [0,1]$. A little calculation shows that the formula $\int \limits_{0}^{1} y \cdot dx = \int \limits _{0}^{1} t^2 \cdot dt$ gives an area of $1/3$. This is the area between the curve and the $x$-axis. Similarly $\int \limits _{0} ^{1} x \cdot dy = \int \limits_{0}^{1} t\cdot 2t \cdot dt = 2/3$. We can interpret this as area of the region between the curve and the $y$-axis.
Now consider the region $R$ bounded by the line $y=x$ and $C$. We know from geometry that $R$ has area $1/2 - 1/3 = 1/6$. Another way to get it is to do $$ \dfrac{1}{2} \int \limits _{0}^{1} x \cdot dy - y \cdot dx \tag{1} $$ which gives us the same value for the area between $C'$ and a chord connecting its endpoints. Is this happy coincidence? I am inclined to think not because what we essentially did is halve the area of the leaf like region bounded by $(t,t^2)$ and $(t^2,t)$.
So I thought I had a formula for the area enclosed by a curve when it subtends a simple chord across its endpoints (by simple here I mean the chord doesn't intersect the curve itself ... so we are at this stage avoiding signed areas). But it does not work for $\left(\cos(t), \sin(t)\right), t \in [0, \pi/2]$.
$$1/2 \cdot \int \limits _{0}^{\pi/2} x dy - y dx = 1/2 \cdot \int \limits _{0}^{\pi/2} \cos^2(t) + \sin^2(t) = \pi/4 $$
We seem to be simply adding up the area between the quarter circle & the $x$-axis and the quarter circle & $y$-axis and halving it.
So my question is: Is there a general formula for the area subtended by a curve $C = \left( f(t), g(t) \right), t \in [a, b]$ and a chord connecting the point $\left(f(a), g(a)\right)$ with $\left(f(b), g(b)\right)$ (maybe by way of Green's Theorem, e.g. we use the chord to close up the curve with a natural orientation)?