I'm trying to understand the complex line integral equation as given in these notes. It's given as
$$\int_{\gamma} f(z) dz = \int_a^b f(\gamma(t)) \gamma'(t) dt$$
where $\gamma(t)$ is a paremeterization of the curve $\gamma$ with $t \in [a, b]$.
For the purposes of understanding the math, I'm trying to use it to find the area of a circle of radius $r$ centered at the origin using the paramerization $\gamma(t) = r e^{it}$.
Intuitively I would expect the area of the circle to be the sum of the area of the triangles with altitude $|z|$ and width $dz$.
However, I end up with $0$:
$$ \int_\gamma f(z) dz = \int_\gamma \frac{1}{2}|z| dz = \\ \frac{1}{2} \int_0^{2\pi} |r e^{it}| r i e^{it} dt = \\ \frac{1}{2} r^2 i \int_0^{2\pi} |e^{it}| e^{it} dt = \\ \frac{1}{2} r^2 i \int_0^{2\pi} e^{it} dt = \\ \frac{1}{2} r^2 i \cdot 0 = 0$$
Obviously the area of a cirle isn't $0$. What am I doing wrong?
Edit:
By way of comparison, in $\mathbb{R}^2$, the line integral is expressed as
$$ \int_C f(x) ds = \int_a^b f(x(t)) \cdot |x'(t)| dt$$
where $x(t)$ is an arbitrariy parameterization of $C$ and $t \in [a, b]$.
With the same logic as above, if we let $f(x) = |x|$ and $x(t) = <r\cos(t), r\sin(t)>$ we get:
$$ \int_a^b f(r(t)) \cdot |r'(t)| dt \\ = \int_0^{2\pi} \frac{1}{2} |\sqrt{r^2 \cos^2(t) + r^2 \sin^2(t)}| \cdot |\sqrt{r^2 (-\sin(t))^2 + cos^2(t)}| dt \\ = \frac{1}{2} r^2 \int_0^{2\pi} |1| \cdot |1| dt \\ = \frac{1}{2} r^2 (2 \pi - 0) \\ = \pi r^2$$
Which is correct for the area of a circle.
I don't understand why this rough idea seems to work in $\mathbb{R}^2$ and not in $\mathbb{C}$.
The correct formula is that the (oriented) area of the domain (circle here) enclosed by (simple rectifiable Jordan) $\gamma$ is $\frac{1}{2i}\int_{\gamma}\bar w dw=\frac{1}{2i}\int_0^{2\pi}re^{-it}ire^{it}dt=\pi r^2$ so we are good
(here by abuse of notation $\gamma=w$ of course so if you want the integrand is $\bar \gamma \gamma'$)