Say I have a semi circle, oriented counter clockwise, $C_R$ that goes from R on the positive real axis to -R on the negative real axis. I am looking for -
$\int_{C_R}dz$
I have the answer from class as $\pi R$ and I can see from intuition that that is correct, as it is basically the perimeter of a semi circle.
But I when I try to work out the integral I am having trouble. I would like to be able to do it by two separate methods, parameterizing it, and using an antiderivative. Here's my attempts using both methods -
As f(z) = dz has an antiderivative of z I have
$\int_{C_R}dz = z$ ("evaluated from -R to R") $= R - (-R) = 2R$, which is not equal to $\pi R$
This has me confused, I thought there is supposed to be path independence when you do it using an antiderivative. But I cant see how their can be path independence because if f(z) = 1, as it in in this case, the integral of f(z) is the length of the contour. But if you go from -R to R along the axis the length is 2R, yet if you go along the curve it is $\pi R$. So path independence doesn't seem applicable here?
Using the parameterization method I parameterized the curve with $z(t) = Re^{it}, ( 0 <= t <= \pi)$
Then using the integration formula $\int f(z(t))z'(t)$ with
$f(z(t)) = 1$ and $ z'(t) = (R)(it)e^{it}$ I have
$\int$ (from 0 to $\pi$) of $(R)(it)e^{it}$
I dont know where to go from here...
Is anyone able to do this integral using the antiderivative method and the parameterization method? $\int_{C_R}dz$