Evaluating the complex analysis integral by parametrization Evaluating the integral below
$\int_\gamma \sqrt{z}dz \;$ where $\gamma =\{z\in C : \lvert z\rvert = 3$}
I think I've some mistakes. Can you check my process and tell me where do I have a mistake or mistakes?
I parameterized by thinking it is a circle;
z(t) = 3$e^{it}$ and i thought t: 0 $\le t\le2\pi$
But I do not know the direction is counterclockwise or not that would change the parameterization sign
$$\int_\gamma f(z)\mathrm dz:=\int_a^b \gamma'(t) f(\gamma(t))\mathrm dt.$$
$$\int_0^{2\pi}\sqrt{3e^{it}}3\mathrm i3e^{it}dt$$
I did not write the rest of the integral calculations. I just need to find out where do I have a mistake So I can move on by that. Thank you.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\oint_{\verts{z}\ =\ 3}\root{z}\,\dd z & =
-\int_{-3}^{0}\root{-x}\expo{\ic\pi/2}\dd x
-\int_{0}^{-3}\root{-x}\expo{-\ic\pi/2}\dd x
\\[5mm] & =
-\ic\int_{0}^{3}\root{x}\dd x
+\pars{-\ic}\int_{0}^{3}\root{x}\dd x =
\left. -2\ic\,{x^{3/2} \over 3/2}\,\right\vert_{\ 0}^{\ 3} =
\\[5mm] & = \bbx{-4\root{3}\ic} \\ &
\end{align}
