Given a sphere with: $$F := \{(x,y,z) \in \mathbb{R}^3 \mid x^2+y^2+z^2 = 1, x\le0\}$$ $$ \Rightarrow r = 1, \varphi = [\frac{\pi}{2}, \frac{3\pi}{2}], \theta = [0, \pi] $$
My Task is to calculate the surface area by using surface integral.
The default sphere paramterization: $$ X\left(\varphi,\theta\right)= \begin{pmatrix} \cos \varphi\sin \theta\\ \sin \varphi\sin \theta\\ \cos \theta \end{pmatrix} $$
And it's derivates: $$ X_{\varphi} \times X_{\theta} = \begin{pmatrix} -\sin \varphi\sin\theta\\ \cos \varphi\sin\theta\\ 0 \end{pmatrix} \ \ \times \ \ \begin{pmatrix} \cos \varphi\cos\theta\\ \sin \varphi\cos\theta\\ -\sin\theta \end{pmatrix}\ = \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix}$$
The euclidean norm: $$ \| \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix} \| = \sin^2\theta $$
$$S(F)=\int_0^{\pi}\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\sin^2\theta \; d\varphi d\theta = \frac{\pi^2}{2}$$
Now my question is, where did I make a mistake? Since the surface should the surface of half a sphere, being $1/2 * 4\pi * 1^2 = 2\pi$
Your euclidean norm is wrong, you most likely forgot about the square root. It should be:
$$ \left\| \begin{pmatrix} -\cos\varphi\sin^2\theta\\ -\sin\varphi\sin^2\theta\\ -\sin\theta\cos\theta \end{pmatrix} \right\| = \sqrt{(\cos^2\varphi+\sin^2\varphi)\sin^4\theta + \sin^2\theta\cos^2\theta} = \sqrt{(\sin^2\theta + \cos^2\theta)\sin^2\theta} = \sqrt{\sin^2\theta} = \sin\theta$$
And the integral:
$$S(F)= \left(\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}}d\varphi\right) \cdot \left( \int_0^{\pi} \sin\theta d\theta\right) = \pi \cdot 2 = 2\pi$$
By the way, writing an independent variable as the innermost integral and with such weird boundaries is really bad style and confusing for the reader.