I'm trying to calculate the vector integral over a hemi-sphere, i.e. the integral of all vectors from the origin to a hemisphere. Let's say the $xy$ cuts the sphere in two. I thought the following rough idea could work:
I calculate the vector integral over a half circle with an analogy to complex numbers and then rotate the result from $0$ to $\pi$.
For the first part I get: $$ \int_0^{\pi} \exp(i\phi) d\phi=\left[\frac{\exp(i\phi)}i \right]_0^{\pi}=\frac{-1}i-\frac1i=2i $$ Reinterpreted in my real world scenario this means a vector of lenght 2 along the $z$ axis.
For the second part I get: $$ \int_0^{\pi} 2d\theta = 2\pi $$
So I conclude that the vector integral results in a vector of length $2\pi$ along the $z$-axis, right?
A direct computation of the vector integral shows the correctness of my result:
$$ \int_0^{2\pi} \int_0^{\pi/2} \pmatrix{\cos\alpha&-\sin\alpha& \\ \sin\alpha& \cos\alpha& \\ & & 1} \pmatrix{\cos\beta& &\sin\beta\\ & 1 & \\ -\sin\beta& &\cos\beta} \pmatrix{1\\ 0\\ 0} d\beta d\alpha = \dots = \pmatrix{0 \\ 0 \\ 2\pi} $$