Dgdrxrt

So, I'm a bit rusty when it comes to multidimensional integrals, but a friend said in an exam protocol there was the question "calculate the flow(flux?) of a constant vector field through the boundary of a closed semisphere." (vague translation).

So my thought was, a constant vector field (in 3d) looks like this:
$F=(x_0,y_0,z_0)^T$, in particular the divergence is zero, so
$$int_partial Semisphere F cdot nu dA=int_Semispherediv F d V=int_Semisphere0dV=0.$$

Then I wanted to actually put some work into it and show that it's zero without Gauss' Theorem. So I said the boundary of the Semisphere can be written as the union of the upper hemisphere (where by "up" I mean $z$-Direction) $U$ with

$$U=(x,y,z)^T$$
and a bottom disc $D$ defined as
$$D= z=0,x^2+y^2<1$$
and their boundary, which is a 2-d-nullset so it can be ignored when integrating. (We assume the Radius to be 1, centered at origin).

As we're pointing from the inside to the outside, the normal Field $nu_D$ surely is $(0,0,-1)^T$ so we have
$$int_D Fcdot nu_D dA=int_D -z_0 dA=-z_0Vol_2(D)=-z_0pi.$$

So far so good. Now the hemisphere $U$ has a normal field of precisely $(x,y,z)^T$, as this point on that sphere is always perpendicular to the sphere and has length 1. With the help of the image on Wikipedia

https://de.wikipedia.org/wiki/Datei:Kugelkoord-def.svg

and the coordinate transform

$$x=sin(theta)cos(varphi)$$
$$y=sin(theta)sin(varphi)$$
$$z=cos(theta)$$

I thought it would be reasonable to get the desired hemisphere by either letting $varphi in [0,2pi]$ and $theta in [0,pi/2]$, which I'll call Version 1, or letting $varphi in [0,pi]$ and $theta in [-pi/2,pi/2]$, which I'll call version 2. If I've done nothing wrong so far, the result should be $pi*z_0 $ in both cases, since they must add up to 0, right? So let's start with version 1:

$$int_U F cdot nu_U dS=int_U x x_0 + y y_0 + z z_0 dS=$$
$$=int_theta in [0,pi/2],varphi in [0,2pi]sin(varphi)[x_0sin(theta)cos(varphi) + y_0sin(theta)sin(varphi)+z_0cos(theta)]dvarphi dtheta.$$

Now $sin(x)*cos(x)$ as well as $sin(x)$ have $2pi$ periodic antiderivatives, so these terms vanish (already bad) and we're left with

$$y_0int_0^pi/2sin(theta)dtheta int_0^2pisin(varphi)sin(varphi)dvarphi=y_0 cdot 1 cdotpi$$

because I can't be bothered not to put it into wolfram alpha. Close, but no cigar.

So, let's try version 2:

$$int_theta in [-pi/2,pi/2],varphi in [0,pi]sin(varphi)[x_0sin(theta)cos(varphi) + y_0sin(theta)sin(varphi)+z_0cos(theta)]dvarphi dtheta$$

where we integrate the odd sin-Function around the origin (in the $theta$ integral), so only the $z$-Term remains. So now we're stuck with

$$z_0 int_-pi/2^pi/2cos(theta)dtheta int_0^pi sin(varphi)dvarphi=z_0 cdot 2 cdot 2= 4z_0.$$

Funnily enough, that is numerically quite close, but still not right. I must have messed up several times. What am I doing wrong?