Flux through a section of a sphere Given $\textbf{F} = \langle 0,x,0 \rangle $ and $x^2 + y^2 + z^2 = a^2 $ where $ x\geq0, y\geq 0, z\geq0 $, calculate the flux through the surface
If I parameterize the surface as:
$$x=\cos u\sin v$$
$$y=a\sin u\sin v$$
$$z=a\cos v$$
with $0\leq u\leq \frac{\pi}{2}$ and $0\leq v\leq \frac{\pi}{2}$
I get a flux of $-a^3/3$
OR if I parameterize the surface as:
$$x=\cos u\cos v$$
$$y=a\sin u\cos v$$
$$z=a\sin v$$
with $0\leq u \leq \frac{\pi}{2}$ and $-\frac{\pi}{2} \leq v \leq 0$
I get a flux of $a^3/3$
Why are they opposite in sign when the flux should be invariant under any coord system as a physical quantity? The answer in the book is positive.
 A: Flux depends on the designated normal direction.  From the other side of the surface the flux has the opposite sign.  For non-closed surfaces in space, we usually follow the convention that the “positive” normal vector is the one that points upwards, i.e., has positive $z$-component.
If you are computing the surface integral in the usual manner, you are probably forming the vector-valued function $\mathbf{r} = \left<a \cos(u) \sin(v),a\sin(u)\sin(v),a\cos(v)\right>$.  Then
$$
\iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_D \mathbf{F}(\mathbf{r}(u,v))\cdot (\mathbf{r}_u \times \mathbf{r}_v)\,dA
$$
where $S$ is the surface in question (an eighth of the sphere, BTW), and $D=[0,\pi/2]\times[0,\pi/2]$.  But notice
$$\mathbf{r}_u =\left<-a\sin(u)\sin(v),a\cos(u)\sin(v),0\right>$$
and
$$\mathbf{r}_v =\left<a \cos(u)\cos(v),a\sin(u)\cos(v),-a\sin(v)\right>$$
So
$$
\mathbf{r}_u\times\mathbf{r}_v= \left<-a^2\cos(u)\sin^2(v),-a^2\sin(u)\sin^2(v),-a^2 \sin(v)\cos(v)\right>
$$
Notice this vector points downwards on $S$.  So we take its opposite vector, or take the cross product in the opposite order, to correctly orient the surface.
Since
$$
\mathbf{F}(\mathbf{r}(u,v))\cdot (\mathbf{r}_v \times \mathbf{r}_u)
= \left<0,a \cos(u) \sin(v),0\right>\cdot\left<a^2\cos(u)\sin^2(v),a^2\sin(u)\sin^2(v),a^2 \sin(v)\cos(v)\right>
= a^3 \sin(u)\cos(u)\sin^3(v)
$$
The integral is equal to
$$
\int_0^{\pi/2}\int_0^{\pi/2} a^3 \sin(u)\cos(u)\sin^3(v)\,dv\,du = \frac{a^3}{3}
$$
