Negative flux when vector arrows are pointing outwards I'm doing some homework for my calculus class and I came across an exercise where I shall calculate the flux of a vectorfield out of a elliptic cylinder $\frac{x^2}{9}+\frac{y^2}{4} \leq 1$, limited by a parabolic function $z = x^2 + y^2$.
I drew a graph to help me out understand the vector field arrows and this is my plot.

I calculated the total flux using $\int\int\int_D \nabla \cdot F\ dv$ over the vectorfield $F(x,y,z) = (3x,2y,z)$ and I get a total flux of $117\pi$. Now when I calculate the flux out of the cylinder I get $\frac{405\pi}{2}$. 
Now I am asked to find the flux out of the top of sylinder limited by the parabolic function. Since the bottom does not have any flux going out of it (since $z \geq 0$). I simply use the total minus the cylinder, but then I get a negative flux. How come? When the arrows are clearly pointing outwards at any point inside the given volume?


I simply use the total minus the cylinder, but then I get a negative flux. How come? When the arrows are clearly pointing outwards at any point inside the given volume?

Your calculations look alright, so I think you're wrong in your geometric interpretation or intuition. Notice that although the paraboloid is the 'top' of the surface, it goes all the way down to the origin where it meets the bottom surface at $z=0$.
The field lines near the origin but above the $xy$-plane (e.g. imagine points $(x,y,z)$ with $z>0$ and with $x$ and $y$ sufficiently small so that the points lie on the inside of the paraboloid, so outside the region enclosed by the paraboloid and the cylinder), enter the enclosed region via the paraboloid-side of the surface to exit it again a bit further through the cylinder-side.
If you find this hard to picture in three dimensions, take a look in two dimensions. Draw some vector field arrows of the vector field $(3x,0,z)$ in the $xz$-plane and draw the parabola $z=x^2$:

Now imagine rotating this thing around the $z$-axis to get a three-dimensional feel.

For completeness, the contributions to the net flux of $\color{blue}{117\pi}$ are:


*

*$\color{green}{\tfrac{405}{2}\pi}$ through the cylinder;

*$\color{red}{-\tfrac{171}{2}\pi}$ through the paraboloid;

*$\color{purple}{0}$ through the bottom surface ($xy$-plane);


so in total you indeed have:
$$\color{blue}{117\pi}=\color{green}{\tfrac{405}{2}\pi}\color{red}{-\tfrac{171}{2}\pi}+\color{purple}{0}$$
