# Calculate the flux through a closed surface

While studying for a test I have encountered such a task:

Calculate the flux through a closed surface, where $S$ is a boundary of area $V$ with an outward orientation. The data:

$$\vec{F}(x,y,z)=(\arctan(y), \frac{z^2}{1+x^2}, 2xy)$$ $$V: x^2+y^4+z^6 \le 1$$ $$\iint\limits_S \vec{F} \vec{ds} =\text{ ?}$$

I know, that such examples are supposed to be done using the divergence (Gauss-Ostrogradsky) theorem. So I begin with calculating $P_x, Q_y, R_z$. As can be easily seen, each of them is equal $0$. Besides, $V$ is very unusual here, with $z$, for example, raised to the sixth power.

1. Is there anything wrong with my reasoning?

2. Is there any other way to calculate this flux?

3. What for do I need the info that the surface is oriented outwards?

• You are right, the divergence of F is zero and therefore the flux is zero. – user204299 Aug 30 '15 at 21:47
• @JakeLebovic Thank you. So apparently after calculating $P_{x}, Q_{y}, R_{z}$ the exercise is finished? – Peter Cerba Aug 31 '15 at 8:02

## 1 Answer

After calculating $P_x,Q_y,R_z$, deduce that $\nabla \cdot F$ (the sum of these) is zero, which means (by the divergence theorem) that the total flux (which is an integral of $\nabla \cdot F$) is zero.

There's nothing wrong with your reasoning. You could calculate the flux without the theorem with a parametrized surface integral, but this approach would be much more difficult. Switching the orientation of the boundary switches the sign of your answer, but that makes no difference in this case.