# 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

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.