# Sign of a flux surface integral

Use a parametrization to find the flux $$\iint_S F \cdot n \, d\sigma$$ across the surface in a given direction: $$F=xy\overrightarrow i-z\overrightarrow k$$ outward (normal away from the z-axis) through the cone $$z=\sqrt{x^2+y^2} \qquad 0\le z\le 1 .$$

I managed to parametrize the cone using cylindrical coordinates:

$$r(r,\theta)=(r\cos\theta)\overrightarrow i+(r\sin\theta)\overrightarrow j+ r\overrightarrow k, \qquad 0\le r\le 1, 0\le \theta \le 2\pi .$$

However, the surface integral flux result I got was $-2\pi /3$ while the solution manual got $2\pi /3$ and while looking through the answer, I saw that this was due to the way we computed $r_{r}\times r_{\theta}$, I computed it as written while they did $r_{\theta}\times r_{r}$.

Since we can do the calculation both ways, does the sign matter in the result? If it does, how am I supposed to know which way to do it?

Yes, it matters, since the question specifies finding the "flux... in a given direction". Drawing a diagram and using the Right-Hand Rule (or just computing directly) shows that $\partial_r \times \partial_{\theta}$ points toward the $z$-axis at each (nonsingular) point, not away from it as specified.
• Thanks for your answer! Can you elaborate on the diagram method so I can understand why $r_{r}\times r_{\theta}$ points towards the z-axis? – Omrane Apr 30 '16 at 12:06
• Let me see if I get it: when I use $r_{r}\times r_{\theta}$ the k component of the final vector is $r$ which is positive so it points towards the z-axis? – Omrane Apr 30 '16 at 12:09
• Sure: Using your parameterization, if we fix a point ${\bf r}(r_0, \theta_0)$ on the cone, the constant-$\theta$ curve $t \mapsto {\bf r}(r(t), \theta_0)$ through this point is a ray from the origin, so $\partial_r$ points along that ray. The constant-$r$ curve traces out the circle of points on the cone a distance $r_0$ from the origin anticlockwise, so the $\partial_{\theta}$ points anticlockwise and parallel to the $xy$-plane. Then, the R.-H. Rule gives that the normal defined by this parameterization points up and toward the $z$-axis. – Travis Willse Apr 30 '16 at 12:11