I've got a conceptual problem regarding inward and outward normals. The textbook question (2nd year vector calculus) is as follows:
A uniform fluid that flows vertically downward is described by the vector field $\mathbf F(x,y,z) = -\mathbf k$. Find the flux through the cone $z = \sqrt{x^2+y^2} \ , \ \ \ x^2+y^2\le1$
So I've parametrised the surface in cylindrical coordinates as $$\Phi(\rho,\phi) = (\rho\cos(\phi),\rho\sin(\phi),\rho)$$ with $0\le\rho\le 1$ and $0\le\phi\le2\pi$
Then
$$\vec {\mathbf T_\rho} = (\cos(\phi),\sin(\phi),1)
\\ \vec {\mathbf T_\phi} = (-\rho\sin(\phi),\rho\cos(\phi),0)$$
$$\vec {\mathbf T_\rho} \ \ \times \ \ \vec {\mathbf T_\phi} =
\begin{vmatrix} \mathbf i & \mathbf j &\mathbf k \\
\cos(\phi) & \sin(\phi) & 1 \\
-\rho\sin(\phi) & \rho\cos(\phi) & 0
\end{vmatrix} \\ = \ (-\rho\cos(\phi), -\rho\sin(\phi), \rho)
$$
But then I try to calculate $$\iint_s \mathbf F \cdot \mathrm d \mathbf s \\ = \int_{0}^1 \int_{0}^{2\pi} (0,0,-1) \ \cdot \ (-\rho\cos(\phi), -\rho\sin(\phi), \rho) \ \mathrm d\phi \ \mathrm d\rho \\ = \int_{0}^1 \int_{0}^{2\pi} -\rho \ \mathrm d\phi \mathrm d\rho \\ = \int_0^1 -2\pi\rho \ \mathrm d\rho \\ = -\pi$$
which is the negative of the actual answer of $\pi$. I've got a feeling it's to do with the order of the cross product so my question is:
How can I fix this problem, and why is it that I should take the reverse cross product?
Thanks in advance for any help!
Best Answer
A $\rho$-curve i.e. fixing some $\rho$ and vary $\phi$ is vertical translation of some circle in $XY$ plane and it's tangent is $\overrightarrow T_\phi$ which is something looks like vertical translation of some tangent to a circle of $XY$ plane.
Now for a $\phi$-curve if you increase $\rho$ then you traverse this $\phi$-curve upward direction i.e. tangent to this $\phi$-curve has direction upwards, in some sense same direction as of $\widehat k$.
Now if you apply right-handed rule to find the direction of $\overrightarrow T_\rho\ ×\ \overrightarrow T_\phi$ which direct towards OUTSIDE of the solid region inside the cone , but then $\overrightarrow T_\phi\ ×\ \overrightarrow T_\rho\ = -\ \overrightarrow T_\rho\ ×\ \overrightarrow T_\phi$ directs towards the INSIDE of the solid region of the cone.
Therefore $\overrightarrow T_\phi\ ×\ \overrightarrow T_\rho$ is the INWARD drawn normal to the surface. And $\overrightarrow T_\rho\ ×\ \overrightarrow T_\phi$ is OUTWARD drawn normal to the surface.