TITLE: Finding out whether normal points outwards or inwards
QUESTION [1 upvotes]: Consider $F(x,y,z)=(x,x^2y,0)$ and $$\Omega=\{(x,y,z)\in\mathbb{R}^3\mid(x^2+y^2)^2<z<\sqrt{x^2+y^2}\}$$
I want to compute $\iint_{\partial\Omega}(F\cdot\nu)\text{ds}$ where $\nu$ is the normal pointing outwards.
I have that $\partial\Omega = \Sigma^1\cup\Sigma^2$ where
$$\Sigma^1=\{\sigma^1(r,\theta)=(r\cos\theta,r\sin\theta,r^4)\mid (r,\theta)\in(0,1)\times(0,2\pi)\}$$
$$\Sigma^2=\{\sigma^2(r,\theta)=(r\cos\theta,r\sin\theta,r)\mid (r,\theta)\in(0,1)\times(0,2\pi)\}$$
So when I compute the normals I get:
$$\sigma^1_r\times\sigma^1_\theta=(-4r^4\cos\theta,-4r^4\sin\theta,r)$$
$$\sigma^2_r\times\sigma^2_\theta=(-r\cos\theta, -r\sin\theta,r)\, $$
I know that the first one points inwards and the second one outwards (it's written in the textbook's solution) but I don't know how they find that.
Thanks in advance for your help.

REPLY [0 votes]: Drawing the space $\Omega$ in the $(r,z)$ plane for $\theta=0$ can help visualize thing, see this wolfram alpha plot. When $\theta=0$, your normal vectors become:
$$
\sigma^1_r\times\sigma^1_\theta=(-4r^4,0,r) \\
\sigma^2_r\times\sigma^2_\theta=(-r,0,r)
$$
and both point upwards towards the left. Consequently, the lower one (on $\Sigma_1$) is pointing inwards whereas the upper one (on $\Sigma_2$) is pointing outwards.
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