Lets say I've a sphere $x^2+y^2+z^2=1$ and I need to solve an integral $\iint_S\vec F\cdot\vec ndS$. while $S$ is the sphere.
And I can't use gauss law because $\vec F$ is not continuous at some point inside $S$. I had
$$\vec F=\left(\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}}\right).$$
So I went and tried to go for it normally using ball coordinates $$\vec r(\theta,\phi)=(\cos\theta \sin\phi, \sin\theta \sin\phi, \cos\phi).$$
And I found that $$r_{\theta}\times r_{\phi}=(-\cos\theta \sin^2\phi,-\sin\theta \sin^2\phi,-\sin\phi \cos\phi).$$
Now I know that this could be the normal pointing inside the ball or outside of it.
In order to try to check, I tried to reach a point $(1,0,0)$ on it, and check if the "$x$" part of the normal is positive or negative in that point.
But I got lost trying to find $\phi,\theta$ which satisfy that, and wanted to know if there's any better way to decide in what direction the normal I found is pointing.
Any feedback is appreciated, thanks in advance!
Best Answer
One general rule for a region $\ R\ $ defined by $\ R=\big\{x\in\mathbb{R}^3\,\big|\,f(x)< C\,\big\}$, where $\ f:\mathbb{R}^3\rightarrow\mathbb{R}\ $ is a differentiable function, and $\ C\in\mathbb{R}\ $ a constant, is that when $\ f(x)=C\ $ and $\ \nabla f(x)\ne\vec{0}\ $, then $\ \nabla f(x)\ $ is an outward normal to the boundary of $\ R\ $ at the point $\ x\ $. If you get the normal at a given point in some other way, as you've done by taking the cross product of two tangents to the boundary, then the direction of the normal is outward if the directional derivative of $\ f\ $ at that point in the direction of the normal is positive, and it's an inward normal if the directional derivative is negative. If the directional derivative is zero, then you need to look at higher order directional derivatives to tell whether $\ f\ $ is increasing or decreasing in the given direction.
In your case, with $\ S\ $ being the unit sphere, of which the inside is the open unit ball, $\ R=\big\{\,(x,y,z)\,|\,x^2+y^2+z^2<1\,\big\}\ $, you have $\ f(x,y,z)=x^2+y^2+z^2\ $, and $\ \nabla f(x,y,z)=2\big(x\,\vec{\mathbf{i}}+y\,\vec{\mathbf{j}}+z\,\vec{\mathbf{k}}\big)=$$2\vec{r}\ $, so $\ \vec{r}\ $ itself, as a positive multiple $\ \frac{1}{2}\nabla f\ $ of $\ \nabla f\ $ is an outward (unit) normal. Note that your cross product, $$ r_\theta\times r_\phi=-\sin\phi\,\vec{r}(\theta,\phi)\ , $$ as a negative multiple of $\ \vec{r}\ $ (for $\ 0<\phi<\pi\ $) , must be an inward normal.