Surface integrals were always confusing for me, mainly because I never studied them carefully and in depth. After some research on several sources (not sufficient enough) I came across with the following questions/statements which I would highly appreciate if someone could explain/verify.
Let's consider the unit sphere $\mathbb S^2=\{(x,y,z):\;x^2+y^2+z^2=1\}$.
- A function $f$ defined on $\mathbb S^2$ is, in general, a function of $3-$variables.
- The surface integral $\int_{\mathbb S^2} f\;dS$ though, can be calculated with an appropriate parametrization as a double integral (with respect to $2-$coordinates). Is this correct? If yes, then how is this justified?
- Assume now that $f$ is a function defined on $\mathbb S^2$ as before but depends only on variable $x$. If I want to integrate $f$ over the set $\{x>x_0\}$, how would this integral look like? I've seen this formula
$\int_{\{x>x_0\}} f\;dS=2\pi \int_{x_0}^1 f(x)\;dx$
which implies (if not mistaken) that there was some extra integration from $0$ to $2\pi$ that resulted in the $2\pi$ factor but I can't understand how this came up. I tried to use cylindrical coordinates but I don't obtain the desired formula.
I'm having a really hard time getting my head around these notions and arguments. Any help, hint or comments are very much welcome.
Thanks a lot in advance!
Suppose $f : \mathbb S^2 \to \mathbb R$ is a function which only depends on the $x$ component of the input point. To compute $\int_{x>x_0} f\ dS$, let us first describe a parametrization of $\mathbb S^2$ to fit our purposes.
Suppose $(x,y,z)\in \mathbb S^2$ and $x = x_0$ is fixed. Then $(y,z)$ lies on a circle in the $yz$-plane of radius $r_x = \sqrt{1 - x^2}$, and thus both $y$ and $z$ can be specified by a single angle $\theta \in [0, 2\pi)$.
We thus have a coordinate system/a parametrization of the unit sphere, given by \begin{align*} &x = x \\ &y = r_x\cos \theta \\ &z = r_x\sin \theta \end{align*} for $-1 < x < 1$, and $0 < \theta < 2\pi$. The jacobian of this transformation is given by $||(1,r_x^\prime \cos \theta,r_x^\prime \sin \theta) \times (0, -r_x \sin \theta, r_x \cos \theta)|| = 1$ (amazingly), therefore \begin{align*} \int_{x > x_0} f(x)\ dS &= \iint_R f(x)\ d\theta dx \\ &= 2\pi \int_{x_0}^1 f(x)\ dx \end{align*} Where $R = \{ 0 < \theta < 2\pi, \; x_0 < x < 1\}$