The divergence theorem allows us to go between surface and volume (in some sense), so a natural example would be to compute the surface area of the unit sphere $U$ assuming we know the sphere volume.
I've tried the following:
\begin{align}
\int_{\partial U} |x| \ dS & = \int_{\partial U} \frac{|x|^2}{|x|} \ dS
\\ &= \int_{\partial U} \frac{\langle x,x \rangle}{|x|} \ dS
\\ &= \int_{\partial U} \langle x, \frac{x}{|x|} \rangle \ dS
\\ &= \int_{\partial U} \langle f(x), n(x) \rangle \ dS \ \ \ \text{($f=$id and $n(x)$ is the normal vector)}
\\ &= \int_U \operatorname{div}f \ dV \ \ \ \text{(divergence theorem)}
\\ &= \int_U 3 \ dV
\\ &= 3 \int_U dV
\\ &= 3 \frac{4}{3}\pi
\\ &= 4\pi
\end{align}
And here is my question (if the calculation makes sense): why does $\int_{\partial U} |x| \ dS$ equal to the surface area? I see intuitively why it holds (especially in two dimensions) yet it's not very convincing.
Best Answer
It holds because in the unit sphere, $|x|$ is always 1, so you are just integrating 1, which gives the area. This won't work for any other shape.