Surface area of a slightly deformed sphere Consider the unit sphere, which can either be described by $x^2+y^2+z^2=1$ or by the equation $r(\theta,\phi)=1$, where $(r,\theta,\phi)$ are spherical polar coordinates.
I define a deformed sphere by the equation, $r(\theta,\phi) = 1+\delta r(\theta,\phi)$, where $\delta r (\theta,\phi)$ is a small smooth deformation of the surface. How would I go about writing a formal expression for the surface area of this deformed sphere, in a perturbative expansion in $\delta r$.
 A: There are two conventions for spherical polar coordinates. For concreteness, I will use the physicist version here:
$$[0,2\pi] \times [0,2\pi) \ni (\theta,\phi) \quad\mapsto\quad
\hat{n}(\theta,\phi) = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \in \mathbb{R}^3$$
When we deform the unit sphere slightly, we can continue to use $(\theta,\phi)$
as a parametrization of the deformed sphere. The position vector corresponding to
$(\theta,\phi)$ will have the form
$$\vec{x}(\theta,\phi) = r(\theta,\phi) \hat{n}(\theta,\phi) =
( 1 + \epsilon \eta(\theta,\phi) )\hat{n}(\theta,\phi)$$
where $\epsilon$ is a small parameter controling the amount of deformation.
For any function $f$, let $f_\theta$ and $f_\phi$ be the partial derivative of $f$ with respect to $\theta$ and $\phi$ respectively. The area element of the deformed sphere is given by the formula
$$d\sigma = | \vec{x}_\theta \times \vec{x}_\phi | d\theta d\phi$$
Notice
$$\begin{cases}
\vec{x}_\theta = r_\theta \hat{n} + r \hat{n}_\theta\\
\vec{x}_\phi   = r_\phi   \hat{n} + r \hat{n}_\phi\\
|\hat{n}| = |\hat{n}_\theta| = 1, |\hat{n}_\phi| = \sin\theta\\
\hat{n} \cdot \hat{n}_\theta = \hat{n} \cdot \hat{n}_\phi =
\hat{n}_\theta \cdot \hat{n}_\phi = 0
\end{cases}
\quad\implies\quad
\begin{cases}
|\vec{x}_\theta|^2 &= |r_\theta|^2 + r^2\\
|\vec{x}_\phi|^2   &= |r_\phi|^2 + r^2\sin^2\theta\\
\vec{x}_\theta \cdot \vec{x}_\phi &= r_\theta r_\phi
\end{cases}
$$
We have
$$|\vec{x}_\theta \times \vec{x}_\phi| = \sqrt{ |\vec{x}_\theta|^2 |\vec{x}_\phi|^2 - |\vec{x}_\theta \cdot \vec{x}_\phi|^2}
= r\sin\theta \sqrt{ r^2 + r_\theta^2 + \frac{r_\phi^2}{\sin^2\theta} }
$$
This leads to
$$d\sigma = \sin\theta d\theta d\phi  \left[
(1 + \epsilon \eta )\sqrt{
(1 + \epsilon \eta)^2 + \epsilon^2 \Delta }\right]
\quad\text{ where }\quad 
\Delta = \eta_\theta^2 + \frac{\eta_\phi^2}{\sin^2\theta}
$$
Throw the RHS to a CAS and ask it to Taylor expand in $\epsilon$, we get
$$\frac{d\sigma}{\sin\theta d\theta d\phi} = 
1 
+ 2\eta\epsilon
+ \left(\eta^2 + \frac{\Delta}{2}\right)\epsilon^2
- \frac{\Delta^2}{8} \epsilon^4
+ \frac{\eta\Delta^2}{4} \epsilon^5
+ \cdots
$$
If you need more terms, you can ask a CAS to crank that out for you.
