I'm working through the proof of the mean value inequality (1.15) of Colding-Minicozzi's A Course on Minimal Surfaces, and I'm stuck on this subproblem. Let $\Sigma$ be a $k$-dimensional minimal submanifold of $\mathbb{R}^n$ (I don't think the minimality property will be relevant here, but it might), and let $B_s$ denote $n$-ball of radius $s$ centered at the origin. For $p \in \Sigma$ and $x \in T_p\mathbb{R}^n$, let $x^T$ denote the orthogonal projection onto $T_p \Sigma$. For a real valued function $f$ defined on $\Sigma$, I want to show $\frac{d}{ds} \int_{B_s \cap \Sigma} f = \int_{\partial B_s \cap \Sigma} f \frac{|x|}{|x^T|}$. Ultimately this comes down to applying the Reynold Transport Theorem about the derivative of an integral over a varying domain of integration, but I'm stuck computing the velocity of the area element.
The outward pointing unit normal on $x \in \partial B_s \cap \Sigma$ is $\frac{x^T}{|x^T|}$. I want to compute the derivative of this outward pointing normal with respect to the radius $s$. I believe the answer is something like $\frac{|x|}{|x^T|^2}x^T$, as the proof uses that the dot product of this radial derivative with $\frac{x^T}{|x^T|}$ is $\frac{|x|}{|x^T|}$, but I'm not sure how to show this.
It turns out the right approach is to use the co-area formula instead of trying to compute the velocity of the area element, though I'd still be interested to see how that computation plays out. The text said to use the co-area formula, but I mistakenly thought that was coming in the subsequent equation!
For $x \in \Sigma$, let $h(x) = |x|$. Then
$\begin{eqnarray*} \frac{d}{ds} \int_{B_s \cap \Sigma} f &= \frac{d}{ds} \int_{B_s \cap \Sigma} f |\nabla_\Sigma h|^{-1} |\nabla_\Sigma h| \\ &= \frac{d}{ds} \int_{0}^s \int_{h = \tau} f|\nabla_\Sigma h|^{-1} d\tau \\ &= \int_{\partial B_s \cap \Sigma} f|\nabla_\Sigma h|^{-1} \end{eqnarray*}$
Note that $\nabla_{\mathbb{R}^n} |x| = \frac{x}{|x|}$, so $\nabla_\Sigma |x| = ( \nabla_{\mathbb{R}^n} |x| )^T = \frac{x^T}{|x|}$.
Hence, we get the result that $\frac{d}{ds} \int_{B_s \cap \Sigma} f = \int_{\partial B_s \cap \Sigma} f \frac{|x|}{|x^T|}$.