Differentiation under the integral sign in $R^3$ I'm trying to take derivative from an integral. I know about the Reynolds transport theorem, but I do not know how to obtain the unit normal and the velocity. I'm going to take the derivate from the volume of a region $\Omega(t)$ that varies with $t$.
$\frac{\mathrm{d}}{\mathrm{d}t}\left[
    \int_{\Omega(t)}
    \mathrm{d}{\bf V}
  \right] =
  \int_{\partial\Omega(t)} (v^b.n)\mathrm{d}{\bf S}$.
Specifically, my region $\Omega(t)$ is 
$$\Omega(t)=\{(x_1,x_2,x_3)\geq 0\ | \ (1-t)*x_1x_2+x_2x_3 + x_3 \geq 0 \}$$
Can you please tell me how to derive the unit normal $n$ and the velocity $v$? Do they belong to $R^2$ or $R^3$?
[Wikipedia] https://en.wikipedia.org/wiki/Reynolds_transport_theorem
 A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}$The unit normal and velocity are spatial, i.e., belong to $\Reals^{3}$. Particularly, at time $t$, the "moving portion" of the boundary of $\Omega(t)$ has equation
$$
(1 - t) x_{1} x_{2} + x_{2} x_{3} + x_{3} = 0.
\tag{1}
$$
The left-hand side of the transport theorem can be calculated (in principle, at least; haven't tried) by evaluating a volume integral, then differentiating with respect to $t$.
For the right-hand side, you need a (time-dependent) parametrization
$$
(x_{1}, x_{2}, x_{3}) = X(t, u, v)
$$
of the surface (1), and then
$$
v^{b} = \frac{\dd X}{\dd t},\qquad
N = \frac{\dd X}{\dd u} \times \frac{\dd X}{\dd v},\quad
n = \frac{N}{\|N\|},\qquad
(v^{b} \cdot n)\, dS = (v^{b} \cdot N)\, du\, dv.
$$
The final equation holds because the surface element is $dS = \|N\|\, du\, dv$.
A: For the surface $$(1-t)x_1x_2+x_2x_3 + x_3 =  0, $$ 
you need to parametrize the equation. 
Let $u=x_1$ and $v=x_2$, you get 
$$x_3=f(u,v)=(t-1)\frac{uv}{1+v}.$$
The normal vector is thus $$(f_u,f_v, -1)=((t-1)\frac v{1+v},(t-1)\frac u{(1+v)^2},-1).$$ 
(I didn't convert it to unit vector.)
For velocity, it will be 
$$\frac{dX}{dt}=\frac{d}{dt}(x1,x2,x3)=(0, 0, \frac{uv}{1+v}).$$
