Integral with cross product Given a function $\vec{r}(t)$, is there a way to simplify:
$$\int_0^u \frac{\vec{r}}{\vec{r} \cdot \vec{r}} \times \frac{d\vec{r}}{dt}~dt$$
With an abuse of notation, it's a line integral like this:
$$ \int_\gamma \frac{\vec{r}}{\vec{r} \cdot \vec{r}} \times d\vec{r}$$
but I've only seen dot products used with the $d\vec{r}$, never cross. No idea if it makes sense.
I know that it's path-dependent (consider $\vec{r}$ going in a circle), but in two dimensions, it turns out to be the winding number around $0$ times $2\pi$. Is there a way to extend this correlation to three dimensions?
(My motivation here is: I like angular velocities to be vectors, so I define $\vec{\omega}$ not as $\frac{d\theta}{dt}$, but as $\frac{\vec{r} \times \vec{v}}{|\vec{r}|^2}$. It points along the instantaneous axis of rotation, and gives the angular velocity around that point. But I couldn't come up with a physical interpretation for $\theta$ as a vector, so I'm curious to see what comes out of this integral)
 A: Let $\vec{J}$ be your integral.
  Using the identity $\vec{A} \cdot (\vec{B} \times \vec{C}) = \vec{B} \cdot(\vec{C} \times \vec{A}) = \vec{C} \cdot (\vec{A} \times \vec{B})$ with $\vec{A}$ a constant vector, $\vec{B} = \vec{r}/(\vec{r} \cdot \vec{r})$
and $\vec{C} = \dfrac{d{\vec r}}{dt}$, 
$$ \vec{A}  \cdot \vec{J} = \int_0^u \vec{A} \cdot (\vec{B} \times \vec{C})\ dt = \int_0^u (\vec{A} \times \vec{B})\cdot  d{\vec r}$$
In particular, for the case of a closed curve which is the boundary of a surface $S$,  Stokes's theorem says this is the flux of the vector field $\vec{A} \times \vec{B}$ through 
$S$.
A: The analogous integral for 3d would be
$$\oint_S \frac{\vec r}{r^3} \cdot \hat n \, dS$$
If the surface $S$ is a simple, closed surface enclosing the origin, then the result is $4\pi$.  It could be that the surface is more complicated.

Why is this the analogous form?  Well, a couple things are being hidden from you because vector calculus is kinda lame.  First, because vector calculus only lets you deal with vectors and scalars (and associated fields), the structural similarity between these two integrals is obscured.  In geometric calculus, they both look quite similar:
$$\begin{align*} \text{2d:} & \oint_C \frac{\vec r}{r^2} \wedge d\vec \ell  \\ \text{3d:} & \oint_S \frac{\vec r}{r^3} \wedge \, d\vec S\end{align*} $$
What I've written as $d\vec S$ here is not $\hat n \, dS$.  It is a 2-vector, or a bivector.  Think of it as the plane orthogonal to $\hat n$--in other words, the plane tangent to the surface of integration.  Bivectors can have magnitudes just like vectors.  Writing these integrals using their tangent elements--the tangent vector for a curve, the tangent bivector for a surface--demonstrates manifestly how similar these integrals are, using the wedge product.
In short, both integrals are structurally the same, but because vector calculus insists you use only scalars and vectors, you have to fudge it and introduce an arbitrary notational difference between them.  This is silly.

Now, why is the 2d integral $\vec r/r^2$, when the 3d integral is $\vec r/r^3$?
This has to do with Green's functions.  In vector calculus, it's nonsensical to talk about the Green's function of $\nabla$.  Instead, one talks only about Green's functions for the Laplacian, for instance--a scalar differential operator.
Geometric calculus allows divergence and curl to be unified, and it gives a meaningful Green's function for $\nabla$:  $G(\vec r) = \hat r/S_n(r)$, where $S_n(r)$ is the surface area of an $n$-dimensional ball of radius $r$.  In 2d, the "surface area" is a circumference and is $2\pi r$.  In 3d, it's $4\pi r^2$.  This makes the Green's function in 2d and 3d:
$$\begin{align*} \text{2d:} & G(\vec r) = \frac{\hat r}{2\pi r} = \frac{\vec r}{2\pi r^2} \\ \text{3d:} & G(\vec r) = \frac{\hat r}{4\pi r^2} = \frac{\vec r}{4\pi r^3} \end{align*} $$
The Green's function obeys
$$\nabla G = \delta(r)$$
Its derivative is just a delta function.
What follows, then, is that all you need to do is use some form of the fundamental theorem of calculus (Stokes' theroem, divergence theorem, etc.).
$$\begin{align*} \text{2d:} & \oint_{\partial S} G(\vec r) \times d\vec \ell = \int_S \delta(\vec r) \hat z \, dS  \\ \text{3d:} & \oint_{\partial V} G(\vec r) \cdot \hat n \, dS = \int_V \delta(\vec r) \, dV \end{align*}$$
If the origin is contained, then these integrals come out to 1 for simple curves and surfaces.  The 2d integral is the winding number (by pulling the factor of $1/2\pi$ into the integral, I've restored symmetry between the cases and also fixed the scaling problem that you had, where you were getting winding number multiplied by $2\pi$) times $\hat z$.  I'm not aware what the analogous 3d concept is called.
