Let S be a piecewise smooth oriented surface in $R^3$ with positive oriented piecewise smooth boundary curve $\Gamma:=\partial S$ and $\Gamma : X=\gamma(t), t\in [a,b]$ a rectifiable parametrization of $\Gamma$. Imagine $\Gamma$ is a wire in which a current I flows through. Then
$m:=\frac{I}{2}\int_a^b\gamma(t)\times \dot{\gamma}(t)dt$
is the magnetic moment of the current.
a) Show that for the given orientation of $\Gamma$ the magnetization m is independent of the choice of the parametrization of $\gamma$.
b) For $b\in R^3$ let $k_b$ be the vector field $k_b(x):=\frac{1}{2}b\times x$. Show that
$\text{rot}\, k_b = b$.
Describe all vector fields $k\in C^1(R^3,R^3)$, which fulfill $rot ~k~=~b$ on $R^3$.
c) Show that for an arbitrary $b\in R^3$
$m\cdot b=I\int_Sb\cdot d\Sigma$ is true.
I don't know how to solve a) because I never dealt with oriented surfaces before and I don't know how to start.
The first part of b) is relatively easy. I just applied the general formula for the rotation of a vector field and got the desired result. But I don't know how to describe all vector field which fulfill this condition. Any tips?
I'm pretty much lost on c). I don't know how to even start here.
Suppose that there is another way to parameterize the curve using $s$ and that $t=\phi(s)$.
Let's $a'\leq s \leq b'$ let the interval correspond to the interval $a\leq t \leq b$. The new parameterization, $\beta(s)=\alpha(\phi(s))$, so the chain rule gives us $ds=(\dfrac{ds}{dt})dt$, and
$$m=\frac{I}{2}\int_{a'}^{b'}\beta(s)\times \dot{\beta}(s)ds$$
$$=\frac{I}{2}\int_{a'}^{b'}\alpha(\phi(s))\times \dot{\alpha(\phi(s))}ds$$
$$=\frac{I}{2}\int_{a'}^{b'}\alpha(\phi(s))\times \dot{\alpha(\phi(s))}(\dfrac{d\phi}{ds})ds$$
$$=\frac{I}{2}\int_a^b\alpha(t)\times \dot{\alpha(t)}(\dfrac{dt}{ds})ds$$
$$=\frac{I}{2}\int_a^b\alpha(t)\times \dot{\alpha(t)}dt=m$$
See that changing the parameterization had no effect at all on the line integral! This property is referred to as independence of parameterization