I've got a problem where I come a little unprepared not knowing how to tackle:
Given a Vector field $$\textbf{W}: \mathbb{R^3} \rightarrow\mathbb{R^3}, \quad \textbf{W(x)} = \vec{b} \times \vec{x}$$
In other words $\vec{b}$ is a fixed vector, $\vec{x}$ is variable
Now, it is claimed that the magnetic moment $\textbf{M}$ of a circuit, that circles around the boundary $\partial S$ of a surface $S$ defined by $$\begin{align} &M = \frac{I}{2}\int_0^1\gamma(t)\times \gamma'(t)\,\mathrm{dt} \quad\text{where $\gamma: [0,1]\rightarrow \mathbb{R}^3$ is the parametrization of $\partial S$} \\\\ &\text{should have the property} \\\\ &\langle\textbf{M},\vec{b}\rangle = I\,\int\int_s \langle \vec{b},N\rangle\,\mathrm{dA} \quad \text{where $N$ is the unit normal field } \end{align}$$
First thing associated with this problems is stokes theorem: $$\int \int_s \langle \text{curl}\,F,N\rangle\,\mathrm{dA} = \int_{\partial S}\langle F, T\rangle\,\mathrm{ds} \quad \text{where $F$ is a vector field and $T$ the unit tangent field} $$
Now I really don't know how to proceed, for instance what vectorfield to chose for $F$: $\textbf{W}$ or $\textbf{M}$? Prior to that I had to show that $\text{curl} \, \textbf{W} = 2\,\vec{b}$ what could fit in here, but probably has no relations at all. Also having difficulties untangling $ \langle \textbf{M}, \vec{b}\rangle $: $\langle M = \frac{I}{2}\int_0^1\gamma(t)\times \gamma'(t)\,\mathrm{dt},\left(\begin{array}{cc}b_1 \\b_2 \\b_3\end{array}\right) \rangle$ What's all of that? I'd be thankful for any hints.
$\textbf{Addendum}$
After a fruitful discussion, we came up with the solution:
$$\begin{align}{} &\langle \textbf{M}, \vec{b}\rangle = \frac{I}{2} \int_{0}^{1} \langle \gamma(t)\times \gamma'(t),\vec{b}\rangle \,\mathrm{dt} \\\\ & \text{because $\langle a \times b,c \rangle = \langle c \times a,b \rangle$} \\\\ & \langle \textbf{M}, \vec{b}\rangle = \frac{I}{2} \int_{0}^{1} \langle \vec{b}\times \gamma(t),\gamma'(t) \rangle \,\mathrm{dt} \\\\ &\text{because $\textbf{W}(\gamma(t)) = \vec{b}\times \gamma(t)$} \\\\ &\langle \textbf{M}, \vec{b}\rangle = \frac{I}{2} \int_{0}^{1} \langle \textbf{W},\gamma'(t) \rangle \,\mathrm{dt} \\\\ &\text{because $\mathrm{dt} = \dfrac{\mathrm{ds}}{\vert \gamma'(t) \vert}, \quad T = \dfrac{\gamma'(t)}{\vert \gamma'(t)\vert}$} \\\\ &\langle \textbf{M}, \vec{b}\rangle = \frac{I}{2} \int_{0}^{1} \langle \textbf{W},T \rangle \,\mathrm{ds} = \frac{I}{2} \int_{\partial S}\langle \textbf{W},T \rangle \,\mathrm{ds} \\\\ &\text{Now by Stokes Theorem}: \\\\ &\frac{I}{2} \int_{\partial S} \langle \textbf{W},T \rangle \,\mathrm{ds} = \frac{I}{2} \int\int_{S} \langle \text{curl}\textbf{W},N \rangle \,\mathrm{dA} \\\\ &\text{lastly as shown: $ \text{curl}\textbf{W} = 2\,\vec{b}$} \\\\ &\frac{I}{2} \int\int_{S} \langle \text{curl}\textbf{W},N \rangle \,\mathrm{dA} = I \int\int_{S} \langle \vec{b},N \rangle \end{align}$$
Hint: The integral is a linear transformation, so the inner product with $b$ can be moved in: \begin{align} M\cdot b &= \frac{I}{2}\int_{\partial S}[(\xi\times d\xi)\cdot b] \\ &= \frac{I}{2}\int_{\partial S}[b\times \xi]\cdot d\xi\\ &=\frac{I}{2}\int_{\partial S}W(\xi)\cdot d\xi \end{align} The first line means $\frac{I}{2}\int_0^1[(\gamma(t)\times \gamma'(t))\cdot b]\,dt$ and so on. The middle equality is a standard vector identity). Therefore...
Edit:
Filling in more details, and conforming more to your notation, we have \begin{align} \langle M,b\rangle&=\frac{I}{2}\int_0^1\langle \gamma(t)\times \gamma'(t),b\rangle\,dt\tag{linearity}\\ &=\frac{I}{2}\int_0^1\langle b\times \gamma(t),\gamma'(t)\rangle\,dt\tag{vector identity}\\ &=\frac{I}{2}\int_0^1\langle W(\gamma(t)),|\gamma'(t)|T(\gamma(t))\rangle\,dt\tag{definition}\\ &=\frac{I}{2}\int_{\partial S}\langle W,T\rangle\,ds\tag{definition}\\ &=\frac{I}{2}\int_S\langle \text{curl}(W),N\rangle\,dA\tag{Stokes} \end{align} Note that once we make the parametrization, $ds$ becomes $|\gamma'(t)|\,dt$, and the unit tangent vector is $\frac{\gamma(t)}{|\gamma'(t)|}$. So, there's a double occurrence of $|\gamma'(t)|$, and they both cancel out, and I believe this is par of what you're missing so far.