How to simplify the integral $\int_C \mathbf r\times (d\mathbf r \times \mathbf f)$ given that the field $\mathbf f$ is uniform? How to simplify the integral $\int_C \mathbf r\times (d\mathbf r \times \mathbf f)$ given that


*

*$\mathbf f=a\mathbf k$, where $\mathbf k$ is the unit vector in $z$ direction and $a$ is  constant;

*$C$ is a simple closed curve that lies on a plane, which has normal vector $\mathbf n$, and $\mathbf n.\mathbf k=\cos \phi$;

*$C$ encloses an area $A$?


(The last item above...well, I am not sure if it helps us compute the integral, but I vaguely feel that the shape of $C$ doesn't affect the value of the integral - it seems only dependent on the area.)
I am not sure if I will get something nice, because this is not a homework problem, but I really wonder whether the integral can be simplified, because this is very important in physics - examples of this include force on electric motors. 
I have tried to use ABC-ACB rule, but it doesn't work. I am not sure if I need to do the boring work of writing out every component. Cross-products are very restrictive; they are not associative, which makes it difficult. In the end, I have made little progress apart from the well-known formula
$$
A\mathbf n=\frac12 \int_C \mathbf r \times d\mathbf r.
$$
Any help?
 A: Let's take the simple closed curve $C$ to be oriented counterclockwise with respect to $\mathbf{N}$, and let $\Sigma$ be the oriented flat surface in the plane $\{\mathbf{N}\}^\perp$ bounded by $C$ with unit normal $\mathbf{N}$. First, by BAC-CAB,
$$
\begin{align}
 \mathbf{r} \times (d\mathbf{r} \times \mathbf{f}) &= (\mathbf{r} \cdot \mathbf{f})\,d\mathbf{r} - (\mathbf{r} \cdot d\mathbf{r})\mathbf{f}\\
&= (az\mathbf{i} \cdot d\mathbf{r})\mathbf{i}+(az\mathbf{j} \cdot d\mathbf{r})\mathbf{j} + (az\mathbf{k} \cdot d\mathbf{r})\mathbf{k} - d\left(\tfrac{1}{2}a\mathbf{r}\cdot\mathbf{r}\right)\mathbf{k},
\end{align}
$$
so that
$$
 \int_C \mathbf{r} \times (d\mathbf{r} \times \mathbf{f}) = \left(a\int_C z\mathbf{i} \cdot d\mathbf{r}\right) \mathbf{i} + \left(a\int_C z\mathbf{j} \cdot d\mathbf{r}\right) \mathbf{j} + \left(a\int_C z\mathbf{k} \cdot d\mathbf{r}\right) \mathbf{k},
$$
where $\int_C d\left(\tfrac{1}{2}a\mathbf{r}\cdot\mathbf{r}\right) = 0$ since $C$ is closed. Now, let's apply Stokes's theorem to the three remaining line integrals: since
$$
 \nabla \times (z \mathbf{i}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ z & 0 & 0 \end{vmatrix} = \mathbf{j}, \quad \nabla \times (z \mathbf{j}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 0 & z & 0 \end{vmatrix} = -\mathbf{i}, \quad \nabla \times (z \mathbf{k}) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \partial_x & \partial_y & \partial_z \\ 0 & 0 & z \end{vmatrix} = \mathbf{0},
$$
it follows that
$$
\begin{align}
\int_C z\mathbf{i} \cdot d\mathbf{r} &= \int_\Sigma \mathbf{j} \cdot \mathbf{N} \, dA = (\mathbf{N} \cdot \mathbf{j})\operatorname{Area}(\Sigma),\\
\int_C z\mathbf{j} \cdot d\mathbf{r} &= \int_\Sigma (-\mathbf{i}) \cdot \mathbf{N} \, dA = -(\mathbf{N} \cdot \mathbf{i})\operatorname{Area}(\Sigma),\\
\int_C z\mathbf{k} \cdot d\mathbf{r} &= \int_\Sigma \mathbf{0} \cdot \mathbf{N} \, dA = 0,
\end{align}
$$
and hence
$$
\begin{align}
\int_C \mathbf{r} \times (d\mathbf{r} \times \mathbf{f}) &= a(\mathbf{N} \cdot \mathbf{j})\operatorname{Area}(\Sigma)\mathbf{i} - a(\mathbf{N} \cdot \mathbf{i})\operatorname{Area}(\Sigma)\mathbf{j} = a \operatorname{Area}(\Sigma)(\mathbf{N} \times \mathbf{k}).
\end{align}
$$
