Define a vector field $ \vec{f}(\vec{R}) = \oint_C{|\vec{r} - \vec{R}|^2 d\vec{r} }$ where C is a simple closed curve.
show that there are constant vectors $ \vec{P} $ and $ \vec{Q} $ such that $ \vec{f}(\vec{R}) = \vec{R} \times \vec{P} + \vec{Q} $
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i have tried expanding, which gives a possible term for Q, but i cannot see how the remaining terms can be written in the form R x P – user66419 Mar 12 '13 at 12:25
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@Muphrid what does the notation $\oint_C f d\vec r$ mean? What kind of integral is this? It is not a type of line integral that I am familiar with. – nullUser Mar 12 '13 at 15:59
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1@nullUser it is $ \oint{ f \frac{d\vec{r}}{dt} dt} $ – user66419 Mar 12 '13 at 16:32
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$\def\VP{{\bf P}} \def\VQ{{\bf Q}} \def\VR{{\bf R}} \def\VS{{\bf S}} \def\ve{{\bf e}} \def\vf{{\bf f}} \def\vr{{\bf r}} \def\o{\cdot}$We have \begin{align*} \vf(\VR) &= \oint_C |\vr-\VR|^2 d\vr \\ &= \oint_C (r^2 - 2\VR\o\vr+R^2)d\vr \\ &= \oint_C r^2 d\vr -2\oint_C (\VR\o\vr) d\vr +R^2\underbrace{\oint_C d\vr }_{{\bf 0}}. \end{align*} Note that $d\vr = \ve_i(\ve_i\o d\vr)$, where a sum over the index $i$ is implied and where $\{\ve_i\}_{i=1}^3$ is a (constant) orthonormal basis of $\mathbb{R}^3$. Thus, \begin{align*} \oint_C (\VR\o\vr) d\vr &= \ve_i\left\{\oint_C [(\VR\o\vr)\ve_i]\o d\vr\right\} \\ &= \ve_i \left(\iint_S \left\{\nabla\times[(\VR\o\vr)\ve_i]\right\}\o d\VS\right), \end{align*} by Stokes' theorem. It is a straightforward exercise to show that $$\nabla\times[(\VR\o\vr)\ve_i] = \VR\times\ve_i.$$ Thus, \begin{align*} \oint_C (\VR\o\vr) d\vr &= \ve_i\left[\iint_S(\VR\times\ve_i)\o d\VS\right] \\ &= \ve_i\left[\iint_S(d\VS\times\VR)\o\ve_i\right] \\ &= \iint_S d\VS\times\VR \\ &= -\VR\times\iint_S d\VS. \end{align*} Therefore $$\vf(\VR) = \VR\times\VP + \VQ,$$ where $\VQ = \oint_C r^2 d\vr$ and $\VP = 2\iint_S d\VS$.
Addendum
There appear to be at least two partial duplicates which I found after answering this question:
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