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The vector field $\vec F(\vec R)$ is defined by

$$\vec F(\vec R) = \oint_C \|\vec r − \vec R\|^2 d\vec r$$

where $\vec r$ lies on the simple closed curve $C$. Show that there are constant vectors $\vec A$ and $\vec B$ such that $\vec F(\vec R) = \vec R \times \vec A + \vec B$. Deduce that $\nabla\times \vec F = -4\iint_S d\vec S $, Where $S$ is any smooth surface spanning $C$.

David H
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Tom King
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  • What do you mean by smooth surface spanning $C$? – Daniel Donnelly Mar 29 '14 at 09:45
  • @EnjoysMath A surface $S$ whose bounding curve is $C$: $\partial S = C$. – David H Mar 29 '14 at 09:49
  • Duplicate answer: http://math.stackexchange.com/questions/724459/line-integrals-cross-products-surface-integrals-and-stokes-theorem-related-pr

    Somehow I can't flag it because it doesn't find this thread.

     – orion Mar 29 '14 at 10:43

1 Answers1

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For the first part,

$$\vec F(\vec R) = \oint_{\partial S} \|\vec r − \vec R\|^2 d\vec\ell\\ =\iint_Sd\vec a\times\nabla\left(\|\vec r − \vec R\|^2\right)\\ =\iint_Sd\vec a\times\left(2(\vec r - \vec R)\right)\\ =2\iint_Sd\vec a\times\left(\vec r - \vec R\right)\\ =2\iint_Sd\vec a\times\vec r -2\iint_Sd\vec a\times\vec R\\ =2\iint_Sd\vec a\times\vec r+2\vec R\times\iint_Sd\vec a\\ =\vec R \times \vec A + \vec B,$$

where $\vec A := 2\iint_Sd\vec a$ and $\vec B := 2\iint_Sd\vec a\times\vec r$.

For the second part,

$$\nabla\times\vec F = \nabla\times\left(\vec R \times \vec A + \vec B\right)\\ =\nabla\times\left(\vec R \times \vec A\right)\\ =-\vec A(\nabla\cdot\vec R)+(\vec A \cdot \nabla)\vec R\\ =3\vec A + \vec A\\ =-2\vec A\\ =-4\iint_Sd\vec a.$$

Notes:

$$\nabla\left(\|\vec r − \vec R\|^2\right)=2((\vec r - \vec R)\cdot\nabla)(\vec r - \vec R)=2(\vec r - \vec R)$$

David H
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