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To solve this question, how do I define any vector field $F$, in order to solve it? I called $F = (ax,by,cz)$, in which case already $\nabla\times F = 0$. How would i go about proving this? Many thanks!

stellarhawk 34
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    The most brutally simple approach: Write out the curl of a generic $\vec{F}=(F_x,F_y,F_z)$, and then take its divergence. The only assumption required is that all partial derivatives commute, e.g. $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}F_z=\frac{\partial}{\partial y}\frac{\partial}{\partial x}F_z.$$ – Semiclassical Apr 22 '16 at 20:18
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    http://www.mathpages.com/home/kmath103/kmath103_files/image007.gif – Matt Apr 22 '16 at 20:18

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It's better if you define $F$ in terms of smooth functions in each coordinate. For instance I would write $F = (F_x, F_y, F_z) = F_x\hat{i} + F_y \hat{j} + F_z \hat{k}$ and compute each quantity one at a time. First you'll compute the curl:

$$ \nabla \times F \;\; =\;\; \left | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ \partial_x & \partial_y & \partial_z \\ F_x & F_y & F_z \\ \end{array} \right | \;\; =\;\; G_x\hat{i} + G_y \hat{j} + G_z\hat{k} $$

where the functions $G_x, G_y, G_z$ are obtained by computing the determinant. Then you will want to compute

$$ \nabla\cdot (\nabla\times F) \;\; =\;\; \frac{\partial G_x}{\partial x} + \frac{\partial G_y}{\partial y} + \frac{\partial G_z}{\partial z}. $$

You should find that the last equation yields zero.

Mnifldz
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