I've been using Stewart's Calculus: Early Transcendentals in my Calculus class, and one of the definitions of curl offered by the book was $$\text{curl } \vec F = \nabla\times \vec F$$ where $\,\nabla=\left\langle\dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z}\right\rangle$. My question is, why are we allowed to put operations into a vector and then "multiply" them with components of $\vec F$ in that way?

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    $\begingroup$ For your purposes, you should think of it as a mnemonic, not as a definition. $\endgroup$ – Ian Nov 16 '15 at 2:08
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    $\begingroup$ No, this is simply a shorthand. $\endgroup$ – MathMajor Nov 16 '15 at 2:08

I agree with @Ian, this notation is used for memorizing formula rather than as a strict and formal definition of mathematical concept. Let me quote article about curl from Wikipedia:

The notation $\,\nabla \times \overrightarrow{F}\,$ has its origins in the similarities to the $3$ dimensional cross product, and it is useful as a mnemonic in Cartesian coordinates if $\,\nabla\,$ is taken as a vector differential operator del. Such notation involving operators is common in physics and algebra. However, in certain coordinate systems, such as polar-toroidal coordinates (common in plasma physics), using the notation $\,\nabla \times \overrightarrow{F}\,$ will yield an incorrect result.

Expanded in Cartesian coordinates ... , $\,\nabla \times \overrightarrow{F}\,$ is, for $\,\overrightarrow{F}\,$ composed of $\,\left[\begin{array}{3} F_x,& F_y, &F_z\end{array}\right]\,$:

\begin{align} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ {\frac{\partial}{\partial x}} & {\frac{\partial}{\partial y}} & {\frac{\partial}{\partial z}} \\ F_x & F_y & F_z \end{vmatrix} \end{align}

where $\,\mathbf{i},\, \mathbf{j},\,$ and $\,\mathbf{k}\,$ are the unit vectors for the $\,x,\, y,\,$ and $\,z\,$ axes, respectively. This expands as follows:

\begin{align} \left(\frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}\right) \mathbf{i} + \left(\frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}\right) \mathbf{j} + \left(\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y}\right) \mathbf{k} \end{align}

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    $\begingroup$ If the unit vectors $\vec F$ in the general curvilinear coordinate system are kept in side the differential operations, then actually one can proceed to evaluate straightforward. So, for example in cylindrical coordinates $$\nabla \times \vec F=\nabla \times \left(\hat \rho F_{\rho}+\hat \phi F_{\phi}+\hat z F_z\right)$$ with $\nabla$ given by the $Del$ operator in cylindrical coordinates. This is not convenient, but yields the correct result for the curl after painstaking care. $\endgroup$ – Mark Viola Nov 16 '15 at 4:39

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