<s>[INST] Can anyone tell me the formula for this vector field? 3Blue1Brown's video on curl

$$V=\bigg[ \begin{array}{ccc}
P\left(x,y\right) \\
Q\left(x,y\right) \end{array} \bigg] $$


*

*$P$ gives you the x component at all points in space.

*$Q$ gives you the y component at any point in space.
What is the explicit formula for the above vector field?


*

*vortexes at $(6,0)$, $(-6,0)$, $(0,6)$, $(0,-6)$

*and a source and a sink at the origin.

*symmetrical
Hints, links welcome
 [/INST] Working through Andrew D. Hwang's answer above:
$$F\left(x,y\right)=\left(\frac{2(-y)}{1+\sqrt{x^2+y^2}},\frac{2(x)}{1+\sqrt{x^2+y^2}}\right)$$

$$G\left(x,y\right)=\left(\frac{-2(y)}{1+\sqrt{(x-6)^2+(y)^2}}+\frac{-2(y)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(y-6)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(y+6)}{1+\sqrt{(x)^2+(y+6)^2}},\\
\\
\\
\frac{-2(x-6)}{1+\sqrt{(x-6)^2+(y)^2}}\\
\\
+\frac{-2(x+6)}{1+\sqrt{(x+6)^2+(y)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y-6)^2}}-\frac{-2(x)}{1+\sqrt{(x)^2+(y+6)^2}}\right)$$



*

*first four terms give the x component 

*the second four terms compute the y component



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