Is there a vector field F such that Curl(F) = ($xy^2$, $yz^2$, $zx^2$)? Explain.
Ive been testing it out myself, coordinate by coordinate, and once determining what $F_3$ or $F_1$ or $F_2$ would need to be I realize that i'm pretty sure it is not possible to have a vector field that produces that curl. I am not absolute in my answer and im having a tough time figuring out how to explain this. Any help would be appreciated.
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$\begingroup$You have a vector field $\mathbf v=(xy^2,yz^2,zx^2)$ and you are searching if this field admits a vector potential $\mathbf F$ such that $ \nabla \times \mathbf F=\mathbf v$ .
A necessary condition for this is that the divergence of $\mathbf v$ is null $$ \nabla \cdot \mathbf v= \nabla \cdot (\nabla \times \mathbf F)=0 $$
This is a consequence of the fact that $rot$ and $div$ are exterior derivatives, and a fundamental theorem states that the exterior derivative of an exterior derivative (of a smooth function) is zero. (you can see here for an introduction to n-forms and exterior derivatives)
In this case $\nabla \cdot \mathbf v= y^2+z^2+x^2 \ne 0$ for $x,y,z \ne 0$ , so the fileld $\mathbf v$ does not admits a vector potential.
$\endgroup$ $\begingroup$Let $X$ a vector field in $\Bbb{R}$, thus the followinf always is true: $$\mbox{div}(\mbox{curl} (X))=0$$ But if you suppose there exists some vector field $F$ such that $curl(F)=(xy^2,yz^2,zx^2)$ then $\mbox{div}(\mbox{curl} (F))=0$, thus $$y^2+z^2+x^2=0$$ But this is possible iff, $x=y=z=0$. Thus $F\equiv0$.
So, there is not vector field with those coordinates in the curl.
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