Will $vec{f}(x,y,z)=nabla times vec{g}(x,y,z)$ here?











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Suppose we have a vector field $vec{f}(x,y,z)$ defined everywhere in space except a certain region say $(y=3)$. We also know divergence of $vec{f}(x,y,z)$ is zero at all points where $vec{f}$ is defined. From this information, is it valid to say that: $$vec{f}(x,y,z)=nabla times vec{g}(x,y,z)$$ at all points wherever $vec{f}(x,y,z)$ is defined? Why? Why not?






multivariable-calculus vectors divergence curl







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    Suppose we have a vector field $vec{f}(x,y,z)$ defined everywhere in space except a certain region say $(y=3)$. We also know divergence of $vec{f}(x,y,z)$ is zero at all points where $vec{f}$ is defined. From this information, is it valid to say that: $$vec{f}(x,y,z)=nabla times vec{g}(x,y,z)$$ at all points wherever $vec{f}(x,y,z)$ is defined? Why? Why not?






    multivariable-calculus vectors divergence curl







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      Suppose we have a vector field $vec{f}(x,y,z)$ defined everywhere in space except a certain region say $(y=3)$. We also know divergence of $vec{f}(x,y,z)$ is zero at all points where $vec{f}$ is defined. From this information, is it valid to say that: $$vec{f}(x,y,z)=nabla times vec{g}(x,y,z)$$ at all points wherever $vec{f}(x,y,z)$ is defined? Why? Why not?






      multivariable-calculus vectors divergence curl







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      Alec is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      Suppose we have a vector field $vec{f}(x,y,z)$ defined everywhere in space except a certain region say $(y=3)$. We also know divergence of $vec{f}(x,y,z)$ is zero at all points where $vec{f}$ is defined. From this information, is it valid to say that: $$vec{f}(x,y,z)=nabla times vec{g}(x,y,z)$$ at all points wherever $vec{f}(x,y,z)$ is defined? Why? Why not?






      multivariable-calculus vectors divergence curl




      multivariable-calculus vectors divergence curl






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