If the divergence of a vector field is zero

adamabel
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Homework Statement


If the divergence of a vector field is zero, I know that that means that it is the curl of some vector. How do I find that vector?


Homework Equations


Just the equations for divergence and curl. In TeX:
\nabla\cdot u=\frac{\partial u_x}{\partial x}+\frac{\partial u_y}{\partial y}+\frac{\partial u_z}{\partial z}
and the equivalent for curl.


The Attempt at a Solution


I really don't know at all how to find an answer.
 
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The divergence of the curl of ANY vector is =0. You can't find that "vector" without some more information, eg boundary conditions.
 
adamabel said:

Homework Statement


If the divergence of a vector field is zero, I know that that means that it is the curl of some vector. How do I find that vector?
the statement: \nabla\cdot(\nabla\times A)=0 is true for all vector field A. So without any additional info, you just have an arbitrary vector field.
 
So when a problem gives a vector field where it's divergence is zero, and it asks to find a vector field such that the curl of the vector field is the given vector field, I can just choose any vector field?
 
No, those responses were to what you had posted before- that all you knew about the vector field was that its divergence was equal to 0. You did not say you were given a vector field that happened to have divergence equal to 0!

If you are given a vector field, say, u(x,y,z)i+ v(x,y,z)j+ w(x,y,z)k with divergence 0, Then write out the formula for curl of a vector field and set the components equal:
\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}= u
\frac{\partial f}{\partial z}- \frac{\partial h}{\partial x}= v
\frac{\partial g}{\partial x}- \frac{\partial f}{\partial x}= w

Solve those for f, g, h,
 
I already knew that; I suppose I just didn't write it out clearly enough. But what was confusing me was how to solve for those. It seems like that is a system of PDEs, and I have no idea how to solve those.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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