<s>[INST] How to find the div(curl $v$)? 
let $v= ( v_1 ,v_2,v_3)$ be  a  vector field on $\mathbb{R}^3$  where $v_1 = \sqrt {1+x^2+y^2} , v_2= \sqrt {1+z^2} $and $v_3= \sqrt{ 1 +x^2y^2z^2}$. Evaluate div(curl $v$)

My attempt : $$Cur v= \begin{bmatrix} i &j&k\\ \frac{dv_1}{dx}&  \frac{dv_2}{dy} & \frac{dv_3}{dz} \\ \sqrt {1+x^2+y^2} &\sqrt {1+z^2}  &  \sqrt{ 1 +x^2y^2z^2} \end{bmatrix}$$
$$Cur v= \begin{bmatrix} i &j&k\\ \frac{x}{\sqrt {1+x^2+y^2}}&  0 & \frac{zx^2y^2}{ \sqrt{ 1 +x^2y^2z^2}} \\ \sqrt {1+x^2+y^2} &\sqrt {1+z^2}  &  \sqrt{ 1 +x^2y^2z^2} \end{bmatrix}$$
= $i(\frac{2zx^2y^2}{ \sqrt{ 1 +x^2y^2z^2}})(\sqrt {1+z^2}) - j((\sqrt {1+x^2+y^2})(\frac{2zx^2y^2}{ \sqrt{ 1 +x^2y^2z^2}})-(\frac{x}{\sqrt {1+x^2+y^2}})\sqrt{ 1 +x^2y^2z^2})) + k(\sqrt {1+z^2} )(\frac{x}{\sqrt {1+x^2+y^2}})$
After that im not able to proceed  further 
 [/INST] It is a general fact that $\operatorname{div}(\operatorname{curl}(F))=0$. You can verify this by expanding. Let $F=(P,Q,R)$. Then 
$$\operatorname{curl}(F)=(R_y-Q_z)i-(R_x-P_z)j+(Q_x-P_y)k$$ 
by definition. Then 
$$ \operatorname{div}(\operatorname{curl}(F))=R_{xy}-Q_{xz}-R_{yx}+P_{yz}+Q_{zx}-P_{zy}$$
$$=(R_{xy}-R_{yx})+(P_{yz}-P_{zy})+(Q_{zx}-Q_{xz})=0+0+0.$$
Here, I used the fact that mixed second partials agree for $\mathscr{C}^2$ functions. The only thing that one should be careful about here is whether or not your functions are $\mathscr{C}^2$, but I am fairly certain they are by inspection.
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