Calc. 3 Conservative Vector Field question

Question

Question: The following vector field is conservative in $\mathbb{R}^{3}$

$\hat{F}(x,y,z) = ye^x\hat{\imath} + (e^{x}+Azcos(y))\hat{\jmath}+(sin(y)+Bz^{2})\hat{k}$ Find the constants $A$ and $B$.

By the neccessary condition rule and using partial derivatives; $f_{yz} = f_{zy} \rightarrow A = 1$.

But it seems that when I use the same technique, the $Bz^{2}$ term is always treated as a constant. So I think that $B$ can be any value? How do I find $B$ in this case?

Thanks.

Answer 1

Notice that $F-Bz^2\hat{k}$ is also a conservative vector field because $Bz^2\hat{k}$ is a conservative field and the sum of conservative fields is conservative. In other words the conservative condition isn't enough to specify $B$.

Answer 2

You're right. Whatever value $B$ takes, the field is always conservative. So $B$ is arbitrary.