Let $$ \mathbf G(x,y,z)=(b\sin y)\mathbf i+(x\cos y+\cos z)\mathbf j+(cy\sin z)\mathbf k,\text{ when }\Phi(0,0,0)=1 $$ where $a,b,c \in \mathbb R$ a,b,c are constants.
How do I prove that the following vector field has a potential only for certain values of the constants involved in its definition?
I know that: $$ \begin{split} F_1 & = b\sin y\\ F_2 & = x\cos y+\cos z\\ F_3 & = cy\sin z \end{split} $$ and $$ \mathbf G(x,y,z)=\nabla\Phi(x,y,z)=\frac{\delta\Phi(x,y,z)}{\delta x}\mathbf i + \frac{\delta\Phi(x,y,z)}{\delta y}\mathbf j + \frac{\delta\Phi(x,y,z)}{\delta z}\mathbf k$$ thus $$ \frac{\delta\Phi(x,y,z)}{\delta x}\mathbf i=F_1=b\sin y\implies \Phi(x,y,z)=\int b\sin y \, dx = bx\sin y + C_1(y,z) $$
this is as far as I get how do I solve this?
If $G=\nabla\Phi$ then, since $\partial_i\partial_j\Phi=\partial_j\partial_i\Phi$, $\partial_i G_j=\partial_j G_i$. For$$G_1=b\sin x_2,\,G_2=x_1\cos x_2+\cos x_3,\,G_3=cx_2\sin x_3,$$we have$$\cos x_2=\partial_1G_2=\partial_2G_1=b\cos x_2\implies b=1$$and$$c\sin x_3=\partial_2G_3=\partial_3G_2=-\sin x_3\implies c=-1.$$We can then choose $\Phi=x_1\sin x_2-x_2\cos x_3=x\sin y-y\cos z$.