vector field question Consider the vector field $$F(x,y,z)=(zy+\sin x, zx-2y, yx-z)$$ (a) Is there a scalar field $f:\mathbb{R}^3 \rightarrow \mathbb{R}$ whose gradient is $F$?
(b) Compute $\int _C F\cdot dr \neq 0$ where the curve $C$ is given by $x=y=z^2$ between $(0,0,0)$ and $(0,0,1)$.
I have no idea how to do the first one and for the second one, is there any typo on the curve equation because I have no idea how to parameterize it...

Let $\phi(x,y,z)$ be the scalar field. If we find one, there exists one.
We know:$$\frac{\partial \phi}{\partial x}=F_x=zy+\sin x$$
Integrating with respect to $x$:
$$\phi(x,y,z) = zyx-\cos x+M(y,z) \: \: \: (1)$$
$M(y,z)$ is the integration constant. Because we only integrate with respect to $x$, we can let $M$ depends on $y$ and $z$. Now differentiating $(1)$ with respect to $y$ gives.
$$\frac{\partial \phi}{\partial y}=F_y=zx+\frac{\partial M}{\partial y}$$
And $F_y=zx-2y$. So:
$$zx-2y=zx+\frac{\partial M}{\partial y}$$
$$-2y=\frac{\partial M}{\partial y}$$
$$M(y,z) =-y^2+N(z)$$
For the same reasons as before $N$ depends on $z$.
Plugging this in to $(1)$, we get:
$$\phi(x,y,z) = zyx-\cos x-y^2+N(z)$$
Differentiating this to $z$ gives us:
$$\frac{\partial \phi}{\partial z}=F_z=yx+\frac{dN}{dz}=yx-z$$
So $$N(z)=-\frac{1}{2}z^2$$
So we get:
$$\phi(x,y,z) = zyx-\cos x-y^2-\frac{1}{2}z^2$$
Can you go further? Maybe I solved too much, but there are plenty of these kind of exercises.
