vector field question

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...

Answer 1

Hints: Note that ${\rm Dom}(F) = \Bbb R^3$ is simply connected, so said $f$ exists if and only if $\nabla \times F = {\bf 0}$.

On item $(b)$ there is a typo: the point $(0,0,1)$ does not satisfies $x=y=z^2$. Maybe it meant $(1,1,1)$, on which case the parametrization ${\bf r}(t) = (t^2,t^2,t), \,0 \leq t \leq 1$ will work (try to convince yourself of that). However, if the answer to item $(a)$ is affirmative and $f$ does exist, then by the Fundamental Theorem of Calculus: $$\int_C F\cdot{\rm d}{\bf r} = f(1,1,1)-f(0,0,0).$$

Answer 2

The way to approach this is to consider $F(x,y,z) = F_x \hat{x} + F_y \hat{y} + F_z \hat{z}$, and compute the following integrals:

$$ \int F_x(x,y,z)dx \hspace{3pc} \int F_y(x,y,z) dy \hspace{3pc} \int F_z(x,y,z)dz $$

I'll compute the first one for you: $\int (yz + \sin x)dx = xyz - \cos x + c(y,z)$ where we observe that the "constant" term is actually a function of $y$ and $z$ since if we take the partial derivative of this function with respect to $x$ the $c$ term will drop out. You need to compute the corresponding integrals for $F_y$ and $F_z$ and ask yourself can the constant terms I get from integrating each equation fit into the other equations? For instance, integrating the second equation will yield $xyz - y^2 + d(x,z)$. We therefore see that the first and second integrals have $xyz$ in common, and it's completely reasonable that $-y^2$ might be a part of $c(x,y)$. Additionally, it's also quite possible that $-\cos x$ is a part of $d(x,z)$.

Once you compute the scalar field, the second part can be done by the fundamental theorem of line integrals: $\int F\cdot \textbf{dr} = f(x,y,z) |_{\textbf{r}_0}^{\textbf{r}_1}$.

Answer 3

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.