I want to prove if $y^2 i − z^2 j + x^2k$ is a conservative vector field. Is it sufficient to say that it is 'simply connected', and show that the curl is $0$? Or do I need to prove it by attempting to find a potential function?
Is $y^2 i − z^2 j + x^2 k$ conservative vector field
calculusmultivariable-calculusvector analysisvectors
Related Solutions
To begin, the reason the gradient of a scalar-valued function is a vector field is: $$ \nabla \phi = \langle \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y},\frac{\partial \phi}{\partial z} \rangle $$ which is the formula for a vector field. For each point $p$ in some region of $\mathbb{R}^3$ we assign $\nabla \phi (p)$. This makes $\nabla \phi$ a vector field.
Now, if $\vec{F} = \nabla \phi$ then $\nabla \times \vec{F} = \nabla \times \nabla \phi = 0$. However, the converse is not necessarily true. It is possible to have $\nabla \times \vec{F} =0$ throughout some domain $U \subset \mathbb{R}^3$ and yet there does not exist $\phi$ on all of $U$ such that $\nabla \phi = \vec{F}$. If you want the curl vanishing to imply the existence of the potential function $\phi$ then you also need to have a topological condition on $U$. In particular, if $U$ is simply connected then we can apply Stokes' Theorem to arbitrary surfaces in $U$ and for each surface $S$: $$ \int_{\partial S} \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = 0.$$ Hence integrals of $\vec{F}$ around loops in $U$ vanish hence $\vec{F}$ is path-independent and then we can prove $\displaystyle \phi(\vec{r}) = \int_{p}^{\vec{r}} \vec{F} \cdot d\vec{r}$ is a valid formula for the construction of a potential in $U$.
In any event, if you are currently taking multivariate calculus and you were just introduced to conservative vector fields then relax. In a week or two these things should all gel together. There are a few moving pieces, but, once you see how they all fit it's really pretty. We have the following equivalence: Suppose $U$ is an open connected subset of $\mathbb{R}^n$ then the following are equivalent
- $\vec{F}$ is conservative; $\vec{F}=\nabla \phi$ on all of $U$
- $\vec{F}$ is path-independent on $U$
- $\oint_C \vec{F} \cdot d\vec{r} =0$ for all closed curves $C$ in $U$
- (add condition $U$ be simply connected) $\nabla \times \vec{F}=0$ on $U$.
Any mapping, be it a vector field or a scalar function or something else, requires a domain.
It is true that where $\phi(x,y)$ is defined, $\nabla \phi = \vec F$. But $\vec F$'s domain is the plane minus the origin, while $\phi$'s domain is the plane minus a line (the $y$-axis).
Since there's no function with the same domain as $\vec F$ whose gradient is $\vec F$, $\vec F$ is not conservative.
Notice that the right half of the plane is simply connected, and as you've shown, $\vec F$ restricted to that domain is conservative. $\phi$ works as a potential on that domain.
The upshot is that the question of whether $\vec F$ is conservative on $U$ is a question not just about the component functions of $\vec F$ but the “shape” (we say topology) of $U$.
Best Answer
Our vector field is $$\mathbf{F}(x,y,z)=y^2\hat{\mathbf{i}}-z^2\hat{\mathbf{j}}+x^2\hat{\mathbf{k}}$$ It is conservative if $\nabla \times\mathbf{F}=0~~\forall (x,y,z)\in\mathbb{R}^3$. So you need to check if $$\nabla \times \mathbf{F}=\det\left(\begin{bmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ y^{2} & -z^{2} & x^{2} \end{bmatrix}\right)=0$$ Which is in fact not the case.