Proving that $\vec F=yz(2x+y+z)\hat i+zx(x+2y+z)\hat j+xy(x+y+2z)\hat k$ is conservative field

Question

I need to prove that $\vec F$ is conservative field

$$\vec F=yz(2x+y+z)\hat i+zx(x+2y+z)\hat j+xy(x+y+2z)\hat k$$

My attempt: $\vec{F}$ is conservative iff $\nabla \times \vec{F} = 0$ $$ \begin{vmatrix} &\hat i&\hat j &\hat k\\ &\frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ &yz(2x+y+z)&zx(x+2y+z)&xy(x+y+2z) \end{vmatrix} $$

$\color{blue}\Longrightarrow$

$$ =\begin{vmatrix} &\hat i&\hat j &\hat k\\ &2yz&2xz&2xy\\ &yz(2x+y+z)&zx(x+2y+z)&xy(x+y+2z) \end{vmatrix} $$

$$=[x+y+2z-2]\hat i -[x+y+2z-2]\hat j +[x+2y+z-2]\hat k\color{red}{\neq 0}$$

Where am I wrong?

Answer 1

Note that you have a 'potential'

$$ \phi = x^2yz + xy^2z + xyz^2 $$

so that

$$ \vec{F} = \vec\nabla \phi, $$

whence

$$ \vec\nabla \times \vec{F} = \vec\nabla \times \vec\nabla \phi = 0, $$

so it is conservative.

Answer 2

It's just as easy to use the definition:

$\vec F$ is conservative $\Leftrightarrow $ there is a $\varphi :\mathbb R^{3}\to \mathbb R$ such that $\nabla \varphi =\vec F$. The method for finding $\varphi $ is straightforward and has the advantage that it is consistent; it will tell you if such a $\varphi $ exists.

If there is a $\varphi $ with the advertised property, then

$\frac{\partial \varphi }{\partial x}=yz(2x+y+z)$ which implies that

$\varphi =yz(x^{2}+yx+zx)+g(y,z)$. Then,

$\frac{\partial \varphi }{\partial y}=zx^{2}+2zxy+z^{2}x+\frac{\partial g(y,z)}{\partial y}$ and this must be equal to

$zx(x+2y+z)$ so that

$\frac{\partial g(y,z)}{\partial y}=0$. This means that

$g(y,z)=f(z)+c$ for some constant $c\in \mathbb R$. So we have, up to now

$\varphi =yz(x^{2}+yx+zx)+f(z)+c$ from which

$\frac{\partial g}{\partial z}=yx^{2}+y^{2}x+2zyx+f'(z)$ and this must be equal to

$xy(x+y+2z)$ so that

$f'(z)=0\Rightarrow f(z)=c_{1}$. Then, finally we have

$\varphi =yz(x^{2}+yx+zx)+C$ where $C=c+c_{1}$.