Examine whether $b: \mathbb R ^3 → \mathbb R^3 $ is a vortex field and determine a vector potential $a$.
$b=\begin{pmatrix} 4x^4y^3z^2-3x^4z^2 \\ -4x^3y^4z^2 \\ 4z^3x^3 \end{pmatrix} $
$a=?$
Hint: Take $a_1$ as $0$.
My idea:
I calculated $curl$ $b$ and because of the fact that $curl$ $b$ $\neq0$ I came to the conclusion that this vector field is "swirly" (vortex field).
Is this correct?
How do I calculate vector potential of this vector field? I found in the book that vector potential is equal the gradient when the vector field is not swirly. But here it is, and how to calculate it in this case?
According to the definition of the vector potential (given here: https://en.wikipedia.org/wiki/Vector_potential) $a$ is the vector potential of $b$ if
$$b=\nabla \times a.$$
Then
$$a(x,y,z)=\begin{bmatrix} 0\\ x^4z^3\\ x^4y^4z^2 \end{bmatrix}.$$
Indeed,
$$b=\begin{bmatrix} \frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}\\ \end{bmatrix} \times\begin{bmatrix} 0\\ x^4z^3\\ x^4y^4z^2 \end{bmatrix}= \begin{vmatrix} i&j&k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ 0&x^4z^3&x^4y^4z^2 \end{vmatrix}= $$
$$=\begin{bmatrix} 4x^4y^3z^2-3x^4z^2\\ -4x^3y^4z^2\\ 4x^3z^3 \end{bmatrix}.$$