Question

Find the values of p, q and r such that the force

F~ = p y2z3 ˆi − q xyz3jˆ + r xy2z2 ˆk is a conservative force. What is the value of the potential?

Answer 1

We know that force is conservative if and only if $\vec\nabla \times F\equiv0$ (strictly speaking, we also need to suppose that domain of F is simply connected, but in this case it is defined on the whole $\mathbb{R}^3$). Let's calculate $\text{curl }\vec F$ :

$\text{curl } \vec F = \begin{pmatrix} \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z} \\ \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x} \\ \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \end{pmatrix} \equiv 0$

This condition gives us 3 equations :

1. $2rxyz^2 + 3 q xyz^2 = 0$ that gives $2r+3q=0$
2. $3py^2 z^2-ry^2 z^2=0$ that gives $3p-r=0$
3. $-qyz^3-2pyz^3=0$ that gives $2p+q=0$

This system of equation gives us a family of solutions of a form $q = -2p, r = 3p, p\in \mathbb{R}$.

Now to find the potential, we can either "guess" it from the expressions $-\frac{\partial V}{\partial x} = py^2 z^3, -\frac{\partial V}{\partial y} = 2pxyz^3$, $-\frac{\partial V}{\partial z} = 3pxy^2 z^2$ and thus we get $V = -pxy^2 z^3+const$, or for a stricter approach define $V(x,y,z) = V(0) - \int_\gamma \vec{F}(x,y,z) \cdot \vec{\tau} dt$, where $\gamma$ is any curve joining the origin and the point (x,y,z), $\tau$ is the tangent vector to $\gamma$ and t is a parametrization of $\gamma$. As $\text{curl } \vec F \equiv 0$ the choice of $\gamma$ doesn't change the integral and thus the definition is consistent. We can choose, for example, a straight line joining the origin with the point (x,y,z) and thus we get ($\gamma(t)=(xt,yt,zt), t\in[0;1], \tau = (x,y,z)$ :

$V(x,y,z) = V(0) - p \int_0^1 (xy^2 z^3 t^5+ 2xy^2z^3 t^5+ 3xy^2z^3 t^5)dt=V(0) - pxy^2z^3$, we get the same expression as earlier.