# A force field is described by <F_x,F_y,F_z> = < xy , 2z-y^2 +x, 2y -zx > . Is this force field conservative?

Feb 25, 2017

The force field is not conservative; $c u r l \left(\vec{F}\right) \ne 0$.

#### Explanation:

If $\vec{F}$ is a vector (force) field in ${\mathbb{R}}^{3}$, then the curl of $\vec{F}$ is the vector $c u r l \left(\vec{F}\right)$ and is written as $\nabla \times \vec{F}$. The vector field $\vec{F}$ is conservative if and only if $\vec{F} = \nabla f$ for some potential function. If $\vec{F}$ is conservative, $c u r l \left(\vec{F}\right) = \vec{0}$. However, this does not mean that a vector field with $c u r l = \vec{0}$ is absolutely conservative; the vector field must also have a potential function to be conservative, but the first prerequisite is that $c u r l \left(\vec{F}\right) = \vec{0}$.

As stated above, the curl is given by the cross product of the gradient of $\vec{F}$ (in the form $< P , Q , R >$) and itself. We have $\vec{F} = < x y , 2 z - {y}^{2} + x , 2 y - z x >$ where

$P = x y$

$Q = 2 z - {y}^{2} + x$

$R = 2 y - z x$

The curl of the vector field is then given as:

$\left\mid \begin{matrix}\vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ x y & 2 z - {y}^{2} + x & 2 y - z x\end{matrix} \right\mid$

We take the cross product as we usually would with vectors, except we'll be taking partial derivatives each time we multiply by a partial differential.

For the $\vec{i}$ component, we have $\frac{\partial}{\partial y} \left(2 y - z x\right) - \frac{\partial}{\partial z} \left(2 z - {y}^{2} + x\right)$

$\implies \left(2 - 2\right) = 0$

So far, so good.

For the $\vec{j}$ component, we have $\frac{\partial}{\partial x} \left(2 y - z x\right) - \frac{\partial}{\partial z} \left(x y\right)$

$\implies - \left(- z - 0\right) = z$

Remember that if we take the partial of a function with respect to some variable which is not present, the partial derivative is $0$ as we treat all other variables as constants.

For the $\vec{k}$component, we have $\frac{\partial}{\partial x} \left(2 z - {y}^{2} + x\right) - \frac{\partial}{\partial y} \left(x y\right)$

$\implies \left(1 - x\right) = 1 - x$

This gives a final answer of $c u r l \left(\vec{F}\right) = < 0 , z , 1 - x > \ne \vec{0}$

$\therefore$ Force field is not conservative