basic potential problem in mechanics

Question

I am new to this field and I'm having trouble with this question where it asks me to calculate the potential from the basis of the origin. The question is:

$\overrightarrow{F} = (0, -2y + 3y^2,0)$

in $(x, y , z)$

I have no idea what should I do.

Answer 1

You want to find $\phi$ so that $$(0, -2y+3y^2, 0) = \vec F = - \nabla \phi = - \left( \frac{\partial\phi}{\partial x}, \frac{\partial\phi}{\partial y}, \frac{\partial\phi}{\partial z} \right) = \left( -\frac{\partial\phi}{\partial x}, -\frac{\partial\phi}{\partial y}, -\frac{\partial\phi}{\partial z} \right),$$ i.e. $$\begin{align} \frac{\partial\phi}{\partial x} &= 0, \\ \frac{\partial\phi}{\partial y} &= 2y-3y^2, \\ \frac{\partial\phi}{\partial z} &= 0. \end{align}$$ Can you solve this? I recommend starting with $\frac{\partial\phi}{\partial y} = 2y-3y^2.$

Answer 2

We have that $$\underline{F}=-\text{grad}U$$ $$(0, -2y+3x^2, 0)^T=-(\partial_xU, \partial_yU, \partial_zU)^T$$ They are equal if and only if all of their components are equal: $$0=-\partial_xU$$ $$-2y+3y^2=-\partial_yU$$ $$0=-\partial_zU$$ Where $U$ is the function of the position: $U(x, y, z)$ Now let's pick $1$ out of the $3$ equations, for example the first one: $$0=\partial_xU$$ If we integrate both sides with respect to $x$ we will get that $$U=f(y, z)$$ Where $f(y, z)$ can be any function of $y$ and $z$. Now let's substitute this $U$ into another equation, for example into the second one: $$\partial_y U=\partial_y f(y, z)=2y-3y^2$$ Integrating both sides with respect to $y$ we will get that $$f(y, z)=y^2-y^3+g(z)$$ Where $g(z)$ can be any function of $z$. Now let's substitute it into the third equation: $$\partial_zU=\partial_z(y^2-y^3+g(z))=g'(z)=0$$ And I think you can solve it for $g$, and you can put everything together to get $U$.