Getting the Potential Energy from a Conservative Force

[Trade]

Homework Statement

The problem basically asked me to check if a given force was conservative and if it was conservative, also find the potential energy.

F = k(x,2y,3z)

Homework Equations

(\nabla X F) = Curl of F
U = Integral of F

3. attempt

So the force is clearly conservative as the curl is equal to zero. I know that I basically need to take the integral of the force, but I'm a bit confused as to how to set up an integral given how the force was given, or even what to integrate with respect to. Any point in the right direction would be awesome. Thanks.

 

[Simon Bridge]

Set it up as a partial DE that you have to solve.

$$\vec{F}=-\vec{\nabla} U$$

 

[Trade]

Set it up as a partial DE that you have to solve.

$$\vec{F}=-\vec{\nabla} U$$

Okay, so I end up with something along the lines of...

K [x, 2y, 3z] = - [dU/dx, dU/dy, dU/dz]

and then we integrate

U = -k [ (1/2)x², y²,(3/2)z²]
Would it be proper to add a constant at the end of each part, noting that it's terms of the other two variables? Something like

U = -k [ (1/2)x² + C_y,z, y² + C_x,z , (3/2)z² + C_x,y]EDIT:

I think I'm being silly. I should just put all the components together right? So that the other components are the Constant for each other right? So it looks like

U = -k [ (1/2)x² + y²+ (3/2)z² ]

 

[Simon Bridge]

You should end up with just one arbitrary constant.

i.e. You need to evaluate c_y,z etc.
That means solving as simultaneous equations or go back and solve the DEs one at a time instead of all in one go.

$F_x=kx=-\frac{\partial}{\partial x}U(x,y,z)\\ \qquad \Rightarrow U(x,y,z)=-\frac{1}{2}kx^2+c(y,z)$

... i.e. c(y,z) is a function of y and z alone.

$F_y = 2ky=-\frac{\partial}{\partial y}U(x,y,z) = -\frac{\partial}{\partial y}\big(-\frac{1}{2}kx^2+c(y,z)\big) = -\frac{\partial}{\partial y}c(y,z)\\ \qquad \Rightarrow c(y,z)=\cdots +d(z) \\ \qquad \qquad \Rightarrow U(x,y,z)=\cdots$

... you should be able to complete it from here.

 

[Trade]

Awesome, I think I have it from here. Thanks.