How Do I Find the Potential Function for This Vector Field?

[xz5x]

I have to found a potential. So, I don't know how to integrate this:
F = (x+2y+4z)i + (2x-3y-z)j + (4x-y+2z)k (bold are vectors)

I think it goes like this:
Potential...F = -grad U
(x+2y+4z, 2x-3y-z, 4x-y+2z) = (-dU/dx, -dU/dy, -dU/dz)
(-dU/dx,-dU/dy,-dU/dz are partial deriv.)


-dU/dx=x+2y+4z
-U=x^2/2+2y+4z+C(y,z)
What know? Please help!

 

[Galileo]

You've integrated -dU/dx with respect to x. The ideas good, but you did not integrate correctly.

You know the x dependence of U and you know dU/dy and dU/dz as well.
This is enough info to find U up to a constant.

Good luck.

 

[HallsofIvy]

dU/dx= x+2y+4z so U= (1/2)x²+ 2xy+ 4xz+ g(y,z)

Given that, dU/dy= 2x+ dg/dy= 2x-3y-z so dg/dy= -3y- z

Must have g= -(3/2)y²- yz+ h(z) and so
U= (1/2)x²+ 2xy+ 4xz+ (3/2)y²- yz+ h(z).

Given that, dU/dz= 4x- y+ h'= 4x-y+2z so h'= 2z.

Now, what is h(z) and what is U?

 

[xz5x]

Oh, yes..now I see my mistake
U is Potential.
So, the result is:
U = -1/2*x^2-2xy-4xz+3/2*y^2+yz-2+C
Thank you for your help!

 

[HallsofIvy]

No, You don't see. You copied the part I gave you did the tiny part I left incorrectly. You differentiated h'= 2z rather than finding the anti-derivative.

 

[xz5x]

But this result is correct. Like Galileo said - I had a good idea but I didn't integrate correctly

 

[HallsofIvy]

No, the result is NOT correct.

If U= -1/2*x^2-2xy-4xz+3/2*y^2+yz-2+C

Then U_z= -4x+ 1. The k component of grad U would be -4x+1, not -4x+y-2z as you want.

I said, earlier, "U= (1/2)x²+ 2xy+ 4xz+ (3/2)y^2-- yz+ h(z)
Given that, dU/dz= 4x- y+ h'= 4x-y+2z so h'= 2z.

Now, what is h(z) and what is U?"

If h'= 2z then z= z²+ C (since h depends only on z, C is a constant.

Then U(x,y,z)= (1/2)x²+ 2xy+ 4xz+ (3/2)y²- yz+ z²+ C.