# Recovering function from its gradient

1. Dec 10, 2012

### Nero26

Hi all,
I've to recover a function ∅ such that F=∇∅,F=3x^2y i +(x^3+2yz)j+y^2k.
So, ∂∅/∂x=3x^2y, Integrating w.r.t. x
∅=x^3y+f(y,z) ,Assuming there may be function of y and z.----------1
∴∂∅/∂y=x^3+f'(y,z)
now, ∂∅/∂y=(x^3+2yz)=x^3+f'(y,z)
∴f'(y,z)=2yz
To find ∅ from 1 I've to find f(y,z).
My problem is how to get f(y,z) from f'(y,z),should I integrate it w.r.t. y or z?

Last edited: Dec 10, 2012
2. Dec 10, 2012

### LCKurtz

OK so far

You shouldn't use ' for a partial derivative. What you have is $\phi_y = x^3 +f_y(y,z)$ which must equal $x^3+2yz$.
And that illustrates perfectly why you should use the partial symbol instead of the '. So what you actually have to solve at this step is $f_y(y,z) = 2yz$. Can you finish it From there?

Last edited: Dec 10, 2012
3. Dec 10, 2012

### Nero26

Thanks a lot for your help.
Now I got it, I've to integrate it w.r.t. y to get $f_y(y,z)$ to get f(y,z),Right?
Btw, I've got a new problem,F=i(x3z-2xyz)+j(xy-3x2yz)+k(yz2-zx) is a solenoidal vector field.I've to find a V such that F=∇XV
So,now it comes to find a vector field from its curl.How can I find it? In my book only from gradient was explained but not from curl So please help me out.

4. Dec 10, 2012

### haruspex

Don't know if there's a smarter way, but the obvious is to write out three PDEs, one for each component of F.

5. Dec 10, 2012

### Nero26

Sorry,I exactly didn't get you.
I think PDE=partial differential equations.If so with respect to which variable x,y,or z or all of them?