Potential function of a vector feild

[kekal6]

Homework Statement

find the potential function of the vector field F(x,y,z0=<1+y^2z^3,1+2xyz^3,1+3xy^2z^2>

Homework Equations

The Attempt at a Solution

for some reason i can't figure out what to do with the third point.
i took the integrals of each point and got
integral(1+y^2z^3) Dc/dx =x+xy^2z^3+D(y,z)
integral(1+2xyz^3) Dc/dy =y+xy^2z^3+D(x,z)
integral(1+xy^2z^2) Dc/dz =z+xy^2z^3+D(x,y)

at this point i get stuck. if anyone could give at least a slight nudge in the right direction that would be great

 

[Char. Limit]

Homework Statement

find the potential function of the vector field F(x,y,z0=<1+y^2z^3,1+2xyz^3,1+3xy^2z^2>

Homework Equations

The Attempt at a Solution

for some reason i can't figure out what to do with the third point.
i took the integrals of each point and got
integral(1+y^2z^3) Dc/dx =x+xy^2z^3+D(y,z)
integral(1+2xyz^3) Dc/dy =y+xy^2z^3+D(x,z)
integral(1+xy^2z^2) Dc/dz =z+xy^2z^3+D(x,y)

at this point i get stuck. if anyone could give at least a slight nudge in the right direction that would be great

Try differentiating your integral with respect to something else.

For example, you know that f(x,y,z) = x + x y^2 z^3 + D(y,z). Now try differentiating that with respect to y... you know that it will equal F(x,y,z) in the j direction. so, \frac{\partial f}{\partial y} = 1 + 2 x y z^3. Use that to find D(y,z).

 

[kekal6]

ok so i did that and then set the whole thing to df/dy and got the c(z)=z. so do i just keep doing that all the way thru to get c(y) and C(x) and just throw that on the end? the part that's really throwing me off is the 1+stuff. i guess I am a little more confused then i thought i was.

 

[kekal6]

rather Dz Dy and Dx sry

 

[Char. Limit]

Well, let me walk you through how I did it for y.

We know from partially integrating f wrt x that f(x,y,z) = x + x y^2 z^3 + D(y,z). We also know that f_y = 1 + 2 x y z^3. Differentiating our new f wrt y, we get this:

2 x y z^3 + \frac{\partial D}{\partial y} = f_y = 1 + 2 x y z^3

Subtracting parts from each one, we get:

\frac{\partial D}{\partial y} = 1

So obviously D(y,z) = y + E(z). And then you have f(x,y,z) = x + y + x y^2 z^3 + E(z)[/tex]. Solve for E(z) using similar methods.

 

[kekal6]

alright i did it all the way through and got E(z)=z E(y)=y and E(x)=x because that's the only difference in the equations once i get down to that point for each of them so would my answer then be x+y+z+xy^2z^3 or is my logic off completely on the E's?

 

[Char. Limit]

alright i did it all the way through and got E(z)=z E(y)=y and E(x)=x because that's the only difference in the equations once i get down to that point for each of them so would my answer then be x+y+z+xy^2z^3 or is my logic off completely on the E's?

No, you have that correct. Correct up to a constant, at least.

 

[kekal6]

ok thank you very much! did not expect someone to answer all my questions at 3 am where i am!

 

[Char. Limit]

ok thank you very much! did not expect someone to answer all my questions at 3 am where i am!

Well, it's 1 AM here, but I stay up late.