# Thread: [SOLVED] General potential of a field

1. ## [SOLVED] General potential of a field

Let $\bold K(x,y,z)=(4xyz,2x^2z+2yz^2,2x^2y+2y^2z+4z^3)$ defined in $\mathbb{R}^3$.
Show that the field is irrotational and calculate the general potential of this field.
I've calculated its curl (0), so the field is irrotational.
Now I must find, I believe, a vector field whose curl is $\bold K$. I called it $\bold A=(a_1,a_2,a_3)$ but I've too much unknowns. I don't know how to solve the last part. Can someone help me?

In details I wrote $\frac{\partial a_3}{\partial y}-\frac{\partial a_2}{\partial z}=4xyz$ and 2 other relations of the same kind. At last I ended up with at least 9 unknowns and only 3 equations.

2. Since K is irrotational, it's the gradient of a function.

3. I believe you need to find $\varphi$ such that

$\mathbf{K} = -\nabla \varphi ,$ so you have

$\varphi_x = -4xyz$

$\varphi_y = -2x^2z-2yz^2$

$\varphi_z = -2x^2y-2y^2z-4z^3.$

4. Ah thanks to both. I get it. I forgot the theorem or consequence of a theorem that if a field is irrotational, then it's the gradient of some function.

5. ## Clueless

I thought I had it, but I realize I made a mistake.
Now I got it, but I think I got it by chance. Would someone explain me how to reach $\varphi$ from $\varphi _x$, $\varphi _y$ and $\varphi _z$?
I realized that $\int \varphi _z dz =\varphi$. Which gives $\varphi =-2x^2yz -y^2z^2-z^4$.

I was thinking about integrating $\varphi _x$ with respect to $dx$ and sum up a function $g(y,z)$ then integrating $\varphi _y$ with respect to y and sum up a function $g_1(x,z)$ and integrating $\varphi _z$ with respect to z and sum up a function $g_2 (x,y)$. Lastly, summing all these functions and try to determine $g$, $g_1$ and $g_2$.
I just don't understand how I got the right result with what I've done...

6. That's what I would do.

$\varphi = -2x^{2}yz + g_{1}(y,z) = -2x^{2}yz-y^{2}z^{2} + g_{2}(x,z) = 2x^{2}yz - y^{2}z^{2}-z^{4} + g_{3}(x,y)$

so $g_{1}(y,z) = -y^{2}z^{2}-z^{4}, g_{2}(x,z) = -z^{4}, \text{and} \ g_{3}(x,y) =0$

7. Originally Posted by Random Variable
That's what I would do.

$\varphi = -2x^{2}yz + g_{1}(y,z) = -2x^{2}yz-y^{2}z^{2} + g_{2}(x,z) = 2x^{2}yz - y^{2}z^{2}-z^{4} + g_{3}(x,y)$

so $g_{1}(y,z) = -y^{2}z^{2}-z^{4}, g_{2}(x,z) = -z^{4}, \text{and} \ g_{3}(x,y) =0$
Makes perfectly sense.