# Determining if a vector field is conservative

by 1MileCrash
Tags: conservative, determining, field, vector
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 Your mistake is thinking that you can integrate each component with respect to a single variable and have only an added constant. If you have $$\nabla g= \frac{\partial g}{\partial x}\vec{i}+ \frac{\partial g}{\partial y}\vec{j}+ \frac{\partial g}{\partial z}\vec{k}= F_x\vec{i}+ F_y\vec{j}+ F_z\vec{k}$$ Yes, because the partial derivative with respect to x treats y and z as constants, you can integrate $F_x$ with respect to x only but the "constant of integration" may be a function of y and z because you are treating them as constants. For example, suppose we have $\nabla g= \vec{F}= 2xy\vec{i}+ x^2\vec{j}+ \vec{k}$. Then we can say $$\frac{\partial g}{\partial x}= 2xy$$ so that $$g= x^2y+ \phi(y,z)$$ where $\phi(y,z)$ is an unknown function of y and z only- which is treated as constants when differentiating with respect to x. And we also have $$\frac{\partial g}{\partial y}= x^2$$ so that $$g= x^2y+ \psi(x,z)$$ Comparing those two, its reasonable to think that $g= x^2y+ C$, a constant. But we still have z to handle! From $$\frac{\partial g}{\partial z}= 1$$ we get $g= z+ \chi(x,u)$ where now it is important that we have added a function of z, not a constant! We might be able to see that $\chi(x,y)= x^2+ C$ and so $g(x, y, z)= x^2y+ z+ C$. A simpler way to do this is, starting with the first equation, integrate with respect to x to get $g(x,y,z)= x^2y+ \phi(y,z)$. Now, differentiate that, with respect to y, to get $$\frac{\partial g}{\partial y}= x^2+ \frac{\partial\phi(y,z)}{\partial y}= x^2$$ from which we get $$\frac{\partial \phi}{\partial y}= 0$$ which means that $\phi$ must not be a function of y. Not that is is a constant, because it still might be a function of z. Now we have $g= x^2y+ \phi(x)$ and, differentiating with respect to z, $$\frac{\partial g}{\partial z}= \frac{d\phi}{dz}= 1$$ and from that, $\phi(z)= z+ C$ where C really is a constant because $\phi$ is a function of z only. That gives $g(x, y, z)= x^2y+ z+ C$.
 P: 1,302 Determining if a vector field is conservative Ok, back, and armed with itex.. I would just do: $\int 2xy dx = x^{2}y + n(y,z)$ $\int x^{2} dy = x^{2}y + k(x,z)$ $\int dz = z + q(x,y)$ Which I view as a simple system of equations that can be solved by looking at it, with: n(y,z) = k(x,z) = z or q(x,y) = x^{2}y. Thus f = x^{2}y + z + C And if the 'system' has no solution then the field is nonconservative.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,565 If all you want to do is determine whether or not it is conservative, rather than actually finding the potential, you can use the "cross condition": If there exist g such that $\partial g/\partial x= F_x$, $\partial g/\partial y= F_y$, and $\partial g/\partial z= F_z$. Then $\partial^2g/\partial x\partial y= \partial^2 g/\partial y\partial x$ becomes $\partial F_x/\partial y= \partial F_y/\partial x$ while the other partial derivatives give $\partial F_x/\partial z= \partial F_z/\partial x$ $\partial F_z/\partial y= \partial F_y/\partial z$ For this particular example, $\vec{F}= 2xy\vec{i}+ x^2\vec{j}+ \vec{k}$ we have $$\frac{\partial F_x}{\partial x}= 2y= \frac{\partial F_y}{\partial x}$$ $$\frac{\partial F_y}{\partial z}= 0= \frac{\partial F_z}{\partial y}$$ $$\frac{\partial F_z}{\partial x}= 0= \frac{\partial F_x}{\partial z}$$