Exercise 14

Stokes' Theorem is generally used to turn a line integral into a flux integral. Sometimes it is possible to be given a flux integral and recognize that the given vector field $\vF$ is $\curl(\vG)$ for some vector field $\vG$, however. When this is the case, we call $\vG$ a vector potential for the vector field $\vF$, much like a function $f$ so that $\vF = \grad(f)$ is called a potential function for $\vF$.

1. Find a vector field $\vF$ so that $\curl(\vF) = \langle x-y^2,4xy-y-4,-4xz\rangle$.

2. When finding an anti-derivative of a function of a single variable, you know that there is an infinite family of anti-derivatives, but that any two anti-derivatives differ by a constant. This is why we write expressions such as $\displaystyle\int\cos(x)\, dx = \sin(x) + C$. A similar phenomenon occurs with (scalar) potential functions for gradient vector fields. Find a second vector field $\vG$ with the same curl as in part, and do so in a way that $\vF-\vG$ is not a constant vector. That is, after simplifying fully, $\vF-\vG$ must contain at least one of the variables $x,y,z$.

3. Verify that for the vector fields you found above, $\vF-\vG$ is a gradient vector field. Explain why for every pair $\vF, \vG$ of vector potentials for a vector field $\vH$, you must have that $\vF-\vG$ is a gradient vector field.

4. Explain why if $\vH$ is a vector field with a vector potential $\vF$, $\divg(\vH) = 0$. Such a vector field is called a solenoidal vector field or divergence-free vector field.