CARNEGIE MELLON UNIVERSITY
DEPARTMENT OF MATHEMATICAL SCIENCES
21-259 Calculus in Three Dimensions
Vector Potential

Let

$${\bf F}(x,y,z)=P(x,y,z){\bf i}+Q(x,y,z){\bf j}+R(x,y,z){\bf k}$$

be a smooth field defined on ${\bf\cal R}^3$. Suppose ${\bf F}$ is incompressible, so that

$$\nabla \cdot {\bf F}=0.$$

Then

$${\bf F}=\nabla \times {\bf G}$$

for some field

$${\bf G}(x,y,z)=A(x,y,z){\bf i}+B(x,y,z){\bf j}+C(x,y,z){\bf k}.$$

This is a special case of Poincare's Lemma.

1. Suppose we set
   $$\begin{eqnarray*} A(x,y,z) & = & \int_0^1 t\left[ z Q(tx,ty,tz)-yR(tx,ty,tz) \ri... ...z) & = & \int_0^1 t\left[ y P(tx,ty,tz)-xQ(tx,ty,tz) \right] dt. \end{eqnarray*}$$
   
   
   
   
   
   Show that ${\bf F} = \nabla \times {\bf G}$, where ${\bf G}=\langle A,B,C \rangle$.

   [Note: We may write
   

   $${\bf G}({\bf r})=\int_0^1 t \ \left[ {\bf F}(t {\bf r}) \times {\bf r} \right] \ dt,$$
   
   
   and may therefore also write for incompressible fields ${\bf F}$
   
   $${\bf F}({\bf r}) = \int_0^1 t \ \nabla \times \left[ {\bf F}(t {\bf r}) \times {\bf r} \right] \ dt$$
   
   
   where ${\bf r}=\langle x,y,z \rangle$.]

2. If ${\bf F} = \nabla \times {\bf H}$, for some other field ${\bf H}$, then ${\bf H}={\bf G}+{\bf K}$ for some conservative field ${\bf K}$.

3. Let
   
   $${\bf F}(x,y,z)= \left( 2ye^{xz}+xye^{yz} \right) {\bf i}+ \l... ...x-y^2ze^{xz}\right) {\bf j}+ \left(x^2-e^{yz} \right) {\bf k}.$$
   
   
   Show $\nabla \cdot {\bf F}=0$, and find ${\bf G}$ such that ${\bf F} = \nabla \times {\bf G}$.