Curl and Divergence — Question 6

Question 6

For $$f(x,y,z)=x^2y+yz^2+xz,$$ verify directly that $$\nabla\times(\nabla f)=\mathbf 0.$$

Tasks

1. Compute $\nabla f$.

2. Compute its curl and show each cancellation.

3. State the smoothness principle behind the identity.

Question 6 – Solution

Strategy. Differentiate once to obtain the gradient and again in the curl’s cross-partial pairs.

Step 1: Gradient $$\nabla f =\langle 2xy+z,\,x^2+z^2,\,2yz+x\rangle.$$

Step 2: Curl of the gradient Let the gradient components be $P,Q,R$. Then $$\begin{align*} R_y-Q_z&=2z-2z=0,\\ P_z-R_x&=1-1=0,\\ Q_x-P_y&=2x-2x=0. \end{align*}$$ Therefore $$\boxed{\nabla\times(\nabla f)=\mathbf 0}.$$

Step 3: General principle Each cancellation is equality of a mixed-partial pair: $$f_{zy}=f_{yz},\qquad f_{xz}=f_{zx},\qquad f_{yx}=f_{xy}.$$ Continuous second partial derivatives guarantee these equalities.

Verification The polynomial $f$ has continuous derivatives of every order on $\mathbb R^3$, so there are no domain or differentiability exceptions to the identity here.