Is the gradient theorem equivalent to the curl theorem if F is conservative?

[Meow12]

The gradient theorem states: $\displaystyle\int_C \nabla f\cdot d\mathbf{r}=f(\mathbf{r}(b))-f(\mathbf{r}(a))$
where C is a smooth curve parametrized by the vector-valued function $\textbf{r}(t)$ for $a\le t\le b$.

So, if we consider a closed loop beginning and ending at $t=a$,

$\displaystyle\oint_C \nabla f\cdot d\textbf{r}=f(\textbf{r}(a))-f(\textbf{r}(a))=0$ -------- (1)

The curl theorem states: $\displaystyle\oint_C\textbf{F}\cdot d\textbf{l}=\iint_S \text{curl }\textbf{F}\cdot d\textbf{S}$

If $\mathbf{F}$ is conservative, $\text{curl }\textbf{F}=\mathbf{0}$.

So, the curl theorem implies that $\displaystyle\oint_C\textbf{F}\cdot d\mathbf{l}=\iint_S\mathbf{0}\cdot d\textbf{S}=0$ -------- (2)

Equations (1) and (2) are the same.

 

[blamocur]

Not sure about equivalency of the theorems, but, to me, you've shown that $\nabla f$ is conservative.