By applying Stokes theorem, the Divergence theorem and/or the Gradient Theorem, show that; $$\int_{S}(\nabla T \times \textbf{dA})=-\oint_{P}(Tdl)$$ I've been stuck on this for quite a while, I think largely due to the way the question was presented.
I've never seen the $\nabla T \times \textbf{dA}$, notation before, and couldn't find mention of it anywhere, so I'm quite unsure as to what it actually means. The surface of integration also wasn't mentioned in the original question, but I'm assuming it MUST have a closed boundary, otherwise this could never be true in the first place.
I haven't been able to really make any progress yet, mostly, as I said due to not understanding the notation, but I'm assuming I'll need to somehow apply Stokes Theorem to convert this to a line integral of a gradient, which will thus be conservative and depend only on the end points, leaving us with the result, but yeah, I have no idea how to do thius and any help would be much appreciated.
Choose a vector $\bf{A}$ to be of the form $\bf{T} c$ where $c$ is a constant. Then \begin{align} \nabla \times \bf{A} &= \bf{T} (\nabla \times c) + \nabla T \times c \\ &= \nabla \bf{T} \times c \end{align} Thus \begin{align} \oint_P T c \cdot dl &= \int_C \nabla T \times c \cdot d \bf{S} \\ &= \int_S c \cdot (d\bf{S} \times \nabla T) \end{align} Slight re-arrangement gives \begin{align} c \cdot\oint_P T dl &= - c \cdot \int_S \nabla T \times d \bf{S} \end{align} Since the constant $c$ is arbitrary, the result follows.