Given:
So basically I just need to work with vector operations in order to get the result. In trying to solve this problem, I arrived at this:
$$(\nabla \times \vec{c}T) \cdot d\vec{a}$$ $$=d\vec{a} \cdot (\nabla \times \vec{c}T)$$ $$=(d\vec{a} \times \nabla) \cdot \vec{c}T$$ $$=\vec{c}T \cdot (d\vec{a} \times \nabla)$$
Here's the part I'm wondering about. Am I allowed to distribute the $T$ to only one term of the cross-product? That is, $$=\vec{c} \cdot (d\vec{a} \times T \nabla)$$
Otherwise, if I'm going completely the wrong way with this problem feel free to let me know.
Another attempt:
$$(\nabla \times \vec{c}T) \cdot d\vec{a}$$ $$=(T(\nabla \times \vec{c}) - \vec{c} \times (\nabla T)) \cdot d\vec{a}$$ $$=(T(\nabla \times \vec{c}) + \nabla T \times \vec{c}) \cdot d\vec{a}$$ $$=T(\nabla \times \vec{c} + \nabla \times \vec{c}) \cdot d\vec{a}$$ $$=2T(\nabla \times \vec{c} ) \cdot d\vec{a}$$ $$=(2T\nabla \times \vec{c} ) \cdot d\vec{a}$$ $$=(d\vec{a} \times 2T\nabla) \cdot \vec{c} $$
Right hand side:
$$\vec{c}T \cdot d\vec{l}$$ $$=\vec{c} \cdot Td\vec{l}$$
Putting it all together:
$$\int (d\vec{a} \times 2T\nabla) \cdot \vec{c} = \oint \vec{c} \cdot Td\vec{l}$$
But now there are some problems:
Can I cancel a dot product occurring on both sides even though they are both under an integral sign?
The $2$ on the LHS does not appear in the final answer.
I can't flip $T \nabla \to \nabla T$ because the former is an operator and the latter is a vector.
Any advice??