Faradays Law can be written as
$\int_{C} E\dot{}dl = -\frac{1}{c}\frac{d}{dt}\int_{S} d^2r' n \dot{}B$ ,where left integral is a contour integral, and right is a surface integral.
Using two surfaces $S_1$, $S_2$, show that one gets the same result for the path integral on the left as long as both surfaces share a common boundary C, and $\nabla \dot{} B=0$
Pretty lost on this problem. I've attempted the following so far
$\int_{V} d^3r' \nabla \dot{} B = 0 = \int_{S}d^2r' n \dot{} B =\int_{S_1}d^2r' n_1 \dot{} B + \int_{S_2}d^2r' n_2 \dot{} B $
this implies that
$-\int_{S_1}d^2r' n_1 \dot{} B = \int_{S_2}d^2r' n_2 \dot{} B$
Some additional help on this would be appreciated- not sure how to get the contour integral on the other side of this derivation. .