The integral $\int\vec{B}.\mathrm{d}\vec{s}$(or $\int\vec{B}.\mathrm{d}\vec{l}$ as I like to call it) has no significance what so ever as compared to $\int\vec{E}.\mathrm{d}\vec{l}$. This is because $\vec{E}$ might beis necessarily a gradient with a potential defined, provided $\vec{E}$ is electrostatic (If however $\vec{E}$ is created from a changing magnetic flux, it is not electrostatic and $\vec{E}\neq -\nabla V$ ). It is defined as $\vec{E}=-\nabla V$, where $V$ is the electrostatic potential. Now, the gradient for any function $f(\vec{r})$ is defined as $\nabla f(\vec{l}).\mathrm{d}\vec{l}=\mathrm{d}f(\vec{l})$, hence $\nabla V.\mathrm{d}\vec{l}=\mathrm{d}V$. This implies that $-\nabla V.\mathrm{d}\vec{l}=-\mathrm{d}V; \vec{E}.\mathrm{d}\vec{l}=-\mathrm{d}V; \int\mathrm{d}V=-\int \vec{E}.\mathrm{d}\vec{l}$ So, $-\int\vec{E}.\mathrm{d}\vec{l}$ is the amount of energy needed for a unit positive charge to travel through the said limits of the integration. This is how you define potential. The vector $\vec{B}$ on the other hand is not a gradient. What that means is that you cannot find a function $f(\vec{r})$ such that $\vec{B}=\nabla f$. But, the fact that $\vec{B}\neq \nabla f$ means $\oint\vec{B}.\mathrm{d}\vec{l}\neq 0$(unlike $\oint\vec{E}.\mathrm{d}\vec{l}=0$) and this quantity signifies a current in ampere's law. The law goes $$\oint_D\vec{B}.\mathrm{d}\vec{l}=\iint_S\mu \vec{J}.\mathrm{d}\vec{A}$$. It means that the loop, $D$ parallel to which you integrate your $\vec{B}$ The integration equals the amount of current flowing perpendicular to the surface area of that loop.
