Using Gauss's Theorem in weak formulation

Can anyone see how exactly Gauss's theorem used in the following case: In defining the weak solution to the linear elliptic equation we start with $$-\sum_{j,k=1}^{n}D_{j}(a_{jk}D_{k}u) + cu = f \text{ }\text{ in the bounded domain }\Omega \subset \mathbb{R}^{n}$$ where $(D_{j} = \frac{\partial}{\partial x_{j}})$ with the Dirichlet boundary condition $$u|_{\partial \Omega} = \phi$$ Assuming that $u$ is a sufficiently smooth(for simplicity e.g. $u \in C^{2}(\overline{\Omega})$) solution of the above two equations. $\partial \Omega$ is sufficiently smooth, multiply the first equation by test function $v \in C_{o}^{1}(\Omega)$ and integrate over $\Omega$, then by using Gauss's theorem, we obtain $$\sum_{j,k=1}^{n}\int_{\Omega}a_{jk}(D_{k}u)(D_{j}v) + \int_{\Omega}cuv = \int_{\Omega}fv$$

How exactly is the Gauss's theorem used? Thanks for any assistance. Let me know if something is unclear.

They use $-\int_\Omega D_j(\ldots) v=\int_\Omega (\ldots) D_j v-\int_{\partial\Omega} (\ldots)\,v\,dS$ (the last term being zero because the test function is $0$ on the boundary). All of this kind of integral identities are analogs of the famous one-dimensional rule $$\int_a^b f'g=-\int_a^b fg' + \int_{\{a,b\}} fg$$ called "integration by parts".
It seems to me that there is a mess in terminology; these types of theorems are sometimes called "Gauss", sometimes "Stokes" or "divergence" theorems; however, it should be not hard to derive it from the most well-known $\int_\Omega \nabla\cdot v=\int_{\partial\Omega} v\,dS$ or whatever you might call "Gauss Theorem".