Let $(M^n,g)$ be an oriented Riemannian manifold with boundary $\partial M$. The orientation on $Μ$ defines an orientation on $\partial M$. Locally, on the boundary, choose a positively oriented frame field $\{e\} ^{n}_{i=1} $ such that $e_1 =\nu$ is the unit outward normal. Then the frame field $\{e\} ^{n}_{i=2} $ positively oriented on $\partial M$. Let $\{\omega ^i\} ^{n}_{i=1} $ denote the orthonormal coframe field dual to $\{e\} ^{n}_{i=1} $. The volume form of Μ is
$$d\mu=\omega^1 \wedge \cdots\wedge\omega^n $$
and the volume form of $\partial M$ is
$$d\sigma=\omega^2 \wedge \cdots\wedge\omega^n $$
By using divergence theorem prove that on a compact manifold,
$$\int _{M^n}(u\Delta v-v\Delta u)d\mu=\int_{\partial M^n} (u\frac{\partial v}{\partial \nu }-v\frac{\partial u}{\partial \nu })d\sigma .$$
Divergence theorem:Let $(M^n,g)$ be a compact oriented Riemannian manifola. If $X$ is a vector fiela, then
$$\int_M div(X)d\mu=\int_{\partial M^n} \langle X, \nu \rangle d\sigma$$
where $div(X)= \nabla _i X^i.$
Use the well known formula: $f\Delta g + \langle \nabla f, \nabla g \rangle = div(f \nabla g) $ and the divergence theorem.