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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 $div(X)= \nabla _i X^i.$

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    $\begingroup$ If you can see that $ div(X)d\mu = d(i_X d\mu) $ then all the same things follow as in $\mathbb{R}^n $ $\endgroup$ – smiley06 May 12 '13 at 18:51
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Use the well known formula: $f\Delta g + \langle \nabla f, \nabla g \rangle = div(f \nabla g) $ and the divergence theorem.

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