Derivative of boundary volume element

Question

If $(M,g)$ is a 2-dimensional surface with boundary, now we have a deformation of metric $(g_t)$, suppose $\frac{dg_t}{dt}\big|_{t=0}=h$ which is a symmetric $(0,2)$ tensor, then the derivative of volume is $$\frac{dV(g_t)}{dt}\big|_{t=0}=\frac{1}{2}\int_M tr_g(h) dv_{g}$$ The volume of boundary is $$A(g_t)=\int_{\partial M}dv_{\tilde{g}_t}$$ But I don't know how to calculate the derivative of this, is it same as above?

Now if $M$ is n-dimensional Riemannian manifold with boundary, I think the derivative of volume is still same as above, but what happens to boundary now?

Answer 1

It is just $$\frac{dA(g_t)}{dt}\big|_{t=0}=\frac{1}{2}\int_{\partial M}h(T,T)dV_{\bar{g}}$$ where $T$ is the tangent vector field on the boundary. Since for orthonormal basis $\{e_i\}$ we have $$tr_gh=g^{ij}h_{ij}=h(e_i,e_i)$$ but on the boundary, the basis is just the tangent vector $T$.