Derivation of gravitational wave as quadrupole moment

Question

In  Carroll's Spacetime and geometry section 7.5, it is deriving metric $bar h_{mu nu}$ with Lorenz gauge in terms of quadrupole moment of energy density.
In Eq.[7.135],
begin{equation}
int d^3y tilde T^{ij} = int partial_k(y^i T^{kj}) d^3y - int y^i (partial_k T^{kj}) d^3y
end{equation}
It says "The first term is a surface integral which will vanish since the source is isolated". But I could not see the form of surface integral for the first term. Is it divergence theorem? I know divergence is $partial_k T^k$, how does it relate to $partial_k(y^i T^{kj})$? It seems to me that $y^i$ is redundant. If it is $partial_k(T^{kj})$, it works and converts as $int T^{kj}n^kd^2y$.
Moreover, to my understanding, the source is emitting energy through gravitational wave. How could it said to be isolated?

Alex Ghorbal · Accepted Answer

Yes, it is the divergence theorem. The term is simply $$intpartial_k(y^iT^{kj})d^3y=int y^iT^{kj}n_k d^2y$$
where the integral is over the surface at infinity and $boldsymbol{n}$ is the unit vector away from the surface. We presume that $T^{kj}$ vanishes at infinity (since the source is isolated) and hence the integral is zero.
You are right that the source is emitting energy through the wave, but here we are dealing with linearised gravity, where $T^{kj}$ only includes the contribution from ordinary matter, and not from the wave itself. This matter is isolated and so $T^{kj}$ vanishes at infinity.

Derivation of gravitational wave as quadrupole moment

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