Why doesn't the volume integral vanish?

Question

I am reading Galactic Dynamics by Binney and Termaine, and I'm struggling with the following passage.

According to Poisson's equation the resulting change in potential $\delta\Phi(\mathbf x)$ satisfies $\nabla^2(\delta\Phi)=4\pi G(\delta\rho)$, so $$\delta W=\frac{1}{4\pi G}\int\mathrm d^3\mathbf x\ \Phi\nabla^2(\delta\Phi) \tag{2.14}$$ Using the divergence theorem in the form $(\mathrm B.45)$, we may write this as $$\delta W= \frac{1}{4\pi G}\int\Phi\nabla(\delta\Phi) \cdot\mathrm d^2\mathbf S -\frac{1}{4\pi G}\int\mathrm d^3\mathbf x\ \nabla\Phi\cdot\nabla(\delta\Phi), \tag{2.15}$$ where the surface integral vanishes because $\Phi\propto r^{-1}$, $|\nabla\delta\Phi|\propto r^{-2}$ as $r\to\infty$, so the integrand $\propto r^{-3}$ while the total surface area $\propto r^2$. But $\nabla\Phi\cdot\nabla(\delta\phi)=\tfrac12\delta(\nabla\Phi\cdot\nabla\Phi)=\tfrac12\delta|(\nabla\Phi)|^2$. Hence $$\delta W= -\frac{1}{8\pi G}\delta\left(\int\mathrm d^3\mathbf x\ |\nabla\Phi|^2\right). \tag{2.16}$$

I am trying to understand the integral being carried out there. I understand why surface integral is vanishing as $r$ tends to infinity. But like same reasoning why volume integral is not vanishing though it has inverse proportionality with $r$.

Similarly there are two instances in the book where he reasons alike.

What is being compromised when volume integral is considered? Any help would be very great.

Answer 1

The surface integral vanishes because there's no integration over $r$ (if $S$ is a sphere). Therefore the surface integral is proportional to $r^{-1}$ and vanishes at $r\rightarrow \infty$. But the volume integral does integrate the function over the entire radius, from $r=0$ to infinity, therefore it is not zero.

Answer 2

Simply put, the surface integral only considers contributions at infinity, while the volume integral considers contributions all the way from the origin to infinity.

The large-$r$ contributions of the volume integral are indeed small, and in fact they are smaller than those for the volume integral. (If $\Phi\propto1/r$ then $|\nabla\Phi|\propto1/r^2$, so $\nabla\Phi\cdot\nabla(\delta\Phi)\propto 1/r^4$, with a volume element $\propto r^2\mathrm dr$, so the overall contribution is $\propto 1/r^2$.) However, in the tail of the volume integral you're adding up lots of small things, instead of looking at their limit as they get smaller and smaller.

Most importantly, though, the volume integral catches contributions for small $r$, where its integrand can be arbitrarily large, and which are completely ignored by the surface integral.

Answer 3

Presumably $\delta W=\frac{1}{4\pi G}\iiint_{\mathbb R^3}d^3x \Phi\nabla^2\delta\Phi.$ Which equals $\lim_{R\rightarrow\infty}\frac{1}{4\pi G}\int_0^R\int_0^\pi\int_0^{2\pi} [\Phi\nabla^2\delta\Phi] (r\sin\theta d\phi)(rd\theta)dr.$

And that finite volume integral can be turned into a surface integral and a different volume integral by integration by parts. And in the limit that surface integral goes to zero, while in the limit the volume integral again becomes an integral over all space.