Suppose that $\Omega\subseteq\mathbb{R}^3$ is a bounded convex region, with the boundary $\partial \Omega$

Question

Suppose that $\Omega\subseteq\mathbb{R}^3$ is a bounded convex region, with the boundary $\partial \Omega$ smooth (i.e. locally $C^\infty$ homeomorphic to the unit disk in $\mathbb{R}^2$). Denote the volume of $\Omega$ as $V$ and the surface area of $\partial\Omega$ as $A$. Denote $\Omega_r=\{x\in \mathbb{R}^3: \exists y\in \Omega, \|x-y\|< r\}$, where the Euclidean norm is taken. Denote its volume as $V_r$. Show that for $r>0$ we have $V_r>V+rA+\frac{4\pi r^3}{3}$.

What I do so far

We have to show that

$V_r > V + rA + \frac{4\pi r^3}{3}$

$V_r = \iiint_{\Omega_r} dx,dy,dz$

$\iiint_{\Omega_r} \nabla \cdot \mathbf{F} ,dx,dy,dz = 3V_r$

$3V_r = \iiint_{\Omega_r} \nabla \cdot \mathbf{F} ,dx,dy,dz = \iint_{\partial\Omega_r} \mathbf{F} \cdot d\mathbf{S}$

$$\iint_{\partial\Omega_r} \mathbf{F} \cdot d\mathbf{S} = \iint_{\partial\Omega} \mathbf{F} \cdot d\mathbf{S} + \iint_{\text{sphere}} \mathbf{F} \cdot d\mathbf{S}$$

Answer 1

Suppose the origin is in the interior of $\Omega$ (if not, just translate $\Omega$). Using spherical coordinates $x=\rho\sin\theta\cos\varphi,\ y=\rho\sin\theta\sin\varphi,\ z=\rho\cos\theta$ where $\theta\in[0,\pi]$ and $\varphi\in[0,2\pi)$, we have

\begin{align*} V_r&\geq\frac43\pi r^3+\iiint_\Omega(\rho+r)^2\sin\theta\,d\rho\,d\theta\,d\varphi\\ &=\frac43\pi r^3+\underbrace{\iiint_\Omega\rho^2\sin\theta\,d\rho\,d\theta\,d\varphi}_{=I_1}+\underbrace{2r\iiint_\Omega\rho\sin\theta\,d\rho\,d\theta\,d\varphi}_{=I_2}+\underbrace{r^2\iiint_\Omega\sin\theta\,d\rho\,d\theta\,d\varphi}_{=I_3}. \end{align*}

It follows $I_1=V$ and $I_3>0$ since $\theta\in[0,\pi]$, and

$$I_2=r\iint_{\partial\Omega}\rho^2\sin\theta\,d\theta\,d\varphi=rA.$$

Collecting all the terms, we get $V_r>\frac43\pi r^3+V+rA$.

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Some remarks:

• Why add $\frac43\pi r^3$: Since $\Omega$ contains origin, $\Omega_r$ always contains $B_r$ (the sphere centred at origin with radius $r$). However, the triple integral cannot cover $B_r$ because all the points being integrated have norm $\rho+r\geq r$, so we need to add back the volume of $B_r$, which is $\frac43\pi r^3$.

• Why $V_r\geq$ but not $V_r=$: First, it is easy to see that any point covered by the triple integral will be in $\Omega_r$, so we at least have $\geq$. Meanwhile, note that given any direction $(\theta,\varphi)$, the integrand $\rho+r$ only covers the $r$-band in the direction of $(\theta,\varphi)$, so we could be missing out some points in $\Omega_r$ that are not in the direction of $(\theta,\varphi)$, so $=$ will not hold in general (e.g. just consider $\Omega$ is a cube centred at origin).