The volumes are equal

Prove that for every constant $c>0$ the set

$\mathcal{B}_{f, g} = \{ (x, y, z) \in \mathbb{R}^3 : (x-f(z))^2 + (y-g(z))^2 \leq c, \quad z \in [a, b] \}$

has the same volume for all continuous functions $f, g: [a, b] \rightarrow \mathbb{R}$ .

Solution

For every $z_0 \in [a, b]$ on the plane $z=z_0$ the set

$\{{(x,y,z_0)\in\mathbb{R}^3\,|\,(x-f(z_0))^2+(y-g(z_0))^2\leqslant c} \}\,, \quad c>0$

is a disk of constant radius $\sqrt{c}$ . Thus the set

$\mathcal{B}_{f,g}=\{{(x,y,z)\in\mathbb{R}^3\,|\,(x-f(z))^2+(y-g(z))^2\leqslant c\,,\;a\leqslant z\leqslant b}\}$

is a “cylinder” which axis is the curve

$\big\{{(f(z),g(z),z)\in\mathbb{R}\;|\;z\in[a,b]}\big\}$

and its radius is $\sqrt{c}$ .More specifically , the set $\mathcal{B}_{f, g}$ is bounded by the planes $z=a$ and $z=b$ and for every $t \in [a, b]$ the intersection of $\mathcal{B}_{f, g}$ with the plane $z=t$ is the disk

$\mathcal{D}_t=\big\{{(x,y,z)\in\mathbb{R}^3\,|\,(x-f(t))^2+(y-g(t))^2\leqslant c\,,\;z=t}\big\}$

The area of this disk is the same with the disk

$\mathcal{D}=\big\{{(x,y,z)\in\mathbb{R}^3\,|\,x^2+y^2\leqslant c\,,\;z=0}\big\}$

The latter one has an area of

$\begin{align*} \oiint \limits_{\mathcal{D}} \left \| \bar{N} \right \| \, {\rm d} \sigma &= \oiint \limits_{\mathcal{D}} \left \| \bar{e}_3 \right \|\, {\rm d}\sigma \\ &= \int_{-\sqrt{c}}^{\sqrt{c}} \int_{-\sqrt{c-x^2}}^{\sqrt{c-x^2}} \, {\rm d} (y, x)\\ &=\pi c \end{align*}$

It follows from Cavalieri’s Principal that $\mathcal{B}_{f, g}$ has the same volume and that is equal to

$\displaystyle \mathcal{V}\left ( \mathcal{B}_{f, g} \right ) = \int_{a}^{b} \pi c \, {\rm d}t = \pi c (b-a)$

which is the same for all continuous functions $f, g:[a, b] \rightarrow \mathbb{R}$ .

A somewhat visualization would be the following:

Rendered by QuickLaTeX.com

This was an exam’s question somewhere in Greece. The answer was migrated from mathematica.gr .

The volumes are equal

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