Calculate the volume enclosed between $x^2 + y^2=1$, $y^2+z^2=1$, $x^2+z^2=1$

I am supposed to solve this question using multiple integral. I am not able to visualize the resultant figure of whose I am supposed to calculate the volume.


$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \mbox{Note that}\quad V \equiv \iiint_{{\mathbb R}^{3}}\Theta\pars{1 - x^{2} - y^{2}} \Theta\pars{1 - y^{2} - z^{2}}\Theta\pars{1 - z^{2} - x^{2}}\,\dd x\,\dd y\,\dd z $$ which we'll evaluate in cylindrical coordinates: $\ds{x = \rho\cos\pars{\phi}}$, $\ds{y = \rho\sin\pars{\phi}}$ and $\ds{z = z}$ with $\ds{\rho \geq 0}$ and $\ds{0 \leq \phi < 2\pi}$.

\begin{align} V&=\iiint_{{\mathbb R}^{3}}\Theta\pars{1 - \rho^{2}} \Theta\pars{1 - \rho^{2}\sin^{2}\pars{\phi} - z^{2}} \Theta\pars{1 - z^{2} - \rho^{2}\cos^{2}\pars{\phi}}\,\rho\,\dd\rho\,\dd\phi\,\dd z \\[3mm]&=\int_{0}^{1}\dd\rho\int_{0}^{\infty}\dd z\int_{0}^{2\pi} \Theta\pars{1 - \rho\sin^{2}\pars{\phi} - z^{2}} \Theta\pars{1 - \rho\cos^{2}\pars{\phi} - z^{2}}\,\dd\phi \\[3mm]&=2\int_{0}^{1}\dd\rho\int_{0}^{\infty}\dd z\int_{0}^{\pi} \Theta\pars{1 - \rho\sin^{2}\pars{\phi} - z^{2}} \Theta\pars{1 - \rho\cos^{2}\pars{\phi} - z^{2}}\,\dd\phi \\[3mm]&=4\int_{0}^{1}\dd\rho\int_{0}^{\pi/2}\dd\phi\int_{0}^{\infty} \Theta\pars{\root{1 - \rho\cos^{2}\pars{\phi}} - z} \Theta\pars{\root{1 - \rho\sin^{2}\pars{\phi}} - z}\,\dd z \end{align}

\begin{align} V&=4\int_{0}^{1}\dd\rho\bracks{% \int_{0}^{\pi/4}\root{1 - \rho\cos^{2}\pars{\phi}}\,\dd\phi +\int_{\pi/4}^{\pi/2}\root{1 - \rho\sin^{2}\pars{\phi}}\,\dd\phi} \\[3mm]&=4\int_{0}^{1}\dd\rho\bracks{% \int_{0}^{\pi/4}\root{1 - \rho\cos^{2}\pars{\phi}}\,\dd\phi +\int_{-\pi/4}^{0}\root{1 - \rho\cos^{2}\pars{\phi}}\,\dd\phi} \\[3mm]&=8\int_{0}^{\pi/4}\dd\phi\int_{0}^{1} \root{1 - \rho\cos^{2}\pars{\phi}}\,\dd\rho =8\int_{0}^{\pi/4}\dd\phi\ \braces{{2\bracks{1 - \rho\cos^{2}\pars{\phi}}^{3/2} \over -3\cos^{2}\pars{\phi}}} _{\rho\ =\ 0}^{\rho\ =\ 1} \\[3mm]&={16 \over 3}\int_{0}^{\pi/4} \bracks{1 - \sin^{3}\pars{\phi} \over \cos^{2}\pars{\phi}}\,\dd\phi ={16 \over 3}\int_{0}^{\pi/4} \bracks{\sec^{2}\pars{\phi} - {1 - \cos^{2}\pars{\phi} \over \cos^{2}\pars{\phi}}\,\sin\pars{\phi}}\,\dd\phi \\[3mm]&={16 \over 3} \bracks{1 +\ \underbrace{\int_{1}^{\root{2}/2}{1 - t^{2} \over t^{2}}\,\dd t} _{\ds{2 - {3 \over 2}\,\root{2}}}}\quad\imp\quad \color{#66f}{\Large V = 8\pars{2 - \root{2}}} \approx {\tt 4.6863} \end{align}

  • $\begingroup$ +4 from me. $\ddot\smile$ $\endgroup$ – Tunk-Fey Sep 9 '14 at 18:21
  • $\begingroup$ @Tunk-Fey Thanks. I like integration 'without pictures'. $\endgroup$ – Felix Marin Sep 9 '14 at 18:30
  • $\begingroup$ Anyway, what is $\Theta$? I don't get it each step in the grey area. Could please elaborate Felix? Sorry, I am not good at this stuff. $\endgroup$ – Tunk-Fey Sep 9 '14 at 18:37
  • $\begingroup$ @Tunk-Fey $\Theta\left(\,x\,\right)$ is the Heaviside Step Function: $$ \Theta\left(\,x\,\right)\equiv \left\lbrace\begin{array}{lcl} 0 & \mbox{if} & x < 0 \\&& \\ 1 & \mbox{if} & x > 0 \end{array}\right. $$ It's quite useful whenever you don't want to deal with pictures. $\endgroup$ – Felix Marin Sep 9 '14 at 19:31
  • $\begingroup$ @Tunk-Fey In the gray area: The first line just set the boundary with the help of the $\Theta$ function. In the second line I did the map $\rho^{2}\mapsto \rho$ and integrate $z$ over $\left(0,\infty\right)$ because it's an even function of $z$. In the third and fourth lines, I manipulated the $\phi$-integration to move everything to $\left(0,\pi/2\right)$. Also, I integrate $\rho$ over $\left(0,1\right)$ because the factor $\Theta\left(1 - \rho\right)$ limits the integration to $\left(0,1\right)$. $\endgroup$ – Felix Marin Sep 9 '14 at 19:40

The intersection of the three cylinders is a "curvilinear polyhedron" with $14$ vertices: $8$ vertices of the type (i) $$\left(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\right)$$ that simultaneously belong to the boundaries of all the three cylinders, and $6$ vertices of the type (ii) $$\left(\pm 1,0,0\right)\quad \left(0,\pm 1,0\right)\quad \left(0,0,\pm 1\right)$$ that belong to the boundaries of just two cylinders. By symmetry, the volume is given by the volume of the cube that is the convex envelope of type-(i) points, $2\sqrt{2}$, plus $6$ times the volume of the apse that lies above the face with vertices in $$\left(\pm\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\right).$$ We may compute the volume of such an apse through Cavalieri's principle. The intersection between the apse and the plane $y=k\in[1/\sqrt{2},1]$ is a square with vertices in $$\left(\pm\sqrt{1-k^2},k,\pm\sqrt{1-k^2}\right),$$ hence its area is just $4(1-k^2)$ and the volume of the apse is: $$4\int_{1/\sqrt{2}}^{1}(1-k^2)\,dk = \frac{1}{3}(8-5\sqrt{2}), $$ so the volume of the intersection of the three cylinders is:

$$ V = 8 (2-\sqrt{2}). $$


Those are 3 cylinders that intersect, try to sketch it, and then use the symmetry of the cylinders and calculate just the part of the volume that satisfies $x,y,z\geq 0$

  • $\begingroup$ Why only for $x,y,z\geq 0$? $\endgroup$ – Who cares Sep 8 '14 at 10:13
  • $\begingroup$ The volume is symmetric, so you could just calculate that and multiply by 8. $\endgroup$ – Snufsan Sep 8 '14 at 10:16
  • $\begingroup$ Oh! I thought you were suggesting to calculate only that. As per the link given by @ganeshie8, three cylinder intersection would look like-mathmos.net/puzzles/intersect1.png. I am still finding it tough. $\endgroup$ – Who cares Sep 8 '14 at 10:19
  • $\begingroup$ Can you explain why the volume just turns out to be $8(2-\sqrt{2})$? Even exploiting symmetry, the resulting integral is non-trivial. $\endgroup$ – Jack D'Aurizio Sep 8 '14 at 10:54

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