# Calculating volume enclosed using triple integral

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.

• Three cylinders intersection. Visualizing is indeed a bit hard, see mathmos.net/puzzles/intersect1.png – AgentS Sep 8 '14 at 10:08
• This post is chosen to be the target/mother for (abstract) duplicates because it has an existing link while others have none. – Lee David Chung Lin Jan 22 '19 at 12:02
• With three circle equations being the same as a sphere and with a radius of one, we have $$V=\frac{4\pi}{3}\approx 4.19$$. – poetasis Aug 24 '19 at 19:40


\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}

• +4 from me. $\ddot\smile$ – Tunk-Fey Sep 9 '14 at 18:21
• @Tunk-Fey Thanks. I like integration 'without pictures'. – Felix Marin Sep 9 '14 at 18:30
• 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. – Tunk-Fey Sep 9 '14 at 18:37
• @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. – Felix Marin Sep 9 '14 at 19:31
• @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 $\left(0,1\right)$. – 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 $x,y,z\geq 0$

• Why only for $x,y,z\geq 0$? – user88923 Sep 8 '14 at 10:13
• The volume is symmetric, so you could just calculate that and multiply by 8. – Snufsan Sep 8 '14 at 10:16
• 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. – user88923 Sep 8 '14 at 10:19
• Can you explain why the volume just turns out to be $8(2-\sqrt{2})$? Even exploiting symmetry, the resulting integral is non-trivial. – Jack D'Aurizio Sep 8 '14 at 10:54