Calculating volume enclosed using triple integral

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \mbox{Note that}\quad V \equiv \iiint_{{\mathbb R}^{3}}\Theta\pars{1 - x^{2} - y^{2}} \Theta\pars{1 - y^{2} - z^{2}}\Theta\pars{1 - z^{2} - x^{2}}\,\dd x\,\dd y\,\dd z $$ 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}

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:

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

No need to use trig or to integrate things more complicated than 1, $x$, or $x^2$.

First we separate the volume into octants based on the signs of $x$, $y$, $z$, each octant giving 1/8 of the total volume. Then we consider that the variables can be ordered by size in 6 ways, so we can assume $x>y>z$ and get 1/6 of the volume of an octant or 1/48 of the total volume $V$. Thus $x$ goes from 0 to 1, $y$ goes maximally to either $x$ (because in our sector $x>y$) or $\sqrt{1-x^2}$ (which is the border of the object). $z$ alway goes from 0 to $y$ because the other inequalities $x^2+z^2<1$ and $y^2+z^2<1$ are satisfied automatically by having sorted the variables.

The integral to work on is then

$$w=\int_{x=0}^1 \int_{y=0}^{{\rm min}(x,\sqrt{1-x^2})} \int_{z=0}^y 1\ dz\,dy\,dx $$

The innermost integral is $y$, the second goes from 0 to either $x$ if $x<\frac{2}{\sqrt{2}}$ or to $\sqrt{1-x^2}$ otherwise (where I solved the trivial integral along $y$). We split the integral in a sum of the two cases:

$$w_1=\int_{x=0}^{2/\sqrt2} \frac{x^2}{2} dx = \frac{\sqrt2}{24} $$

$$w_2=\int_{x=2/\sqrt2}^1 \frac{1-x^2}{2} dx = \frac{1}{3} - \frac{5\sqrt2}{24}$$

$$w = w_1+w_2 = \frac13 - \frac{\sqrt2}{6}$$

Therefore $V = 48 w = 16-8\sqrt2 $

That's 1.11877 times the volume of the enclosed sphere, which sounds plausible.