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

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

A: 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.
A: 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 $x,y,z\geq 0$
