Error in calculation of intersection of three cylinders Set $A:=\{(x,y,z)\in \mathbb R^{3}: x^2+y^2\leq 1, x^2+z^2\leq 1, y^2+z^2 \leq 1\}$
I want to find the volume. I have to use symmetry and without polar coordinates.
My idea: Using symmetry we can look at the first octant and can restrict $x,y,z \geq 0$
$\lambda^{d}(A)=\int_{A}dxdydz$
And we then set $0\leq z\leq 1$ and $0 \leq x \leq \sqrt{1-z^2}$ and  $0 \leq y \leq \sqrt{1-z^2}$
So $\int_{A}dxdydz=\int_{0}^{1}\int_{0}^{\sqrt{1-z^2}}\int_{0}^{\sqrt{1-z^2}}dxdydz=\int_{0}^{1}2\sqrt{1-z^2}dz$
And then I would use substitution, namely, $z = \sin{x}\Rightarrow dz =\cos{x}dx$, so $\int_{0}^{1}2\sqrt{1-z^2}dz=\int_{0}^{\frac{\pi}{2}}2\sqrt{1-\sin{x}^2}\cos{x}dx=2\int_{0}^{\frac{\pi}{2}}\cos^{2}{x}dx=2\int_{0}^{\frac{\pi}{2}}\frac{1}{2}+\frac{1}{2}\cos{2x}dx=\int_{0}^{\frac{\pi}{2}}1+\cos{2x}dx=x+\frac{\sin{2x}}{2}\vert_{0}^{\frac{\pi}{2}}=\frac{\pi}{2}$
So getting back to eight octants, we'd get $\lambda^{d}(A)=4\pi$ but I have been told this is incorrect. I do not understand where I went wrong

$4\pi$ is quite clearly too much: we should expect a volume close to (and a bit larger than) the volume of a unit sphere, i.e. $\frac{4}{3}\pi$. Assume that the value of $z\in[-1,1]$ has been fixed. The $z$-section of our body is shaped as
$$ \left\{\begin{array}{rcl}x^2+y^2&\leq& 1\\x^2,y^2&\leq &1-z^2\end{array}\right.$$
hence it is a square with side length $2\sqrt{1-z^2}$ for any $|z|\geq \frac{1}{\sqrt{2}}$ and a circle with four circle segments being removed for any $|z|\leq\frac{1}{\sqrt{2}}$. In the former case the area of the section is $4(1-z^2)$, in the latter it is $\pi + 4|z|\sqrt{1-z^2}-4\arcsin(|z|)$. The wanted volume is so
$$ 2\left[\int_{0}^{1/\sqrt{2}}4(1-z^2)\,dz + \int_{1/\sqrt{2}}^{1}\pi + 4z\sqrt{1-z^2}-4\arcsin(z)\,dz\right]$$
i.e. $\color{blue}{8\sqrt{2}-2\pi}\approx 5$.
