How could we express the space in an other way? I want to calculate a triple integral over the space $$D=\{(x,y,z)\mid |x|\leq 1, |y|\leq 1, z\geq 0, x^2+y^2+z^2\leq 1\}$$
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We have the following:

*

*$|x|\leq 1\Rightarrow -1\leq x\leq 1$


*$|y|\leq 1\Rightarrow -1\leq y\leq 1$


*$x^2+y^2+z^2\leq 1 \Rightarrow z^2\leq 1-x^2-y^2\Rightarrow -\sqrt{1-x^2-y^2}\leq z\leq \sqrt{1-x^2-y^2}$, since $z\geq 0$ we get $0 \leq z\leq \sqrt{1-x^2-y^2}$.
So that the square root is defined it must hold that $1-x^2-y^2 \geq 0 \Rightarrow x^2+y^2\leq 1 \Rightarrow x^2\leq 1-y^2 \Rightarrow -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}$, right?
So, how could we express $D$ in an other way?
$$D=\{(x,y,z)\mid -1\leq x\leq 1, -1\leq y\leq 1, 0\leq z\leq \sqrt{1-x^2-y^2}\}$$ or $$D=\{(x,y,z)\mid -\sqrt{1-y^2}\leq x\leq \sqrt{1-y^2}, -1\leq y\leq 1, 0\leq z\leq \sqrt{1-x^2-y^2}\}$$ ?

Note that $x^2+y^2+z^2\leq 1$ implies  $|x|\leq 1, |y|\leq 1$: 
$$|x|^2=x^2\leq x^2+y^2+z^2\leq 1 \implies |x|\leq 1.$$
Hence, $D$ is the upper-half unit ball centered at the origin
$$D=\{(x,y,z)\in\mathbb{R}^3\;:\; z\geq 0, x^2+y^2+z^2\leq 1\}.$$
By using the cartesian coordinates, we have (see your last line)
$$\iiint_D f(x,y,z)\, dxdydz =\int_{y=-1}^1\int_{x=-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\int_{z=0}^{\sqrt{1-x^2-y^2}} f(x,y,z)\, dxdydz.$$
By using the cylindrical coordinates, we get
$$\iiint_D f(x,y,z)\, dxdydz =\int_{z=0}^1\int_{\theta=0}^{2\pi}\int_{\rho=0}^{\sqrt{1-z^2}} f(\rho\cos\theta,\rho\sin\theta,z) \rho d\rho  d\theta dz.$$
By using the spherical coordinates, we get
$$\iiint_D f(x,y,z)\, dxdydz \\=\int_{\rho=0}^1\int_{\theta=0}^{2\pi}\int_{\varphi=0}^{\pi/2} f(\rho\sin \varphi\cos\theta,\rho\sin \varphi\sin\theta,\rho\cos \varphi) \rho^2\sin(\varphi)d\rho d\varphi d\theta.$$
