$$\ D = \{(x, y, z) ∈ \mathbb{R}^3: x^2 + y^2 + z^2 \leq 1,\space z\geq {(\sqrt{5}-1)/2} \}$$ I do not understand why the two following parametrisations for the integral of this volume do not give the same result $$\sigma_1(r,\theta,\phi)=(r\cos\theta \sin\phi,r\sin\theta \sin\phi,r\cos\phi) $$ with $\space0 \leq r\leq 1, \space 0\leq \theta\leq 2\pi, \space 0\leq \phi\leq\arccos((\sqrt{5}-1)/2)$ $$V_1 = \int_0^{\arccos((\sqrt{5}-1)/2))}\int_0^1\int_{0}^ {2\pi} r^2\sin(\phi) d\theta drd\phi \space = \frac\pi3 (3-\sqrt{5})$$ $$ $$ $$ \sigma_2(r,\theta,z)=(r\cos\theta,r\sin\theta,z) $$ with $ \space0 \leq r\leq \sqrt{1-z^2}, \space 0\leq \theta\leq 2\pi, \space (\sqrt{5}-1)/2\leq z\leq 1 $
$$V_2 = \int_{(\sqrt{5}-1)/2}^1\int_0^{\sqrt{1-z^2}}\int_{0}^ {2\pi}r\space d\theta drdz \space = \frac\pi6 (3-\sqrt{5})$$
Your bounds on $r$ in the first integral is incorrect. This can easily be seen as it ranges from $0$ to $1$, but we shouldn't be including anything of distance less than $(\sqrt{5}-1)/2$ from the origin because $z\geq (\sqrt{5}-1)/2$. The range of $r$ will be even shorter as $\phi$ increases. Try drawing a picture to help you find the exact bounds.
Your second integral is fine.