Define $$B:=\left\{ (x,y,z) \in \mathbb R^3 : \; x,y,z \geq 0, \; x^2+y^2+z^2\in [1,2] \right\}.$$
Using the function/spherical coordinates $\Phi: \mathbb R^3 \to \mathbb R^3$ given by $$\Phi(r, \theta, \eta):=\left(r\sin(\theta)\cos(\eta),r\sin(\theta)\sin(\eta),r\cos(\theta)\right),$$
compute $$x_0 := \int_B x \, d(x,y,z), \; y_0 := \int_B y \, d(x,y,z) \text{ and } z_0 := \int_B z \, d(x,y,z).$$
I understand the main change of variable section of this question and how to calculate $x_0,y_0$ and $z_0$, but in the beginning of the solution for this question it states:
Set $A := [1, 2] \times \left[0,\frac{\pi}{2} \right] \times \left[0, \frac{\pi}{2} \right]$. Then $Φ(A) = B$ and $Φ|_{A^o} : A^o → Φ(A^o)$ is a homeomorphism. To see this, observe that $$Φ(A^o) = \left\{(x, y, z) ∈ \mathbb R^3: x, y, z > 0, \; x^2 + y^2 + z^2 ∈ ]1, 2[ \right\}.$$
and that on this set, the inverse of $Φ$ is given by
$$Ψ(x, y, z) := \left( \sqrt {x^2 + y^2 + z^2}, \cos^{−1} \left( \frac{z}{\sqrt{x^2+y^2+z^2}} \right),\tan^{−1}\left( \frac{y}{x} \right) \right),$$
which is continuous.
Why is $A := [1, 2] × [0,\fracπ2]×[0,\fracπ2]$? Can anyone show me where these values come from?
You want to find a suitable $r$, $\theta$ and $\eta$, so let's look at the region you have. Note that $B$ is the solid portion between the spheres with radii 1 and 2, in the first octant. So you have $1 \leq r\leq 2$ from the radii. The fact that $x, y \geq 0$ means that $0 \leq \theta \leq \frac{\pi}{2}$, and that $z \geq 0$ means that $0 \leq \eta \leq \frac{\pi}{2}$. That's where the domain $[1,2] \times [0, \frac{\pi}{2}] \times [0, \frac{\pi}{2}]$ comes from.