Let $A:=\{(x,y,z) \in \mathbb R^{3}:x^2+y^2+z^2\leq 2\} \cap \{(x,y,z)\in \mathbb R^{3}:x^2+y^2\leq z\}$
I know that $\{(x,y,z) \in \mathbb R^{3}:x^2+y^2+z^2\leq 2\}$ is a sphere with radius $\sqrt{2}$ and
$\{(x,y,z)\in \mathbb R^{3}:x^2+y^2\leq z\}$ is a paraboloid.
In the solution to calculate $\lambda^{3}(A)$, we use polar coordinates, namely:
$P(r, \Phi, z)$ where $r > 0, \Phi \in ]0,2\pi[,z\in \mathbb R$
Problem: We then attempt to find a space $V$ and set $V:=\{0\}\times \mathbb R_{\geq 0}\times\mathbb R$
It is clear that $V$ is just a "half" of the z-y-plane and it is a subset of a hyperplane in $\mathbb R^{3}$, and therefore it has $\lambda^{3}$ measure $0$. I also know that before using the transformation formula we need to have a diffeomorphism that by construction needs to be surjective.
My question: Why is $V$ supposed to be the space we leave out, surely $V \neq \mathbb R^{3} -A$
So why are we using it?
For more context:
We are using the coordinates $p(r, \theta, z)=(r\cos{\theta}, r\sin{\theta}, z)$
and
$\lambda^{3}(A)=\int_{A}1d\lambda^{3}=\displaystyle{\int}_{[0,1]}\displaystyle{\int}_{[0,2\pi]}\displaystyle{\int}_{[r^{2},\sqrt{2-r^2}]}rdzd\theta dr$
In our classes, we have always, when switching to polar coordinates, had to find a $V$ such that $\lambda^{3}(V)=0$ in order to state
$\displaystyle{\int}_{\mathbb R^{3}}f(x,y,z)dxdxdz = \displaystyle{\int}_{R^{3}-V}f(r\cos{\theta},r\sin{\theta},z)rdzd\theta dr$
But I do not see why is it $\{0\}\times \mathbb R_{\geq 0}\times\mathbb R$ in this case?