I recently computed the three-dimensional Lebesgue measure of the set $E := \{(x,y,z)\in\mathbb{R}^3 : x,y,z \geq 0, x+y+z \leq 1\}$
With $\lambda^3(E) = ab\pi r^2.$
I now want to conclude from this result, that the Lebesgue measure $\lambda^3(B_r(0)$ is $\pi r^2$ where $B_r(0) = \{(x,y) \in \mathbb{R}^3 x^2+ y^2< r \}$.
I know that this is true, because $B_r(0)$ is just a circle in the x-y plane, but how can I show that $\lambda^3(E) = ab\pi r^2 \Rightarrow \lambda^3(B_r(0)) = \pi r^2$ without having to compute $\lambda^3(B_r(0))$?
Many thanks in advance!