If I have $B:=\{x\in \mathbb R^3 :||x||_2 \le r\}$ as a closed 3 dimensional ball of radius $r>0$, how can I show that the formula $\lambda^3=\frac4 3\pi r^3$ holds? I was given these figures to use:
the half-ball of radius $r$ is $H:=\{(x,y,z)\in \mathbb R^3:x^2+y^2+z^2\le r^2 \text{ and } z\ge 0\}$,
the cone standing on its peak of height $r$ with circle of radius $r$ as base:
$K:=\{(x,y,z)\in \mathbb R^3:x^2+y^2\le z^2 \text{ and } 0\le z\le r\}$
the cylinder of radius $r$ and height $r$: $Z:=\{(x,y,z)\in \mathbb R^3:x^2+y^2\le r^2 \text{ and } 0\le z\le r\}$
I honestly have no idea where or how to use them, or where to even begin with this, any help on this matter would be amazing!
Also I have an idea to use that Lebesgue-measure $\lambda^2 $ of the circle is $\pi r^2 $ but not really sure how to use it