Ball $x^2+y^2+z^2 \le R^2$ and cone $z \ge \sqrt{x^2+y^2}$ constitute a section. This section is a K $\subset \mathbf R^3$
How do I express this K in polar coordinates and how do I calculate the sections volume?
I know that:
$x = rsin\theta cos\phi$
$y = rsin\theta sin\phi$
$z = rcos\theta$
On
So, apparently, your "section" (not sure what that exactly is in this context) $\;K\;$ is the 3-dimensional body bounded by the (closed and full) sphere $\;x^2+y^2+z^2\le R^2\;$ and the cone $\;z\ge\sqrt{x^2+z^2}\;$, or: the body between the cone from below and the sphere from above.
Using cylindrical coordinates, we get that in order to evaluate the volume of $\;K\;$ (assuming it is what I wrote above) we need:
$$x^2+y^2=R^2-x^2-y^2\implies x^2+y^2=\frac{R^2}2\implies$$
$$\int_0^{R/\sqrt2} dr\int_0^{2\pi}d\theta\int_r^{\sqrt{R^2-r^2}} r\,dz=2\pi\int_0^{R/\sqrt2}r\sqrt{R^2-r^2}\,dr=$$
$$=\left.-\pi\frac23(R^2-r^2)^{3/2}\right|_0^{R/\sqrt2}=-\frac{2\pi}3\left(\left(\frac{R^2}2\right)^{3/2}-(R^2)^{3/2}\right)=-\frac{2\pi}3\left(\frac{R^3}{2\sqrt2}-R^3\right)=\frac{\pi R^3}3\frac{2\sqrt2-1}{\sqrt2}$$
Impossible: $^2+^2+^2 \leq 0$ (except for the single point $(0,0,0)$).