Find the volume of the region between:
$x^2+y^2+z^2\le R^2$ and $z\ge\sqrt{x^2+y^2}$
Am I going to the right direction for this one:
$\int_0^{R/\sqrt2} dr\int_0^{2\pi}d\theta\int_r^{\sqrt{R^2-r^2}} r\,dz$
and I get for answer:
$\frac{\pi R^3}3\frac{2\sqrt2-1}{\sqrt2}$
You have the right expression for the volume integral, but you don't have the right end result.
$$V=\int_0^{R/\sqrt2} rdr\int_0^{2\pi}d\theta\int_r^{\sqrt{R^2-r^2}}\,dz$$ $$=2\pi\int_0^{R/\sqrt2} rdr (\sqrt{R^2-r^2} -r )=\frac{\pi R^3}{3}(2-\sqrt{2})$$