Calculate the volume of $$V={\{(x,y,z)|x^2+y^2\leq z\leq \sqrt{6-x^2-y^2}}\}$$
My attempt:
$x=r\cos \theta$,
$y=r \sin \theta$, $z=z$
From here we can calculate the boundaries of our variables:
$0\leq r\leq \sqrt6, 0\leq \theta \leq 2\pi$, $r^2\leq z\leq \sqrt{6-r^2}$
The jacobian is $J=r$ $$vol(V)=\int\int\int 1{dxdydz}=\int_0^{2\pi}\int_0^\sqrt6\int_{r^2}^{\sqrt{6-r^2}}|J|{dzdrd\theta}=2\pi\int_0^\sqrt6\int_{r^2}^{\sqrt{6-r^2}}rdzdr=$$ $$2\pi\int_0^\sqrt6r(\sqrt{6-r^2}-r^2)dr$$
Now we substitute $$t=r^2, dt=2rdr\implies\pi\int_0^6(\sqrt{6-t}-t)dt$$
But the correct answer should be $$\pi\int_0^2(\sqrt{6-t}-t)dt$$
The intersection is incorrect, by your substitution $x= r\cos(\theta), ~ y = r \sin (\theta) $ we get:
$$\begin{aligned} r^2 &= \sqrt{6-r^2} \\&\implies r^4 = 6-r^2 \\& \implies r = \sqrt{2}, ~ - \sqrt{2} \end{aligned}$$
And not $0 \le r \le \sqrt{6}$