Consider the triple integral $$\iiint_{K} \frac{z}{2+ x^2 + y^2} dV$$, where K is the region defined by $z \geq \sqrt{x^2 + y^2}$ and $x^2 + y^2 + z^2 \leq 9$.
The question then asks me to rewrite the region using cylindrical coordinates which I did but the problem is that my bounds for r is wrong. Here is what I did.
$$\vec{r}(r,\theta, z) = (r\cos(\theta),r\sin(\theta), z)$$ $$0\leq r \leq 3$$ $$r\leq z\leq\sqrt{9-r^2}$$ $$0\leq \theta \leq 2\pi$$
The reason why I gave r the bounds it has is because if you see the equation: $$x^2 + y^2 + z^2 \leq 9$$ Here you can see that the radius needs to be less than 3 and then if you look at: $$z \geq \sqrt{x^2 + y^2} \iff 0 \geq x^2 + y^2 - z^2 $$ Here you can see that the radius is greater than $0$. But is it though? I just assumed because this is not a sphere but a hyperboloid of 2 sheets.
But like said the correct bounds for r is $0 \leq r \leq \frac{3}{\sqrt{2}}$. And I don't see why at all?
As I wrote in comment firstly we should find intersection of $x^2 + y^2 + z^2 \leqslant 9$ and $z \geqslant \sqrt{x^2 + y^2}$. Solving $z$ from both and equating we get for projection on $Oxy$ circle $x^2+y^2=\frac 9 2$. So, your integral in Cartesian coordinates is $$\int\limits_{-\frac{3}{\sqrt{2}}}^{\frac{3}{\sqrt{2}}}\int\limits_{-\sqrt{\frac 9 2-x^2}}^{\sqrt{\frac 9 2-x^2}}\int\limits_{\sqrt{x^2+y^2}}^{\sqrt{9-x^2-y^2}}$$ and here you can apply cylindrical coordinates.
Addition.
From image is easily seen, that volume in question have mentioned projection
the volume between the two given surfaces $\sqrt{x^2+y^2}\leqslant z \leqslant \sqrt{9-x^2-y^2}$ is the small "parachute" in the figure.