Finite region in $R^3$ that is bounded by the three surfaces (cylindrical) Compute the volume of the finite region in $\mathbb R^3$ that is bounded by the following three surfaces:
$$z = 2 - x^2 - y^2$$
$$z = 8 - x^2 - y^2$$
$$z = \sqrt{x^2 + y^2}$$
I think the easiest way is to solve this in cylindrical coordinates. Therefore I changed the surfaces to cylindrical coordinates:
$$z = 2 - r^2$$
$$z = 8 - r^2$$
$$z = \sqrt{r^2}$$
In the picture, I think the orange part is the area that should be used to find the full volume. But I don't know the boundaries for the ones with a ?.
$$\int_{0}^{2\pi} \int_{?}^{?} \int_{?}^{?}rdzdrd\theta$$
Can someone help me with the boundaries?


So by summing the following two integrals, I will be done:
$$\int_{0}^{2\pi} \int_{0}^{1} \int_{2-r^2}^{8-r^2} rdzdrd\theta$$
$$\int_{0}^{2\pi} \int_{1}^{\frac{-1+\sqrt{33}}{2}} \int_{r}^{8-r^2} rdzdrd\theta$$
 A: For these types of problems there are two similar ways to compute the volume. One is that you take long cylinders along $z$ of thickness $dr$. The volume of such cylindrical shell is $$dV=2\pi rh\ dr$$ The other is you take disks and washers (perforated disks) of height $dz$, with volume $$dV=\pi(r_2^2-r_1^2)dz$$
Here $r_1$ is the inner radius (might be $0$), and $r_2$ is the outer radius.
It does not matter which method you choose, first you need to find the "corners" of the orange region in your figure. At $r=0$ you have $z=2$ and $z=8$. Then for the lowest corner:
$$r=z=2-r^2$$
The only positive $r$ solution for this is $r=1$, so $r=z=1$. For the right corner:
$$r=z=8-r^2$$ The positive solution is $z=r=\frac{\sqrt{33}-1}2\approx 2.37$.
Now we go back to the integration. You will need to split the volume calculations in regions. In the washer method, you have three regions: 1. $z$ from $1$ to $2$, with $r_1=\sqrt{2-z}$ and $r_2=z$, 2. $z$ from $2$ to $2.37$, where $r_1=0$ and $r_2=z$, and 3. $z$ from $2.37$ to $8$, with $r_1=0$ and $r_2=\sqrt{8-z}$.
For the cylindrical shells method you have only two regions: 1. $r$ from $0$ to $1$, where the height of the shell is $(8-r^2)-(2-r^2)=6$, and 2. $r$ from $1$ to $2.37$, where the height is $(8-r^2)-r$. 
