Calculate the volume of the solid: $S=\{(x,y,z) \in R^3: x^2+y^2 \le 4; 0 \le z \le x^2+y^2+4\}$ How can I calculate the volume of the solid from the following equation using double integral?
$$S=\{(x,y,z) \in \mathbb R^3: x^2+y^2 \le 4; 0 \le z \le x^2+y^2+4\}$$
I have been able to graph the solid and I do understand the concept working in $\mathbb R^2$ but I am not sure how in $\mathbb R^3$. 
 A: In $\mathbb R^3$ you write the volume of the solid in terms of $dx dy dz$, integrating $x$ between $-2$ and $2$, $y$ between $-\sqrt{4-x^2}$ and   $\sqrt{4-x^2}$, and $z$ between $0$ and $x^2+y^2+4$. The interpretation is that you count infinitesimally small cubes. In $\mathbb R^2$, you just integrate over the disk in the $xy$ plane, but now the volume element is a rectangular parallelepiped, with volume $dx dy (x^2+y^2+4)$. This is equivalent to doing the integration in the $z$ direction before.
A: There's a general formula for finding certain volumes using double integrals. Suppose $D\subseteq\mathbb{R}$ is a closed bounded region in the $xy$-plane, and $z=z_1(x,y)$ and $z=z_2(x,y)$ are two functions such that $z_1(x,y)\le z_2(x,y)$ for all $(x,y)\in D$. Then the volume of the solid enclosed between the graphs of $z=z_1(x,y)$ and $z=z_2(x,y)$ over the region $D$ is given by
$$\text{Volume}=\iint\limits_D \left(z_2(x,y)-z_1(x,y)\right)\,dxdy.$$
In this example the solid cylinder $x^2+y^2\le4$ gives us the vertical walls of the solid and tells us that the domain $D$ in the $xy$-plane is the disk $x^2+y^2\le4$. The "floor" is the function $z_1(x,y)=0$ and the "ceiling" is the function $z_2(x,y)=x^2+y^2+4$. Setting up and evaluating this double integral will give you the desired volume.
Also, you will want to switch to polar coordinates to evaluate this particular double integral most efficiently.
