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 $\mathbb R^3$.
On
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.
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.