For $x^2+y^2=4$, $$z=2\sqrt{4}\Rightarrow z=4$$ Since the radius of the basis is $2$, then the volume of the cone is $$V=\frac{\pi\cdot2^2\cdot4}{3}\Rightarrow V=\frac{16\pi}{3}$$
However, using double integral and polar coordinates, I get $$V=\int_{0}^{2\pi}\int_{0}^{2} 2r^2 dr d\theta=\frac{32\pi}{3}$$
What is wrong?
You're integrating the wrong or the opposite volume. Instead of integrating the volume inside the cone, you're integrating the volume outside the cone. This corresponds to the fact that a cone has $\frac{1}{3}$ of the volume of a cylinder with same base and height.