What is the obvious mistake of this double integration? There is a definite double integral in Cartesian coordinates as follows:
$$\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} 2\sqrt{4-x^2}\ dydx\Rightarrow \int_{-2}^{2} 2y\sqrt{4-x^2}\rvert_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}  \ dx \Rightarrow 4\int_{-2}^{2} (4 - x^2)dx \Rightarrow 4(4x - \frac{x^3}{3})|_{-2}^{2} = \frac{128}{3} $$
and there is another double integral in polar coordinates which is supposed to be the exact same integral as the above, but the final answer is not 128/3. it took me several hours, but I cant figure out which step of this integration is wrong.
$$ \int_{0}^{2\pi}\int_{0}^{2} 2r\sqrt{4-r^2{\cos^2\theta}} drd\theta
 \xrightarrow[]{u = 4-r^2{\cos^2\theta}}  \int_{0}^{2\pi} - \frac{2}{3} \frac{(4-r^2{\cos^2\theta})^\frac{3}{2}}{\cos^2\theta}\Big|_{0}^{2}  \ d\theta \Rightarrow \int_{0}^{2\pi} (-\frac{2}{3}\frac{(4-4\cos^2\theta)^\frac{3}{2}}{\cos^2\theta} + \frac{2}{3}\frac{4^\frac{3}{2}}{\cos^2\theta} ) d\theta \Rightarrow \frac{16}{3}\int_{0}^{2\pi} \frac{1-(1-\cos^2\theta)^\frac{3}{2}}{\cos^2\theta} d\theta \Rightarrow  \frac{16}{3} \int_{0}^{2\pi} \frac{1-{\sin^3\theta}}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3}\int_{0}^{2\pi} \frac{1}{\cos^2\theta} - \frac{\sin^3\theta}{\cos^2\theta} d\theta \Rightarrow \frac{16}{3}(\int_{0}^{2\pi} \frac{1}{\cos^2\theta} d\theta - \int_{0}^{2\pi} \frac{\sin^3\theta}{\cos^2\theta} d\theta) \Rightarrow \frac{16}{3}(\tan\theta - \int_{0}^{2\pi} \frac{\sin^3\theta}{\cos^2\theta} d\theta) \xrightarrow[]{u=\cos\theta} \frac{16}{3}(\tan\theta - \int \frac{1-u^2}{u^2} du) \Rightarrow \frac{16}{3}(\tan\theta - (-\sec\theta - \cos\theta))\Big|_{0}^{2\pi}\Rightarrow \frac{16}{3} ((0+1+1) - (0+1+1)) = 0 $$
I know there is a huge problem with the second one and my apologies if it's an elementary problem. But I would really appreciate if you could help me find the problem.

The problem is that $(1-\cos^2)^\frac{3}{2}=|\sin^3\theta|$ and not $\sin^3\theta$. Then you have to consider intervals for $\theta$ to discard absolute value.
The second problem is at the second to the last line, the change of variable $u=\cos\theta \Rightarrow d\theta = -\frac{du}{\sin\theta}$ and you forgot a negative sign.
It is very nice that you try to verify your answer by considering other possible ways of solving and in this problem it is possible to solve it in both coordinates but sometimes it is not possible to find answer in other coordinates and the coordinate system that you choose is completely dependent on the geometry of problem, like finding the electrostatic potential function in spherical or cylindrical capacitors, just try to solve it with simplest way possible and be done with it dude.
