<s>[INST] Convert the double integral into polar $$\int^{4}_{0}\int_{0}^{\sqrt{4x-x^2}}\sqrt{x^2+y^2}\text{ dy dx}=14.22$$
In polar, we have $0\leq y\leq \sqrt{4x-x^2}$
Simplying, we get $y^2=4x-x^2$, which turns into $(x-2)^2+y^2=4$. Here is the region we are considering.

 Clearly $0\leq\theta\leq\pi$, and we know that
$x^2+y^2=4x$, which means $r^2=4r\cos\theta\to r=4\cos\theta$
Therefore, $0\leq r\leq 4\cos\theta$, and the original function $\sqrt{x^2+y^2}$ simply becomes $r$. So our integral now is:
$$\int^{\pi}_{0}\int^{4\cos\theta}_{0}r\text{ r dr d$\theta$}=\int^{\pi}_{0}\int^{4\cos\theta}_{0}r^2\text{ dr d$\theta$}=\int^{\pi}_{0}\left[\dfrac{r^3}{3}\right]^{4\cos\theta}_{0}\text{ d$\theta$}$$
$$=\dfrac{64}{3}\int^{\pi}_{0}\cos^3\theta\text{ d$\theta$}=0$$
But according to WA^, the cartesian coordinate integral is different? 

 [/INST] You have to integrate $d\theta$ from $0$ to $\pi/2$.
This will give you
$$\int_0^{\pi/2}\cos^3\theta\ d\theta = \frac{2}{3}$$
Hence
$$\frac{128}{9} = 14.22(...)$$
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