Find multiple integral $$\int \int \int y\sqrt(x^2+y^2) \, \mathrm dx\, \mathrm dy \, \mathrm dz  $$
$$V: z^2=4x^2+4y^2, z=2, y\ge\pm x, z\ge0$$

Here I illustrated area V. As I understand, it is all space under the cone $z^2=4x^2+4y^2$
So I write this:
$$\int_D \int\ \mathrm dx\,\mathrm dy \,\int_0^{4x^2+4y^2} y\sqrt(x^2+y^2) \, \mathrm dz=$$, D is area of projection cone on $OXY$
$$=\int_D \int\ \mathrm dx\,\mathrm dy \,(y\sqrt(x^2+y^2))*(4x^2+4y^2-0)$$

Here I say:
$$x = rcos\phi$$
$$y = rsin\phi$$
$$\mathrm dx\mathrm dy = r\mathrm dr \mathrm  d\phi$$
$$=\int_{\pi/4}^{3\pi/4}sin\phi\mathrm d\phi \int_0^1 4r^5 
\mathrm dr=$$
$$-cos\phi|_{\pi/4}^{3\pi/4}*4r^6/6|_0^1=2\sqrt2/3$$
As I understand, this should mean that volume of this space is equal $2\sqrt2/3$, but volume = full_volume - cone_volume = $\pi R^2/4*H - H/3*\pi r^2/4, H=2, R=1$
volume = $\pi/2-2/3*\pi/4 \neq 2\sqrt2/3$

So, where is a mistake? And did I draw $V$ correctly('cause I am not sure about $y \ge \pm x)$
2: https://i.stack.imgur.com/4cnLr.jpg
 A: If you wanted to find the volume of the region (which is what I'm guessing you want to do), you would solve the integral
$$\iiint_V 1  \ dx \ dy \ dz$$
And as of now, your triple integral's bounds do not make sense, as they describe a surface, not a volume. I believe this integral is what you are looking for.
$$\iiint_{z^2 \leq 4x^2+4y^2 \leq 4, \ 0 \leq z \leq2, \ y \geq |x|} 1 \ dx \ dy \ dz$$
I added the bound $4x^2+4y^2 \leq 4$ because otherwise, $x$ and $y$ can be as large as we'd like and the volume would be infinite. This integral can be written more easily in cylindrical coordinates and we can transform the integral to 
$$\iiint_{z^2 \leq 4r^2 \leq 4, \  0 \leq z \leq 2, \ \pi/4\leq\theta \leq 3\pi/4} r  
 \ dr \ d\theta \ dz$$
Now filling in the bounds and solving, we have
$$\int_0^2 \int_{\pi/4}^{3\pi/4} \int_{z/2}^{1} r \ dr \ d\theta \ dz$$
$$=\int_0^2 \int_{\pi/4}^{3\pi/4} \frac{1}{2}-\frac{z^2}{8} \ d\theta \ dz$$
$$=\int_0^2 \frac{\pi}{4}-\frac{\pi z^2}{16} \  dz = \frac{\pi}{2}-\frac{\pi}{6} = \frac{\pi}{3}$$
So $\pi/3$ is the volume of the region you have drawn and found without calculus. Let me know if there is anything else you want answered.
