${\int\int\int}_B dxdydz$ where $B$ is the region delimited by $x²+y²+z² = 4$ and $x²+y²=3z$ Take the following integral over the specified region:
${\int\int\int}_B dxdydz$ where $B$ is the region delimited by $x²+y²+z² = 4$ and $x²+y²=3z$
(i'm answering my own question because I was writing it and then found out my error, so I didn't want to erase all my work)

This is the intersection betweet the paraboloid $z = \frac{x²+y²}{3}$ and the sphere $x²+y²+z²=4$. Their intersection forms a circle with radius $\sqrt{3}$. Therefore we just need to find the height of the region by integrating with respect to $z$, from the paraboloid to the sphere:
$$\int_{\frac{x²+y²}{3}}^{\sqrt{4-x²-y²}}dz = \sqrt(4-x²-y²)-\frac{x²+y²}{3}$$
then we need to integrate this height all over the circle with radius $\sqrt{3}$, which is our region $B_1$:
$${\int\int}_{B_1}\sqrt(4-x²-y²)-\frac{x²+y²}{3}dxdy$$
So to make this easier, I'm gonna integrate in polar coordinates by doing the substitution $x = p\cos(t), y= p\sin(t)$ where $t$ goes from $0$ to $2\pi$ because our circle is centered at the origin, and $p$ goes from $0$ to $\sqrt{3}$. The jacobian determinant of the transformation is $p$, so our integral becomes:
$$\int_0^{2\pi}\int_0^{\sqrt{3}}\left(\sqrt{4-p²}-\frac{1}{3}p²\right)p \ dp \ dt$$
$$\int_0^{2\pi}\int_0^{\sqrt{3}}\left(\sqrt{4-p²}\ p-\frac{1}{3}p^3\right) \ dp \ dt = $$
$$-\frac{1}{2}\int_0^{2\pi}\int_0^{\sqrt{3}}\sqrt{4-p²}\ (-2p) \ dp \ dt - 
\int_0^{2\pi}\int_0^{\sqrt{3}}\frac{1}{3}p^3 \ dp \ dt = $$
$$-\frac{1}{2}\int_0^{2\pi}\int_{4}^{1}\sqrt{u} \ du - \int_0^{2\pi}\frac{\sqrt{3}^4}{12}dt = $$ 
$$\frac{1}{2}\int_0^{2\pi}\int_{1}^{4}\sqrt{u} \ du - \int_0^{2\pi}\frac{9}{12}dt = $$ 
$$\frac{1}{2}\int_0^{2\pi}\frac{2}{3}u^{3/2} \ du - \frac{3}{4}\int_0^{2\pi}dt = $$ 
$$\frac{1}{3}\int_0^{2\pi}(4)^{3/2}-1^{3/2} \ du - \frac{3}{4}2\pi = $$ 
$$\frac{7}{3}2\pi - \frac{3}{4}2\pi  = \frac{19}{12}2\pi = \frac{19\pi}{6}$$ 
