I am trying to find the volume of $E=\{(x,y,z)\in\mathbb{R^3}:\sqrt{\frac{x^2+y^2}{3}}\leq z\leq \sqrt{4-x^2-y^2}\}$ and I would be grateful if someone could check out my work. Comments are welcome. Thanks.
What I have done:
$$\iiint_E dxdydz=\iint_{\{(x,y)\in\mathbb{R^2}:\ 0\leq x^2+y^2\leq 4 \}} dxdy\int_{z=\sqrt{\frac{x^2+y^2}{3}}}^{z=\sqrt{4-x^2-y^2}}dz=\iint_{\{(x,y)\in\mathbb{R^2}:\ 0\leq x^2+y^2\leq 4 \}} \left(\sqrt{4-x^2-y^2}-\sqrt{\frac{x^2+y^2}{3}}\right)dxdy=\int_{\theta=0}^{\theta=2\pi}d\theta\int_{r=0}^{r=2}\left(\sqrt{4-r^2}-\frac{r}{\sqrt{3}}\right)rdr=2\pi\left[\int_{r=0}^{r=2}r\sqrt{4-r^2}dr-\frac{1}{\sqrt{3}}\int_{r=0}^{r=2}r^2dr\right]= 2\pi\left[\int_{u=4}^{0}\sqrt{u}(-\frac{1}{2}du)-\frac{8}{3\sqrt{3}} \right]=2\pi\left[\frac{1}{2}\int_{u=0}^{u=4}\sqrt{u}du-\frac{8}{3\sqrt{3}}\right]=2\pi\left[\frac{1}{2}\cdot\frac{2}{3}\cdot (8-0)-\frac{8}{3\sqrt{3}}\right]=2\pi\left[\frac{8}{3}-\frac{8}{3\sqrt{3}}\right]=\frac{16\pi}{3}\left(1-\frac{1}{\sqrt{3}}\right).$$
$\begin{align*}\iiint_E dxdydz&=\iint_{\{(x,y)\in\mathbb{R^2}:\ 0\leq x^2+y^2\leq 4 \}} dxdy\int_{z=\sqrt{\frac{x^2+y^2}{3}}}^{z=\sqrt{4-x^2-y^2}}dz\\ &=\iint_{\{(x,y)\in\mathbb{R^2}:\ 0\leq x^2+y^2\leq 4 \}} \left(\sqrt{4-x^2-y^2}-\sqrt{\frac{x^2+y^2}{3}}\right)dxdy\\ &=\int_{\theta=0}^{\theta=2\pi}d\theta\int_{r=0}^{r=2}\left(\sqrt{4-r^2}-\frac{r}{\sqrt{3}}\right)rdr\\ &=2\pi\left[\int_{r=0}^{r=2}r\sqrt{4-r^2}dr-\frac{1}{\sqrt3}\int_{r=0}^{r=2}r^2dr\right]\\ &=2\pi\left[\int_{u=4}^{0}\sqrt{u}(-\frac{1}{2}du)-\frac{8}{3\sqrt{3}} \right]\\ &=2\pi\left[\frac{1}{2}\int_{u=0}^{u=4}\sqrt{u}du-\frac{8}{3\sqrt{3}}\right]\\ &=2\pi\left[\frac{1}{2}\cdot\frac{2}{3}\cdot (8-0)-\frac{8}{3\sqrt{3}}\right]\\ &=2\pi\left[\frac{8}{3}-\frac{8}{3\sqrt{3}}\right].\end{align*}$