Q:

Volume of the figure bounded by $x=\sqrt{y^2+z^2}$, $x=\sqrt{\frac}{y^2+z^2}{3}$ and $x=4$ using spherical coordinates

Find the volume of the figure bounded by $x=\sqrt{y^2+z^2}$, $x=\sqrt{\dfrac{y^2+z^2}{3}}$ and the plane $x=4$, using spherical coordinates.

$x=\sqrt{y^2+z^2}$  $x=\sqrt{\frac}{y^2+z^2}{3}$ and the plane $x=4$
How can I turn this ino spherical coordinates and find its volume?

A:

Let $z=0$, on $xy-$plane we have an area between lines $y=x$ and $y=\sqrt{3}x$ which restricted by $x=4$. Rotation this area about $x-$axis gives us the volume
$${\bf V}=\pi\int_0^4(\sqrt{3}x)^2-x^2\,dx=\color{blue}{\dfrac{128}{3}\pi}$$
Edit:
Make $x$ turn to $z$ and $z$ turn to $x$, then according to definition of spherical coordinates in wiki we have:
$${\bf V}=\int_0^{2\pi}\int_{\pi/4}^{\pi/3}\int_0^{4/\cos\theta} \rho^2 \sin\theta  \,d\rho \,d\theta \,d\varphi=\color{blue}{\dfrac{128}{3}\pi}$$

