Integral over set given by $x^2+y^2 \le 1, \frac {1}{\sqrt{3}} \le \frac{y}{x} \le \sqrt 3, z \le 1$ I want to compute the integral
$$\int_V x^2yz \, dx \, dy \, dz$$
over the set
$$V = \left\{ (x,y,z) \in \mathbb{R}^3_{>0} \mid x^2+y^2 \le 1, \frac {1}{\sqrt{3}} \le \frac{y}{x} \le \sqrt 3, z \le 1 \right\}$$
I thought the way the set is given motivates the substitution $u=x^2+y^2$, $v=\frac{y}{x}$ and $w=z$. Because then $V$ becomes a cuboid
$$V' = \left\{(u,v,w) \in \Bbb R^3_{\gt 0} \mid u \le 1, \frac{1}{\sqrt{3}} \le v \le \sqrt{3}, w \le 1 \right\} = [0,1] \times [\frac{1}{\sqrt{3}}, \sqrt{3}] \times [0,1]$$
which we can easily integrate over. However this doesn't work out as the integral in the end doesn't become easy. Can somebody tell me what has gone wrong? My computations are below.

To use the substitution rule we define the diffeomorphism
$$\Psi \left( \begin{bmatrix} x\\y\\z \end{bmatrix} \right) = \begin{bmatrix} u\\v\\w \end{bmatrix} = \begin{bmatrix} x^2+y^2\\ \frac{y}{x} \\z \end{bmatrix}$$
with inverse
$$\Phi \left( \begin{bmatrix} u\\v\\w \end{bmatrix} \right) = \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} \sqrt{\frac{u}{1+v^2}} \\ v\sqrt{\frac{u}{1+v^2}} \\ w \end{bmatrix}$$
We need to compute the correction factor with
$$\begin{aligned} D_{(u,v,w)}\Phi &= \begin{bmatrix} \frac{1}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{1}{1+v^2} & \frac{1}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{-2vu}{(1+v^2)^2} & 0 \\ \frac{v}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{1}{1+v^2} & \sqrt{\frac{u}{1+v^2}} + v \left( \frac{1}{2} \frac{1}{\sqrt{\frac{u}{1+v^2}}} \frac{-2vu}{(1+v^2)^2} \right) & 0 \\ 0 & 0 & 1 \end{bmatrix} \\&= \begin{bmatrix} \frac{1}{2} \frac{1}{\sqrt{u(1+v^2)}} & \frac{-v\sqrt{u}}{(1+v^2)^{3/2}} & 0 \\ \frac{v}{2} \frac{1}{\sqrt{u(1+v^2)}} & \frac{\sqrt{u}}{(1+v^2)^{3/2}} & 0 \\ 0 & 0 & 1 \end{bmatrix} \end{aligned}$$
and
$$\begin{aligned} \det(D_{(u,v)}\Phi) &= \frac{1}{2} \frac{1}{\sqrt{u(1+v^2)}} \frac{\sqrt{u}}{(1+v^2)^{3/2}} - \frac{v}{2} \frac{1}{\sqrt{u(1+v^2)}} \frac{-v\sqrt{u}}{(1+v^2)^{3/2}} \\&= \frac{1}{2} \frac{1}{\sqrt{u(1+v^2)}} \frac{\sqrt{u}(1+v^2)}{(1+v^2)^{3/2}} \\&= \frac{1}{2} \frac{1}{1+v^2}\end{aligned}$$
Thereby with substitution
$$\begin{aligned} \int_V f \, dx \, dy \, dz &= \int_{V'} (f \circ \Phi)(u,v) \, \lvert \det(D_{(u,v)}\Phi) \rvert \, du \, dv \, dw \\&= \int_{V'} \frac{u}{1+v^2} v\sqrt{\frac{u}{1+v^2}}w \, \frac{1}{2} \frac{1}{1+v^2} \, du \, dv \, dw \end{aligned}$$

As suggested if you make a sketch of the domain it seems a good idea use cylindrical coordinates

that is
$$\int_{\pi/6}^{\pi/3} d\theta \int_{0}^{1} dr \int_{0}^{1} r^4\cos^2\theta\sin\theta z \,dz$$
