Triple integration problem

The conditions on $\large \Sigma$ implies that

$x^2 + y^2 \leq z \leq 1 - 4x$

so when $z = 0 \Rightarrow (x+2)^2 + y^2 = 5$

Calling $\gamma$ this space, we can say that it is $\gamma = \begin{cases} x = \rho\cos\theta \\ y = \rho\sin\theta \end{cases}$ where $\rho\in[0,\sqrt{5}]$ and $\theta\in[0,2\pi]$ .

So $I$ becomes $\large \int\int_\gamma\left(\int_{x^2+y^2}^{1-4x}x\,dz\right) = \int\int_\gamma x - 4x^2 - x^3 - xy^2\,dx dy$

Solving now for $\rho$ and $\theta$ :

$I = \int_{0}^{2\pi}\int_{0}^{\sqrt{5}} \left( \rho\cos\theta - 4\rho^2\cos^2\theta - \rho^3\cos^3\theta - \rho^3\cos\theta\sin^2\theta \right)\cdot\rho\,d\rho d\theta =$

$= \int_{0}^{2\pi}\left[ \frac{\rho^3}{3}\cos\theta - \rho^4\cos^2\theta - \frac{\rho^5}{5}\cos^3\theta - \frac{\rho^5}{5}\cos\theta\sin^2\theta \right]_{0}^{\sqrt{5}}\,d\theta =$

$= \int_{0}^{2\pi} \frac{5\sqrt{5}}{3}\cos\theta - 25\cos^2\theta - 5\sqrt{5}\cos\theta(\cos^2\theta+\sin^2\theta) \,d\theta =$

$= -\frac{10\sqrt{5}}{3} \int_{0}^{2\pi} \cos\theta \,d\theta - 25\int_{0}^{2\pi} \cos^2\theta \,d\theta =$

$= 0 - 25(\pi) = \boxed{-25\pi}$