<s>[INST] Find the volume of the set in $\mathbb{R}^3$ defined by the equation $x^2 + y^2 \leq z \leq 1 + x + y$. Let $D=\left\{\left(x,y,z\right) \in \mathbb{R}^3: x^2 + y^2 \leq z \leq 1 + x + y \right\}$. I tried to integrate with cartesian coordinates, so that
$$\iiint_Ddxdydz = \int_{x=\frac{1-\sqrt{5}}{2}}^{{\frac{1+\sqrt{5}}{2}}} \int_{z=x^2}^{1+x} \left( \int_{y=z-(1+x)}^{\sqrt{z-x^2}}dy\right)dxdz$$
but the calculation seems to be quite laborious. Is there a way to solve it easily, maybe with another coordinate system?
Edit: After your suggestions I finally solved the problem. If someone is interested in the calculation itself I'll report the main steps below:
$$\int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{\frac{3}{2}}}\left(\int_{z=r^2+r(\sin\theta+\cos\theta)+\frac{1}{2}}^{r(\sin\theta+\cos\theta)+2}dz\right)rdrd\theta = \int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{\frac{3}{2}}} \left( \frac{3}{2}r-r^3\right)drd\theta = \pi\int_{0}^{\frac{3}{2}}\left(\frac{3}{2}-t\right)dt = \frac{\pi}{2} \left[3t-t^2\right]_{0}^{\frac{3}{2}}=\frac{9\pi}{8}$$
 [/INST] $\textbf{Hint}$: Use this modified cylindrical coordinate system:
$$\begin{cases}x = r\cos\left(\theta-\frac{\pi}{4}\right)\\ y = r\sin\left(\theta-\frac{\pi}{4}\right)\\ z = z \\\end{cases}$$
What does the change do? Does it alter the Jacobian (which measures how much the volume is distorted between coordinate systems) ?
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