I want to find the volume of the following subset of $\mathbb R^3$: \begin{equation} (x^2+y^2)^2\leq x,\quad 0\leq z\leq 2x-\sqrt{x^2+y^2}\text{.} \end{equation}
I tried to draw a picture of the given region in my head, but I couldn't. So I changed it into a cylindrical coordinate system, and the region became \begin{equation} r^4\leq r\cos\theta,\quad 0\leq z\leq 2r\cos\theta-r,\quad r>0,\quad 0<\theta<2\pi\text{.} \end{equation}
But I still don't know what to do.
On
You have written the equations correctly in cylindrical coordinates. Now please note that the cylinder $r^3 = \cos \theta$ is between $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.
But $(2r\cos\theta-r)$ is not always $\geq 0$ in that range of $\theta$. As the lower bound of $z$ is $0$,
we must have $2r\cos\theta-r \geq 0 \implies \cos\theta \geq \frac{1}{2} $ or $|\theta| \leq \frac{\pi}{3}$
So your integral to find the volume is -
$\displaystyle \int_{-\pi/3}^{\pi/3} \int_0^{(\cos\theta)^{1/3}} \int_0^{2rcos\theta-r} r \ dz \ dr \ d\theta$
To define correctly the region you only need to see for what values of $\theta$ the given conditions holds. As $0<r\leqslant (\cos \theta)^{1/3}$ from here you know that valid values of $\theta $ are just possible in the region $[-\pi/2,\pi/2]$ because otherwise $\cos \theta <0$.
Now from the condition $0\leqslant z\leqslant r(2\cos \theta -1)$, as $r>0$, then the valid values of $\theta $ for this condition are these such that $2\cos \theta -1\geqslant 0\Leftrightarrow \cos \theta \geqslant 1/2 \Leftrightarrow \theta \in [-\pi/3,\pi/3]$.
Then, all together, gives that the valid range for $\theta $ for the two conditions is $[-\pi/3,\pi/3]$, so the integral of the volume take the form
$$ \int_{-\pi/3}^{\pi/3}\int_0^{(\cos \theta )^{1/3}}\int_{0}^{r(2\cos \theta -1)}r\mathop{}\!d z\mathop{}\!d r\mathop{}\!d \theta $$