How can I find the volume bounded between $z=(x^2+y^2)^2$ and $z=x$?
My idea so far is to use cylindrical polar coordinates and $z$ limit is from $(x^2+y^2)^2$ to $x$ quite clearly but I am struggling to parametrize the surface $(x^2+y^2)^2<x$ for the other integration limits, could someone please help?
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In cylindrical coordinates, your equations become $z=r^4$ and $z=r\cos\theta$. So, take $\theta\in\left[-\frac\pi2,\frac\pi2\right]$ (so that $\cos\theta\geqslant0$). Now, $r^4\leqslant r\cos\theta\iff r\leqslant\sqrt[3]{\cos\theta}$. So, compute$$\int_{-\frac\pi2}^{\frac\pi2}\int_0^{\sqrt[3]{\cos\theta}}\int_{r^4}^{r\cos\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta.$$You should get $\dfrac\pi{12}$.
Clearly $x>0$, or equivalently $\theta\in(-\frac\pi2,\frac\pi2)$. Then $$ (x^2+y^2)^2<x\iff r^4<r\cos\theta\iff 0<r^3<\cos\theta. $$