# Integration to find the volume

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 for the other integration limits, could someone please help?

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