How do I compute the surface of this solid?

I want to compute the surface of the solid given by $$(x^2+y^2+z^2)^2=z^2\,\,\,\,\,\,\,\,\,\,(1)$$ I thought about spherical coordinates, i.e.$$(x,y,z)=(r\cos\phi\sin\theta,r\sin\phi\sin\theta,r\cos\theta)$$ Then (1) would be equivalent to $$r^4=r^2\cos\theta\,\,\,\,\,\,\,\,\,\,(2)$$. I thought to take $r>\epsilon$ for some $\epsilon >0$ therefore (2) is equivalent to $$r=\pm \cos\theta$$. But can I then only define a parametrisation on $\theta, \phi$ of this solid?

I'm confused how to find it otherwise.

Could someone help me?

Thank you a lot

That surface is the union of the surfaces$$x^2+y^2+z^2=z\tag1$$and$$x^2+y^2+z^2=-z\tag2.$$But$$(1)\iff x^2+y^2+\left(z-\frac12\right)^2=\frac14$$and$$(2)\iff x^2+y^2+\left(z+\frac12\right)^2=\frac14.$$So, it's the union of two spheres.