Cartesian, spherical and cylindrical coordinates

Question

For a spin top, what are the boundaries in cartesian, spherical and cylindrical coordinates for a solid top which lies below the sphere $$x^2+y^2+z^2=2z$$ and above the upper half of the cone $$x^2+y^2=3z^2$$

Answer 1

Let's start with the simplest, spherical coordinates:

From the first equation we have $$\rho^2=2\rho \cos \phi$$ $$\rho=2 \cos \phi$$ The second equation yields $$\rho^2 \sin^2 \phi =3 \rho^2 \cos^2 \phi$$ $$\tan \phi = \sqrt{3}\Rightarrow \phi = \frac{\pi}{3}$$ From that, we conclude that this region is described by \begin{cases} 0\leq\rho\leq 2 \cos \phi\\ 0\leq\phi\leq\frac{\pi}{3}\\ 0\leq\theta\leq2\pi \end{cases} At this point, we realize that this solid needs more than one region to be defined for both the Cartesian and cylindrical coordinate systems. Solving the first equation for z, we obtain $$z = 1\pm\sqrt{1-(x^2+y^2)}$$ the plus sign representing the upper half of the sphere and the minus the lower. We need to find the intersection of the two surfaces. It's quite apparent that it happens for $z=0$ or $z=\frac{1}{2}$. The intersection at the origin isn't of much interest, and applying $z=\frac{1}{2}$ to either of the equations we get, $$x^2+y^2=\frac{3}{4}$$ Meaning that it is a circunference of radius $\frac{\sqrt{3}}{2}$ centered at the $z$-axis in the plane $z=\frac{1}{2}$. Now we are able to determine the desired boundaries. For the cylindrical coordinates we have two disjoint regions:

$$\begin{cases} \frac{r}{\sqrt{3}}\leq z \leq 1+\sqrt{1-r^2} \\ 0 \leq r \leq \frac{\sqrt{3}}{2} \\ 0 \leq \theta \leq 2\pi \end{cases}$$

and

$$\begin{cases} 1-\sqrt{1-r^2}\leq z \leq 1+\sqrt{1-r^2} \\ \frac{\sqrt{3}}{2} \leq r \leq 1 \\ 0 \leq \theta \leq 2\pi \end{cases}$$

Finally, the Cartesian coordinates also requires, at least, 2 different regions defined by

$$\begin{cases} -\sqrt{3z^2-x^2} \leq y \leq \sqrt{3z^2-x^2} \\ -\sqrt{3}z \leq x \leq \sqrt{3}z \\ 0 \leq z \leq \frac{1}{2} \end{cases}$$

and

$$\begin{cases} -\sqrt{(2z-z^2)-x^2} \leq y \leq \sqrt{(2z-z^2)-x^2} \\ -\sqrt{2z-z^2} \leq x \leq \sqrt{2z-z^2} \\ \frac{1}{2} \leq z \leq 2 \end{cases}$$