2-sphere: $\theta$ $\in$ $[0,2\pi)$, $\psi$ $\in$ $[0,\pi)$
$$x=\left(R-r\right)\cos\left(\theta\right)\sin\left(\psi\right)$$
$$y=\left(R-r\right)\sin\left(\theta\right)\sin\left(\psi\right)$$
$$z=\left(R-r\right)\cos \psi$$
Torus: $\theta, \psi$ $\in$ $[0,2\pi)$
$$x=\left(R+r\cos \psi\right)\cos\left(\theta\right)$$
$$y=\left(R+r\cos \psi\right)\sin\left(\theta\right)$$
$$z=r\sin \psi$$
- What must be the relationship between $R$ and $r$ in order for the surface areas of these surfaces to be equal?
If the surface areas are equal then $$4\pi\left(R-r\right)^{2}=4\pi^{2}Rr$$
Answer: $$R=\frac{\left(2+\pi\right)r+r\sqrt{\left(2+\pi\right)^{2}-4}}{2}$$ $$=\frac{2+\pi+\sqrt{\pi^{2}+4\pi}}{2}r$$
- What must be the relationship between $R$ and $r$ in order for the volumes enclosed by these surfaces to be equal?
Here is the progress I've made on question 2
If the enclosed volumes are equal then $$\frac{4}{3} π (R - r)^3 = 2 π^2 r^2 R$$
According to Wolfram, manipulating the above equation gives $$R=\frac{ar^2}{(r^3)^{1/3}} + r + b(r^3)^{1/3}$$
where $a≈1.02014$ and $b≈1.53979$
I know that in order to get the real solution we're after one must choose the principal branch of the multivalued function $x^{1/3}$ when evaluating $$R=\frac{ar^2}{(r^3)^{1/3}} + r + b(r^3)^{1/3}$$
So the next step should be to figure out how to go about deriving an analytic expression for each of $a$ and $b$
Thus, my next concern should be the real zero in $R$ in terms of $r$ of the following cubic polynomial:
$$R^{3}-3rR^{2}+\frac{6-3\pi}{2}r^{2}R-r^{3}$$
Is there a shortcut to finding the real zero of this cubic polynomial? Is there no alternative to using the cubic formula?