Courant John, Introduction To Calculus and Analysis, II/1. Exercises 4.8, 4(a)
Show that if any closed curve (in polar coordinates $r,\theta,\phi$): $\theta=f(\phi)$ is drawn on the surface $r^2=a^2\cos(2\theta)$, the area of the surface so enclosed is equal to the area enclosed by the projection of the curve on the sphere $r=a$, the origin of coordinates being the vertex of projection.
So the projections sends $(a^2\cos(2f(\phi)),f(\phi),\phi)$ to $(a,\;\dfrac{f(\phi)}{a\cos(2f(\phi))}\;,\;\dfrac{\phi}{a\cos(2f(\phi))})$
If I write the projection of the curve as $\theta=g(\phi)$, then $g(\phi)=\dfrac{f(\phi a\cos(2f(\phi)))}{a\cos(2f(\phi a\cos(2f(\phi))))}$
When I differentiate this function with respect to $\phi$, I obtain an extremely disgusting and lengthy expression which depends on f and f'. I'll omit it for now. I have no idea what to do next. Any advice?