Here is the attached problem: [![enter image description here][1]][1]
I have solved part a) and can do c and other also..
But for that I need to solve b) this is where I got stuck. I don't know how to find a parametric surface S from curve C. I think we just need to find the domain of $\phi$ and $\theta$ If I am right then I don't know how to find!
If I am wrong then please let me know how to find it and I don't think so I am able to understand the hint given.
Hint: The parametric curve $C$ is the boundary of $S$, so the boundary of the region in the $\phi\theta$-plane gets mapped to $C$.
Details: I'll assume that the $\phi\theta$-plane here just means the box $[-\tfrac\pi2,\tfrac\pi2]^2$. To find the preimage of the parametrized curve $C$ in the $\phi\theta$-plane let $t,\phi,\theta\in[-\tfrac\pi2,\tfrac\pi2]$ be such that $r(t)=r(\phi,\theta)$. Then $$\cos t=\cos\phi,$$ so $\phi=\pm t$. Then also $\sin\phi=\pm\sin t$ and hence the equality $$\sin t\cos t=\sin\phi\cos\theta,$$ implies that either $t=0$ or $\cos\theta=\pm\cos t$. If $t=0$ then $\phi=\pm t=0$ and then it is easy to verify that for every $\theta\in[-\tfrac\pi2,\tfrac\pi2]$ we have $r(0)=r(0,\theta)$. Otherwise, if $t\neq0$ then $\cos\theta=\pm\cos t$. Of course $\cos\theta\geq0$ for all $\theta\in[-\tfrac\pi2,\tfrac\pi2]$ so we have $\cos\theta=\cos t$ and hence $\theta=\pm t$. So every point in the preimage is certainly of the form $$(\phi,\theta)=(t,\pm t)\qquad\text{ or }\qquad(\phi,\theta)=(0,t),$$ for some $t\in[\tfrac\pi2,\tfrac\pi2]$. But we haven't considered the ${\bf j}$-coordinate yet; we must have $$\sin t\sin t=\sin\phi\sin\theta=\sin t\sin\pm t,$$ and hence if $\phi\neq0$ we must have the $+$-sign (or $t=0$). So the preimage of $C$ in the $\phi\theta$-plane is the union of thet $\theta$-axis and the diagonal; all points of the form $(0,t)$ or $(t,t)$ with $t\in[-\tfrac\pi2,\tfrac\pi2]$.
This boundary divides the box $[-\tfrac\pi2,\tfrac\pi2]^2$ into four regions, that come in two opposing pairs. This boundary is the boundary of each of these two pairs. I leave it to you to figure out which pair maps to the interior $S$, and which pair to the exterior.