I've been struggling on this question for quite some time now and I cannot seem to find anything online to assist.
I've been given the parametrisation:
$p(u,v)=\left (\cos u \cos v, \frac{1}{3}\sin u \cos v, \ \frac{1}{2}\sin v \right )$,
$\pi <u< \pi$,
$-\frac{\pi}{2}<v<\frac{\pi}{2}$.
The question asks to find the set of points $(x,y,z)$ satisfying $x^2+9y^2+4z^2=1$ which do not lie on the image of $p(u,v)$.
I know this is a straightforward question but I'm not the brightest of individuals and I'm struggling to work it out. I suspect it has something to do with $u,v$ belonging to open intervals. Is there an easy/methodical way to approach this question?
the answer is: $\left \{(x,y,z):x^2+9y^2+4z^2=1, x \leq 0, y=0 \right \}$
I am sorry to ask such a trivial question but it's been puzzling me for a while now and I'm not making any progress. Many Thanks :)
Your answer is correct.
Geometrically, if $u$ and $v$ are real, then $$ (x, y, z) = (\cos u \cos v, \sin u \cos v, \sin v) $$ is the point on the unit sphere $x^{2} + y^{2} + z^{2} = 1$ with longitude $u$ and latitude $v$ (in radians). The restrictions $-\pi < u < \pi$ and $-\pi/2 < v < \pi/2$ omit the open half-circle where longitude is $\pi$, and the two poles. To handle your question, scale this picture by $\frac{1}{3}$ in the $y$-direction and by $\frac{1}{2}$ in the $z$-direction.
If you need to be more rigorous than this, I'd suggest letting $U$ denote the sphere with the closed half-circle $u = \pi$ (including the poles) removed, defining $q:U \to (-\pi, \pi) \times (-\pi/2, \pi/2)$ by $$ q(x, y, z) = (\operatorname{atan2}(y, x), \arcsin z), $$ (with $\operatorname{atan2}$ the polar angle function defined on the plane with the non-positive $x$-axis removed and taking values in $(-\pi, \pi)$), and showing/observing that $q$ is a two-sided functional inverse of $p$.