parametrize surface region

Question

S is the elliptic region of the plane $y+z=1$ inside the cylinder $4x^2+4(y-0.5)^2=1$. First parametrize $S$ using $(x,y,z)=G(u,v)$ and then calculate $\displaystyle \frac{dG}{du}\times \frac{dG}{dv}$.

How can I parametrize this crazy region??

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} x&\equiv \half\,\cos\pars{\theta}\,,\quad y \equiv \half + \half\,\sin\pars{\theta}\,,\quad z = 1 - y = \half - \half\,\sin\pars{\theta}\,;\qquad\qquad 0 \leq \theta < 2\pi \end{align}

$$ \vec{\rm r}\pars{\theta} =\half\braces{% \cos\pars{\theta}\,\hat{x} + \bracks{1 + \sin\pars{\theta}}\,\hat{y} + \bracks{1 - \sin\pars{\theta}}\,\hat{z}} $$

Answer 2

Perhaps an additional question: Find the centers of the Dandelin Spheres associated with this construction. Their points of tangency with the plane are the foci of the ellipse.

The parametrization : $ x = u/2 \cos(v) , y = 1/2 + u/2 \sin(v) , z = 1/2 - u/2 \sin(v)$ with $ v =[0,2\pi)$ and $ u = [0,1]$ seems to work.