I have to parametrize the lateral section of the cylinder $\frac{x^2}{4} + \frac{y^2}{9}=1$ between the planes $z = 1-x$ and $z = 0$.
I have $r(u,v) = (2\cos{uv},3\sin{uv},\frac{3v}{2} + \frac{3}{2})$ for $\frac{\pi}{3}\leq u \leq \frac{\pi}{2}$ and $-1 \leq v \leq 1$.
However, when I plug this in Mathematica as
ParametricPlot3D[{2*Cos[u*v],3*Sin[u*v],v*3/2 +3/2},{u,Pi/3,Pi/2},{v,-1,1}]
It produces something completely different to a portion of a cylinder, which makes me think my parametrization has a glaring error I'm not seeing.
Best Answer
Step 1: Parametrizing the elliptical cylinder.
The equation $\tfrac{x^2}{4}+\tfrac{y^2}{9}=1$ can be rewritten $\left(\tfrac{x}{2}\right)^2+\left(\tfrac{y}{3}\right)^2 = 1$. From the main goniometric property $\cos^2 u + \sin^2 u = 1$, we find that $$ \begin{align*} x &= 2 \cos u \\ y &= 3 \sin u \\ z &= v. \end{align*} $$ Note that the $z$-coordinate is a free variable, since $z$ doesn't occur in the equation of the cylinder.
Step 2: Finding the boundaries of $u$ and $v$.
Since we are going all the way around on the cylinder, $u\in [0, 2\pi]$. Note that the cylinder lies between two planes. Algebraically this gives $$ 0 \leq z \leq 1- x, \quad \text{ or } \quad 1 - x \leq z \leq 0.$$ The first inequality holds for points where $1-x \geq 0$, the second inequality when $1-x \leq 0$. One should consider these two cases separately.
Case 1. $1-x \geq 0$. In this case $1-2\cos u\geq 0$, which happens when $u\in[0,\frac{\pi}{3}]\cup[\frac{5\pi}{3}, 2\pi]$. Then $0 \leq v \leq 1- 2 \cos u$.
Case 2. $1-x \leq 0$. In this case $1-2\cos u\leq 0$, which happens as $u\in[\frac{\pi}{3}, \frac{5\pi}{3}]$. Then $ 1- 2 \cos u \leq v \leq 0$.