Why is the 3D plot of $ \vec{r}=\big{\langle} \pm 2 \sqrt{1 - \frac{v^2}{9}}, v, u + 3v \big{\rangle} $ shown as a surface and not a curve?

Question

I have a plane $$z=x+3y$$ and an elliptical cylinder $$\frac{x^2}{4} + \frac{y^2}{9}=1$$ and I want to plot their intersection.

I have $$x = \pm 2 \sqrt{1 - \frac{v^2}{9}}$$ and a parametric equation

$$ \vec{r}=\big{\langle} \pm 2 \sqrt{1 - \frac{v^2}{9}}, v, u + 3v \big{\rangle} $$

If I plot this in maple I get a surface, but I am expecting a curve:

plot3d([[-2*sqrt(1 - v^2/9), v, u + 3*v], [2*sqrt(1 - v^2/9), v, u + 3*v]], u = 0 .. 4, v = -3 .. 3)

I don't see why this shape has any vertical (z-dimension) length.

Answer 1

When you have two parameters, i.e. $u$ and $v$, you should expect a surface ($n$ parameters $\Rightarrow$ $n$ dimensions). So the problem is your formula – you shouldn't introduce $u$ at all. What you do with $x$ and $y$ is good, but then you should get $$ z = x+3y = \pm 2\sqrt{1-\frac{v^2}{9}} + 3v. $$