How to find parametric equation of the intersection $x²+y²+z²=2$ and $y=x$?

Question

I know that if I substitute $y=x$ into

$x²+y²+z²=2$

I get

$$2x²+z²=2$$

which in some way gives me

$$x²+\frac{z²}{2} = 1$$

which is an ellipse.

My parametric equation goes from $(0,0,\sqrt{2})$ to $(1,1,0)$, so I need to find an equation that follows this path.

My idea was to take $x=t$ as a paremeter, and then I'd have:

$\frac{z²}{2} = 1-x² \implies z² = 2(1-x²) \implies z = \sqrt{2}\sqrt{1-x²}$

(since $x\ge 0, y\ge 0, z\ge 0$ in my exercise)

So I end up with the parametrization

$$(t,t,\sqrt{2}\sqrt{1-t²})$$

But $t$ goes from what to what? Also, this will only work if $z\ge 0$. My teacher made something in the class that used $\sin(t)$ and $\cos(t)$ in the equation, how do I get this? Do I need to use spherical coordinates? Because she didn't use and I don't know how to do it.

Answer 1

That parameterization will never get negative $z$. Your $t$ can be as least $-1$ and at most $1$, but it only parameterizes half of the intersection.

To get a complete parameterization, note that $\left(x,\frac{z}{\sqrt{2}}\right)$ is on the unit circle, so write $x=\cos\theta, z=\sqrt{2}\sin\theta$. Then $y=x$. Now what is the range of $\theta$ to parameterize the circle?

Answer 2

The intersection is a planar ellipse in the plane $y=x$. Note that $x^2+y+^2+z^2=2$ is the equation of a sphere in $3$-space. Try the normal parametrization by spherical coordinates.

Answer 3

Parametric equations of an ellipse $$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1 $$ are given by $$ x=a\cos t, \quad y=b\sin t, \quad 0\le t \le 2\pi $$

In your case, the projection of your curve in the $xz$ plane is the ellipse $$ {x}^2+\left(\frac{z}{\sqrt{2}}\right)^2=1, $$

so you can write $$ x=\sin t, \quad z=\sqrt{2}\cos t, \quad 0\le t \le 2\pi $$

And for $y$, you can use the fact that $y=x=\sin t$.

Since you only want to go from $(0,0,\sqrt{2})$ to $(1,1,0)$, you can restrict $t$ from $0$ to $\pi/2$.