How can I parametrize the intersection curve of cone $z=\sqrt{(x^2+y^2)}$ and plane $z=1+x+y$

Question

Usually, I equalize both equation, then complete the square in order to find something that looks like $\cos^2 t+\sin^2 t=1$ to get a value of $x(t)$ and $y(t)$. I then substitute to get $z(t)$. I am unable to apply that technique with this question. I think it has something to do with the square root and the fact that the intersection curve is split in two.

Answer 1

We have $z^2=x^2+y^2$ and $z^2=(1+x+y)^2$, which gives $$x^2+y^2=1+2x+2y+2xy+x^2+y^2$$ which is equivalent to $1+2x+y(2+2x)=0$.

Now we can set $x=t$ and $y=-\frac{1+2x}{2+2x}=-\frac{1+2t}{2+2t}$ and $z=1+x+y=1+t-\frac12\frac{1+2t}{1+t}$.

Furthermore, $z=\sqrt{x^2+y^2}$ gives $z>0$, removing the problem that the curve is split in two. This means that we need $t>-1$.

Answer 2

You can certainly convert to polar/cylindrical coordinates right away. Letting $x=r\cos(\theta)$ and $y=r\sin(\theta)$, where the plane and cone meet we have

$$z=\sqrt{x^2+y^2}=1+x+y \iff \sqrt{r^2\cos^2(\theta)+r^2\sin^2(\theta)} = 1+r\cos(\theta)+r\sin(\theta)$$

Solve this for $r$ and you'll get your parameterization. See the figure for a sketch of the curve (red) showing where the two surfaces (cone in orange, plane in blue) meet. Shown in the $x,y$ plane is the curve $\sqrt{x^2+y^2}=1+x+y$.