[Math] Parametrization of the intersection of a cone and plane.

multivariable-calculusparametric

EDITED with new progress updates.

As the title states, I'm trying to parametrize the intersection of a cone and a plane. The equations are:

$z^2 = 2x^2+2y^2$ and

$2x+y+3z=4\implies z=\frac{1}{3}(4-2x-y)$

If it were either a plane parallel to the x-y plane or a cylinder instead of a cone, I would simply project the intersection onto the x-y plane, parametrize the projection for x and y then insert those equations into the equation of the plane to get the parametrization of z. I'm thinking a similar approach would work here but I can't figure out how to find the equation of the projection.

I now have what I believe is the projected ellipse onto the x-y plane. The ellipse equation is:

$0 = \frac{1}{9}(4-2x-y)^2-2x^2-2y^2$

My problem is now that if I expand out the squared portion, I end up with $4xy$ in the equation. I have no idea how to parametrize an ellipse with a cross term in it. Could someone help with that?

Any help would be greatly appreciated.

Thank you,
Eric

Best Answer

Hint: eliminate one of the variables by solving the plane equation for it and substituting into the cone equation.

EDIT: For an ellipse equation with "cross terms", one thing you can do is "complete the square" in either the $x$ or $y$, so it becomes something like $a (x + b y)^2 + c y^2 = 1$ with $a, c > 0$. Then you can parametrize it as $$\eqalign{y &= \dfrac{\sin(t)}{\sqrt{c}}\cr x &= -b y + \dfrac{\cos(t)}{\sqrt{a}} = - \dfrac{b \sin(t)}{\sqrt{c}} + \dfrac{\cos(t)}{\sqrt{a}}\cr}$$

Related Solutions

Triple integral (intersection of cone and plane)

Given the set:

$$ \Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : \sqrt{2\,x^2+2\,y^2} \le z \le y+1 \right\} $$

it follows that:

$$ I := \iiint\limits_{\Omega} z\,\text{d}x\,\text{d}y\,\text{d}z = \iint\limits_D \frac{(y + 1)^2 - \left(2\,x^2+2\,y^2\right)}{2}\,\text{d}x\,\text{d}y $$

where:

$$ \small D := \left\{ (x,\,y) \in \mathbb{R}^2 : \sqrt{2\,x^2+2\,y^2} \le y+1 \right\} = \left\{ (x,\,y) \in \mathbb{R}^2 : \frac{(x - 0)^2}{(1)^2} + \frac{(y - 1)^2}{\left(\sqrt{2}\right)^2} \le 1 \right\}. $$

So, opting for the following coordinate transformation:

$$ \Phi : \begin{cases} x = 0 + 1\,\rho\,\cos\theta \\ y = 1 + \sqrt{2}\,\rho\,\sin\theta \end{cases} \; \; \; \; \; \; \text{with} \; (\rho,\,\theta) \in [0,\,1] \times [0,\,2\pi] $$

and Jacobian $J_{\Phi} = 1 \cdot \sqrt{2} \cdot \rho$, all this translates into the following calculation:

$$ \small I = \int_0^{2\pi} \int_0^1 \frac{\left(2+\sqrt{2}\,\rho\,\sin\theta\right)^2 - 2\,\left(0+1\,\rho\,\cos\theta\right)^2 - 2\,\left(1+\sqrt{2}\,\rho\,\sin\theta\right)^2}{2}\,\sqrt{2}\,\rho\,\text{d}\rho\,\text{d}\theta = \frac{\pi}{\sqrt{2}}\,. $$

Finally, wanting to verify the result obtained in Wolfram Mathematica:

Ω = ImplicitRegion[Sqrt[2 x^2 + 2 y^2] <= z <= y + 1, {x, y, z}];
Integrate[z, {x, y, z} ∈ Ω] // TraditionalForm

π/√2

Showing that the intersection of a cylinder and a plane is an ellipse

Sometimes a figure really helps craft the algebra:

Best Answer

Related Solutions

Triple integral (intersection of cone and plane)

Showing that the intersection of a cylinder and a plane is an ellipse

Related Question