How to find the intersection of 3d Graphs in the form $f(x,y)=x^2y^2$ & $x^2+4y^2=24$?

Question

I can't for the life of me figure it out. I tried equating them $f(x,y)=x^2y^2=x^2+4y^2=24$, but it doesn't get me any where. I can solve it for $R^2$ but can't when it is $R^3$. It feels like I should be able to do it and glaringly obvious, but I've spend 6hrs trying to recall how to do it and failed. Pls help:

Answer 1

With the help of MATHEMATICA

h = z - x^2 y^2;
g = x^2 + 4 y^2 - 24;
r = 6; 
ContourPlot3D[{h == 0, g == 0}, {x, -r, r}, {y, -r, r}, {z, 0, 36}, MeshFunctions -> {Function[{x, y, z, f}, h - g]}, MeshStyle -> {{Thick, Blue}}, Mesh -> {{0}}, ContourStyle -> Directive[Orange, Opacity[0.3], Specularity[White, 30]]]

Hint.

By solving

$$ \cases{z=a b\\ a+4b=24} $$

we obtain

$$ \cases{a= 2 \left(6-\sqrt{6^2-z}\right),b= \frac{1}{2} \left(\sqrt{6^2-z}+6\right)\\ a=2 \left(\sqrt{6^2-z}+6\right),b= \frac{1}{2} \left(6-\sqrt{6^2-z}\right) } $$

Clear[z, a, b]
sols = Solve[{z == a b, a + 4 b == 24}, {a, b}]
{a, b} = ({a, b} /. sols[[1]])
gr1 = ParametricPlot3D[{Sqrt[a], Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
gr2 = ParametricPlot3D[{-Sqrt[a], Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
gr3 = ParametricPlot3D[{Sqrt[a], -Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
gr4 = ParametricPlot3D[{-Sqrt[a], -Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
{a, b} = ({a, b} /. sols[[2]]);
gr1a = ParametricPlot3D[{Sqrt[a], Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
gr2a = ParametricPlot3D[{-Sqrt[a], Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
gr3a = ParametricPlot3D[{Sqrt[a], -Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
gr4a = ParametricPlot3D[{-Sqrt[a], -Sqrt[b], z}, {z, 0, 36}, PlotStyle -> {Thick, Blue}];
Show[gr1, gr2, gr3, gr4, gr1a, gr2a, gr3a, gr4a]

Answer 2

We want to have a representation of the curve $z=x^2 y^2$ where $x^2+4y^2=24$.

So we could parameterize the curve on $y$:
$y \in [-\sqrt 6, \sqrt 6]$
$x^2 = 24-4y^2$
$z=(24-4y^2)y^2$

We can also parameterize in $\theta$ (beware, $\theta$ is not the angle made by $(x,y)$):
$\theta \in [0, 2\pi[$
$x = 2 \sqrt 6 \cos \theta$
$y = \sqrt 6 \sin \theta$
$z = 144 \cos^2 \theta \sin^2 \theta = 36 \sin^2(2\theta) = 18 (1-\cos(4\theta))$
This explains the $4$ oscillations of $z$ when $x$ and $y$ do a full turn around their ellipse.
On Wolfram Alpha