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So I have to determine and draw the surfaces $$z-2x^2-4y^2 ≥0,\qquad \mbox{and}\qquad 4y^2-x^2+4z^2-1 ≥0$$ so the first one in my opinion should be transformed like this $$z ≥2x^2+4y^2$$ then we multiply by two both sides and we have $$2z ≥x^2/(1/4) +y^2/(1/8)$$ Tadaa! This happens to be elliptic paraboloid ( the form) ...

but how do I draw $$2z ≥x^2/(1/4) +y^2/(1/8) $$

I mean, the problem is how does ≥ affect the drawing? How about the second one? I thought it was an ellipsoid but I cant transform it..

yohBS
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mathlover484
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1 Answers1

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You correctly observed that a first step could be to draw the surfaces for the equalities first and then tackle the inequalities afterwards.

In the first case you have an elliptic paraboloid, the second one is a hyperboloid of one sheet. To visualise them it might be helpful to plug the equalities into Wolfram alpha.

Now to the inequalities: Try to fix one of the coordinates and look what the inequality changes:

For the paraboloid: For a fixed $z$ and equality you have an ellipse $z=2x^2+4y^2$. Now if you change to inequality $z\geq2x^2+4y^2$ you allow all smaller (in the right sense) $(x,y)$, i.e. you fill the ellipse. In three dimesnions you will get the filled paraboloid (in other words, also all points above the paraboloid).

For the hyperboloid: Fix $x$. Again you get an ellipse for equality (in fact a circle) and it is filled up for inequality. So in three dimensions the hyperboloid will be filled as well (all points inside the hyperboloid).

Simon Markett
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