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 March 16th, 2009, 02:47 PM #1 Member   Joined: Mar 2009 From: San Bernardino, California Posts: 50 Thanks: 0 Three Dimensional Surface trick? Right now I am reviewing for a midterm and trying to memorize the equations for the quadric surfaces like the ellipsoid, elliptic cone, and parabolic cylinder but I also recall my professor mentioning a "math trick" that can give you the equation of the three dimensional surface given the equation of its two dimensional xy-plane based counterparts (eg. use the equation of the ellipse to find the equation of an ellipsoid). But unfortunately I can not find it in my notes anywhere. The only thing I can remember from it is something along the lines of replacing x with $\sqrt{x^2+y^2}$ or something like that. Does anyone have an idea of what the trick is? I'd appreciate the help if anyone knows it so that I can just derive these surface formulas rather than memorizing them.
 March 17th, 2009, 12:23 AM #2 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Three Dimensional Surface trick? It sounds like you're talking about a surface of revolution. This can be found by revolving a closed curve and considering the surface swept out. If you're more specific about what you want, we can be more helpful.
 March 17th, 2009, 12:42 AM #3 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Three Dimensional Surface trick? Given an even function y=f(x), you can sometimes quickly make a surface of revolution by replacing y with z, and x with polar r = $\sqrt{x^2+y^2}$. Parabola: $y= x^2$ Elliptical (well, circular) paraboloid: $z= \left(\sqrt{x^2+y^2}\right)^2 = x^2+y^2$ Hyperbola: $y^2= x^2-1$ Hyperboloid of one sheet: $z^2= x^2 + y^2 - 1$ I don't see how this is easier than taking cross-sections though. Set x, y, z to be zero in turn and see what shape you have.
 March 17th, 2009, 11:21 AM #4 Member   Joined: Mar 2009 From: San Bernardino, California Posts: 50 Thanks: 0 Re: Three Dimensional Surface trick? That's exactly what my teacher proposed. As for setting x, y, and z equal to zero, doesn't that just give you the "cylinder" with the base in the shape of the cross section? Or am I missing the point you are trying to make?
 March 17th, 2009, 12:44 PM #5 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Three Dimensional Surface trick? Setting x, y, and z to zero 'in turn' would produce cross sections in the yz, xz, and xy planes respectively.
 March 17th, 2009, 01:21 PM #6 Member   Joined: Mar 2009 From: San Bernardino, California Posts: 50 Thanks: 0 Re: Three Dimensional Surface trick? Well yes I see that the cross section in the coordinate planes would constitute the the equation of the surface with the "missing" variable in relation to the particular coordinate plane but I thought that in three-dimensional space, that those particular cross section equations from 2D represent a cylindrical shape based off of that cross section? For example the equation of an ellipse in the xy-plane (your standard ellipse equation) in three dimensional space represents an elliptical cylinder because with z missing, z varies indefinitely and the shape is technically an ellipse "dragged" up and down the z-axis. Please keep helping me, I really appreciate it I am really trying to get 3D space correct in my head.
 March 17th, 2009, 01:25 PM #7 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Three Dimensional Surface trick? This is why knowing more than one of the cross-sections can be helpful. One possibility for generating a 3 dimensional surface from an ellipse would indeed be to 'drag it up'. You could, instead, spin it around an axis, and this would produce an ellipsoid. Imagine a spinning coin; with a circular cross section, it produces a sphere by revolution.
 March 17th, 2009, 01:48 PM #8 Member   Joined: Mar 2009 From: San Bernardino, California Posts: 50 Thanks: 0 Re: Three Dimensional Surface trick? So correct me if I am wrong but: Elliptic Cylinder: $(\frac{x}{a})^2+(\frac{y}{b})^2=1$, because z varies indefinitely and is therefore and ellipse "dragged" along the z-axis. Ellipsoid: $z=(\frac{x}{a})^2+(\frac{y}{b})^2-1$, because an ellipsoid is and ellipse rotated in space whose size depends on x and y and there fore the size of z depends on x and y. Assuming those are correct thus far, then what is the three-dimensional equation of the cross section of each of these with a plane, an ellipse??? In the case that z is not present (elliptic cylinder) then it must vary indefinitely and therefore form a cylinder with the base of the cross section. In the case that z is limited in terms of x and y (ellipsoid) then it forms a surface analog of the cross section. My guess is that since z=0 for an xy cross section then the equation of an ellipse is $(\frac{x}{a})^2+(\frac{y}{b})^2=1$ but that is an elliptic cylinder.
 March 17th, 2009, 02:38 PM #9 Senior Member   Joined: Jan 2009 From: Japan Posts: 192 Thanks: 0 Re: Three Dimensional Surface trick? You're talking about two different types of equations here and not making the distinction clear. There's a qualitative difference between "for all z, $\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2= 1$" and "for z = 0" the same equation. The first is a relationship on $\mathbb{R} \times \mathbb{R} \times \mathbb{R}$ and the second on $\mathbb{R} \times \mathbb{R}$. The two-dimensional equivalent would be saying that $y= 1 - x^2$ is a horizontal line since at x = 0, y = 1. Given only "when x = 0, y = 1", there are an uncountably infinite number of functions which fulfill this condition. An ellipsoid centered at the origin will have elliptical cross-sections in the xy, xz, and yz planes. A cylinder with an elliptical base centered at the origin in the xy plane will have two parallel lines as a cross-section in the xz and yz planes.
 March 17th, 2009, 03:24 PM #10 Member   Joined: Mar 2009 From: San Bernardino, California Posts: 50 Thanks: 0 Re: Three Dimensional Surface trick? Ah wow thank you so much things are now insta-clear.

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