Talk:Paraboloid

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Field: Geometry

shrimp are tasty

so in nature the ONLY place where a hyperbolic paraboloid is found is a shrimp? I have trouble with this..... —Preceding unsigned comment added by 75.46.90.105 (talk) 20:56, 7 May 2010 (UTC)

well all i can say is that the parabaloid is the shap of a pringle,. pringles are copywrited so does that mean any one who uses a parabaloid is liable to be sued

there have to be other examples of objects with paraboloid than the mantis shrimp mentioned in "in nature"

different equation

Las monday on my algebra class we learned this equation for a paraboloid: x^2+y^2=z

what's up with that?--Fito 01:09, May 27, 2005 (UTC)

Sure it wasn't a parabola? If that's even the equation for a parabola. I'm too lazy to check.

It's a special elliptic paraboloid of revolution with a=b=1. You get it by rotating the parabola z = x^2 around the z axis.--Syd Henderson 05:04, 6 September 2006 (UTC)

"A daily life example of a hyperbolic paraboloid is the shape of a Pringles potato chip." -surely this sentence could be edited to not require a citation? For example "The Pringles potato chip gives a good phyisical approximation to the shape of a (truncated) hyperbolic paraboloid." This is a fairly self evident statement if you've read the introduction and have seen the picture of the paraboloid on the right.

128.250.24.104 (talk) 21:49, 17 December 2007 (UTC)

Idiosyncrasies

I have removed two of them and corrected some minor ones, but the whole sections on "curvature" and the "multiplication table" look very odd. The section on curvature simply states the Gaussian and mean curvatures of the elliptic and hyperbolic paraboloids, but for what purpose, it remains unclear. The other section is even stranger: the header is most confusing and the content can be reduced to two sentences and perhaps a displayed formula, at most. Arcfrk (talk) 01:45, 15 March 2008 (UTC)

Article seems generally poor

For instance the terms A and B in the very first equation are never defined, and it seems that you would have to be very well read in math to understand most of the article. A math prof should have a go at this. —Preceding unsigned comment added by TimGregg (talkcontribs) 23:33, 25 August 2009 (UTC)

I've added some clarification of the symbols in the first couple of equations. Does that help? LightYear (talk) 08:31, 10 September 2009 (UTC)

Paraboloid value proof

First need to divide paraboloid into 10 parts, which have equal height $h=0.1$. Each part is cylinder with height $h=0.1$. It is so, that paraboloid is divided on z axis, without touching any over axis (y or x). So z axis is divided with step 0.1 from 0 to 1, like this: 0.1; 0.2; 0.3; 0.4; 0.5; 0.6; 0.7; 0.8; 0.9; 1. So radius of each of 10 cylinders depending on certain z position, but paraboloid reminds us parabola, so we can take paraboloid projection onto zOy plane/surface and will see parabola $z=y^2$. So now radius of each cylinder depending what is $z_1=0.1$, $z_2=0.2$, $z_3=0.3$, ..., $z_{10}=1$. And radius of each cylinder is $r=\sqrt{z}$, because say at point $z=0.1$ radius (of smallest cylinder) is $r_1=y_1=\sqrt{z_1}=\sqrt{0.1}\approx 0.316$. So $r_2=\sqrt{0.2}$; $r_3=\sqrt{0.3}$; $r_4=\sqrt{0.4}$; $r_5=\sqrt{0.5}$; $r_6=\sqrt{0.6}$; $r_7=\sqrt{0.7}$; $r_8=0.8^{0.5}$; $r_9=0.9^{1/2}$; $r_{10}=\sqrt{1}=1.$ And cylinder value formula is $V_c=\pi r^2 h$. So $h=0.1$ for each cylinder and so each cylinder values are those:

$V_1=r_1^2\pi h=(\sqrt{0.1})^2\pi h=0.1\pi h=0.1\pi \cdot 0.1=0.01\pi;$
$V_2= r_2^2\pi h=(\sqrt{0.2})^2\pi h=0.2\pi 0.1=0.02\pi;$
$V_3=r_3^2\pi h=0.3\pi h=0.03\pi$; $V_4=0.3\pi h;$ $V_5=0.5\pi h$; $V_6=0.6\pi h;$ $V_7=0.7\pi h$; $V_8=0.8\pi h;$ $V_9=0.9\pi h,$ $V_{10}=\pi h.$
So paraboloid value up to $z=1$ is:
$V=\pi h(r_1+r_2+r_3+r_4+r_5+r_6+r_7+r_8+r_9+r_{10})=0.1\pi(0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9+1)=$
$=0.1\pi\cdot 5.5=0.55\pi\approx 1.727876.$
Precise (calculating with integrals method) paraboloid value of function $z=x^2+y^2$ up to $z=1$ is $V={\pi\over 2}=0.5\pi=1.5708$
Dividing into more parts/cilinders you can get paraboloid value arbitrary precise. —Preceding unsigned comment added by 212.59.24.214 (talk) 14:07, 11 February 2010 (UTC)
This thing $V= h(r_1+r_2+r_3+r_4+r_5+r_6+r_7+r_8+r_9+r_{10})=0.1(0.1+0.2+0.3+0.4+0.5+0.6+0.7+0.8+0.9+1)$ can be puted into integral, $r_{max}=z_{max}=z=1$ as $\int_0^1 z dz={z^2\over 2}|_0^1={1^2\over 2}-{0^2\over 2}={1\over 2}$ and this thing need just multiply by $\pi$ and then you can easily calculate paraboloid value if just know height z. $V=0.5\pi$.
For example if z=100, then if x=0, then y=10, because $z=y^2=10^2=100$ and paraboloid formula $z=x^2+y^2$ is like parabola $z=y^2$, only assume x=0. So then calculation is very simple:
$V=\pi\int_0^{100} z dz=\pi{z^2\over 2}|_0^{100}=\pi({100^2\over 2}-{0^2\over 2})=5000\pi.$ This very logical replacing next method of finding paraboloid value if z=100, $\phi=0$ and up to $\phi=2\pi$; $z=x^2+y^2$ <=> $z=\rho^2$, because $x^2+y^2=\rho^2$ in Cylindrical coordinate system, $\rho$ is also maximum possible radius of paraboloid, so value of paraboloid is:
$V=\iiint_V dxdydz=\iiint_T \rho d\rho d\phi dz=\int_0^{2\pi} d\phi\int_0^{10} \rho d\rho \int_{\rho^2}^{100} dz=$

$=\int_0^{2\pi} d\phi\int_0^{10} \rho (100-\rho^2)d\rho =\int_0^{2\pi} ({100\rho^2\over 2}-{\rho^4\over 4})|_0^{10} d\phi=\int_0^{2\pi} (50\cdot 10^2-{10^4\over 4}) d\phi=$ $=\int_0^{2\pi} (5000-2500) d\phi= 2500\int_0^{2\pi}d\phi=5000\pi.$

And if you going to calculate paraboloid $z=x^2+y^2$ value if z=9, then $z=\rho^2=9$, $\rho=3$; then you have to do all this cylindrical coordinates hell:
$V=\iiint_V dxdydz=\iiint_T \rho d\rho d\phi dz=\int_0^{2\pi} d\phi\int_0^3 \rho d\rho \int_{\rho^2}^9 dz=$

$=\int_0^{2\pi} d\phi\int_0^3 \rho (9-\rho^2)d\rho =\int_0^{2\pi} ({9\rho^2\over 2}-{\rho^4\over 4})|_0^3 d\phi=\int_0^{2\pi} ({9\cdot 3^2\over 2}-{3^4\over 4}) d\phi=$ $=\int_0^{2\pi} ({81\over 2}-{81\over 4}) d\phi=\int_0^{2\pi} {162-81\over 4} d\phi= {81\over 4}\int_0^{2\pi}d\phi={81\pi\over 2}.$

And with obvious method (because of this my proof) to calculate paraboloid $z=x^2+y^2$ value it will takes much less time. So z=9 and paraboloid value is:
$V=\pi\int_0^9 z \;dz=\pi{z^2\over 2}|_0^9=\pi({9^2\over 2}-{0^2\over 2})={81\pi\over 2}.$ —Preceding unsigned comment added by 212.59.24.214 (talk) 18:33, 11 February 2010 (UTC)
Here paraboloid value proof again if each cylinder height h=1 and paraboloid divided into 100 such cylinders; each cylinder value can be calculated $\pi r_n^2 h$. So on z axis there steps by 1, like $z_1=1$, $z_2=2$, $z_3=3$,..., $z_{99}$=99, $z_{100}=100$. We projecting paraboloid onto yOz plane, so then $z_n=y_n^2=r_n^2$, so $y_n$ is radius and $y_n=\sqrt{z_n}=r_n$. So total paraboloid value is:
$V=V_1+V_2+V_3+...+V_{99}+V_{100}=\pi r_1^2 h +h\pi r_2^2+h\pi r_3^2+h\pi r_4^2+...+h\pi r_{99}^2+h\pi r_{100}^2=$

$=h\pi((\sqrt{1})^2+(\sqrt{2})^2+(\sqrt{3})^2+(\sqrt{4})^2+...+(\sqrt{99})^2+(\sqrt{100})^2)=$ $=1\cdot\pi(1+2+3+4+...+98+99+100)=5050\pi.$

And for comparision if to divide only into 10 parts and if h=10 then, z=100 still, so then such answer: $V=10\cdot\pi(10+20+30+40+50+60+70+80+90+100)=5500\pi.$

You can imagine this sumations like dividing place under line y=x into peaces of width a=1 and height $b_n$, $b_1=1$, $b_2=2,$ $b_3=3$, ... , $b_{99}=99$, $b_{100}=100$. If you multiply each column height $b_{n}$ with width a=1, then you will get column area. If you sum up all 100 columns (1*1+1*2+1*3+...+1*98+1*99+1*100=5050), then you will get (approximate) area under line y=x and this line from 0 to 100 makes triangle. And this triangle area $S={100^2\over 2}=5000.$ By integrating you geting the same $\int_0^{100} x \; dx={x^2\over 2}|_0^{100}={100^2\over 2}-{0^2\over 2}=5000$. And by multiplying with $\pi$ you getting paraboloid value $V=5000\pi.$

None of this makes sense to a non-Maths person

I came to this page because it was linked to from the Pringles page. The shape of those crisps was described as being a hyperbolic paraboloid. I wondered what this means, so I clicked on the link and... Uh... That's where the comprehensible part of my story finishes. This page has absolutely no meaning whatsoever to someone who doesn't have advanced knowledge of Mathmatics. At least, I think that the knowledge is Maths based - it seems that I don't even know what I'd need to know to understand the info on this page. Basically, not very useful to anyone wondering what shape a Pringle is...86.166.154.225 (talk) 01:21, 9 October 2011 (UTC)

Well, what kind of understanding of the shape did you hope to get without the math? —Tamfang (talk) 02:50, 31 August 2013 (UTC)

Coordinate System

The article states: "In a suitable coordinate system with three axes x, y, and z"

Thus, a reader comes to this article to try model a paraboloid and is left wondering what the coordinate system is; ie the article is effectively useless. — Preceding unsigned comment added by 212.183.140.43 (talk) 11:06, 30 October 2011 (UTC)

A suitable coordinate system is one that allows the mathematical operations required for the discussion. In this case, one must be able to use Addition, Negation, Multiplication, Division, and Exponentiation operations. An angular axes system, Polar or Spherical for instance, does not allow the operation of multiplication or division with the same Translational meaning as in a Cartesian coordinate system.
Should "Suitable coordinate system" be created as an article?
SBaker43 (talk) 12:42, 13 February 2014 (UTC)

Criteria for stackability

One of two "applications" mentioned in the article is this:

"The manufacturers of a widely-sold fried snack food, Pringles ersatz potato chips, utilize a hyperbolic paraboloid shape in order to produce a stackable snack."

But being a hyperbolic paraboloid has virtually nothing to do with stackability. Any surface for the form z = f(x,y), i.e., that is the graph of a function, is stackable. That is, its vertical translates z = f(x,y) + c where c ranges over an interval [a,b] are disjoint surfaces that fill the solid region between z = f(x,y) + a and z = f(x,y) +b.Daqu (talk) 22:55, 16 November 2011 (UTC)

discrimination

For a general paraboloid of the form $z = ax^2+bxy+cx^2+dx+ey+f$, am I right in guessing that it's hyperbolic if $b^2 > 4ac$, elliptic if $b^2 < 4ac$, and a cylinder (ruled parallel to the xy plane) if they're equal? —Tamfang (talk) 05:28, 31 August 2013 (UTC)