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What is the equation for a cylinder SOLVED FOR Z

  1. Apr 21, 2013 #1
    I've tried googling this but I can't find the answer. I'd like all of the common 3d shape equations if someone has them. THEY HAVE TO BE SOLVED FOR Z so they can actually be graphed on a 3d graphing calc. For some crazy, crazy reason, these equations are never listed solved for z or, don't even have z in them.

    The only one I've been able to figure out, with a lot of trouble was the equation for a bowl which is z=[itex]\frac{1}{radius}[/itex](x2+y2)
     
    Last edited: Apr 21, 2013
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  3. Apr 21, 2013 #2

    cepheid

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    Part of the reason why nobody may have answered up to now is that they looked at your thread title and saw "SOLVED" in all caps, and they assumed you had solved the problem for yourself :tongue:. That is usually what that means on the forums.

    The equation for a 2D surface in 3D space is, of course, going to depend on what coordinate system you use in that space. For a cylinder, it's MUCH easier to work in a cylindrical coordinate system, where you have (r, θ, z), rather than (x, y, z). The coordinates (r, θ) are the standard polar coordinates that you can use in a 2D plane, and the z coordinate changes which 2D plane you're talking about. So, for an infinite circular cylinder of radius R oriented along the z-axis, the equation for a cylinder is just going to be r = R. The angle θ can vary to be whatever it wants to be, and so can the z-coordinate. The cylindrical surface is the locus (collection) of all points in space with the same r-coordinate. Some plotting programs allow you to specify equations for plots in 3D polar coordinate systems (e.g. spherical, cylindrical). For example, in Grapher, in Mac OS X, if I plot r = 3, it gives me a sphere, and if I type r0 = 3, it gives me a cylinder. This is because "r" is the symbol that the program has reserved for the radial coordinate in a spherical coordinate system, and r0 is the symbol it has reserved for the radial coordinate in a cylindrical coordinate system. It even lets you write equations in implicit form, so that you don't have to solve for one variable in terms of the others.

    Maybe your plotting program doesn't have this feature, and so you want to use a Cartesian (rectangular) coordinate system so that you can write equations in terms of x, y and z. In that case, you just have to think about it a little bit. What is the intersection of the cylinder with any flat plane corresponding to z = constant? Why, it's just a 1D curve (a circle) of radius R in that plane. So, the equation for that circle is going to be ##\sqrt{x^2 + y^2} = R##. Now imagine "extruding" that circle into a cylinder along the z-axis by letting z vary. So, the equation for a cylinder in Cartesian coordinates isn't going to have z in it, because z is a free parameter. It can vary to be whatever it wants to be. The points are constrained to lie on the 2D surface defined by the equation above. So the equation for a cylinder along the z-axis is indeed just ##\sqrt{x^2 + y^2} = R##.
     
    Last edited: Apr 21, 2013
  4. Apr 21, 2013 #3
    It must be possible to graph a cylinder in Cartesian coordinates..

    I've taken calc 3. I'm familiar with the other coordinate systems and their benefits. I literally want what I was asking for which is the equations, in rectangular form (hence the "entered into a graphing calc" statement), for all of the common solids (cylinders, spheres, cones, bowls, etc.). It seems to be super secret information for some reason. You would think that all of these equations, in rectangular form, solved for z, would be listed somewhere but I haven't been able to find them..
     
    Last edited: Apr 21, 2013
  5. Apr 21, 2013 #4

    LCKurtz

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    Most of the conic surfaces have standard forms for their equations. These forms are generally not solved for any particular variable. That is because the standard form conveys much useful information at a glance. If you have had calc III, you certainly should be able to solve any such equation for ##z## yourself because they are only second degree. But any surface that is a cylindrical surface with axis parallel to the ##z## axis is going to have the ##z## variable missing, so it can't be put in the form ##z = f(x,y)##.

    My suggestion to you would be to become more familiar with the standard forms and what they tell you and less dependent on your graphing calculator.
     
  6. Apr 21, 2013 #5
    ok so center it on the y or x axis instead. Then you will have two equations which combine to graph a horizontal cylinder. What would that equation be?
     
  7. Apr 21, 2013 #6

    LCKurtz

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    If you are replying to someone, quote them so they know it.
     
  8. Apr 21, 2013 #7

    cepheid

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    I don't get it. I already told you how to graph the cylinder in Cartesian coordinates in my first post (third paragraph). Did you read it? A 2D surface in 3D space is necessarily going to have one free (unconstrained) parameter, and in this case that's z, which is why, as LCKurtz said, the equation for the cylinder can't be put in the form z = f(x,y).

    Again, my first post (third paragraph) already gives all the information needed to answer this question. You just have to think about it a bit. In each plane of constant 'blah' (where blah = x, y, or z), you have a circle. What's the equation for that circle?
     
  9. Apr 21, 2013 #8
    (x−h)2+(y−k)2=r2

    That's the equation of a circle and it is of no use in answering my question.

    Let me explain it another way. Lets way you had to make a cylinder show up on a 3d graphing calc that only uses Cartesian coordinates. Ignore everything else and simply accomplish the goal. z=something. Your first step of noticing that it can't be graphed centered on the z axis is no big deal. You simply give another equation that centers it along the y or x axis instead. What would the equations be?
     
  10. Apr 21, 2013 #9

    LCKurtz

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    I get the impression you don't try anything for yourself. So change your equation above to$$
    (x-h)^2 + (z-k)^2 = r^2$$There, that's a different axis. Solve it for ##z##. What is so difficult about that?
     
  11. Apr 21, 2013 #10

    cepheid

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    No, YOU are the one who is not answering MY questions. So the cylinder is centered on the y-axis or x-axis instead. That means the circles are no longer in planes parallel to the x-y plane, are they? They're in some other orthogonal set of planes. What is the equation for THESE circles? That, plus the description in my first post tells you EVERYTHING you need to know to get the equation of these cylinders. You have to do some thinking for yourself, man. We're not going to spoon-feed you information, esp. when we've already told you everything you need to know to do this.
     
  12. Apr 21, 2013 #11
    For anyone's future reference, here's the answer for a cylinder:
    Centered on the x axis:
    z=[itex]\pm[/itex] [itex]\sqrt{radius2-y2}[/itex]
    Centered on the y axis:
    z=[itex]\pm[/itex] [itex]\sqrt{radius2-x2}[/itex]

    Now we just need the other ones..
     
  13. Apr 21, 2013 #12
    I just wanted a quick reference of all of the solids, solved for z in Cartesian coordinates, not a lesson on learning them again. There is no long-term benefit to re-learning these equations. I will forget them again anyway. I just need a reference. I'll eventually have them listed and then some other lucky guy will be able to come along and go, "ah, here they are" like I should've been able to do had they already existed.
     
  14. Apr 21, 2013 #13

    cepheid

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    Well that doesn't make any sense to me at all, and it misses the point I was trying to make. If you really know math at the Calc III level, then you should be able to just sit down and write down these equations after no more than a minute of thinking about it. There is no need to "re-learn" or "memorize" anything, and there is certainly no need to look it up in some reference table.
     
  15. Apr 21, 2013 #14

    SammyS

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    Previously you have stated that those are merely equations for circles. In fact, they're also equations for cylinders of infinite height. They have no base.

    I assume you are looking for equation(s) to describe the surface of a finite cylinder. -- more specifically, a finite right circular cylinder.


    By the way, there is no need to chop-up your LaTeX expressions.
    can be displayed as a single LaTeX expression.
    [itex]z=\pm\,\sqrt{\text{radius}^2-y^2}[/itex]​
    More importantly: Don't use the X2 superscript feature inside or LaTeX tags. Use "^" for superscripts, "_" for subscripts, etc.
     
  16. Apr 21, 2013 #15
    Depends on you and your instructor I guess. I just finished DiffEq this semester. Made A's in all of my math classes since the beginning of college. If you ask me anything from those classes I will have to look it up though. The only benefit the class gave me is that I know it exists and I can better understand it when I reference it again. Can I get a complete reference for the solids or a link to one?
     
  17. Apr 21, 2013 #16
    ok, so for the people reading this in the future, the actual useful info so far is;

    Cylinder:
    [itex]z=\pm\,\sqrt{\text{radius}^2-y^2}[/itex]
    Sphere:
    [itex]z=\pm\,\sqrt{\text{radius}^2-x^2-y^2}[/itex]
    Bowl:
    [itex]z=\frac{1}{radius}[/itex](x2+y2)
    Cone:
    [itex]z=\frac{1}{radius}\sqrt{x^2+y^2}[/itex]

    That pretty much covers it I think.
     
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