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Finding the equation of a 3 dimensional surface

  1. Jan 23, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Find the equation of the surface generated by rotating the line x = 9y about the x-axis.

    3. The attempt at a solution

    I am not sure how to go about this one... I know that x=9y is a line in the xy plane that crosses through the origin, so rotating it about the x-axis will create a double cone, but I don't know how to find its equation.

    And hints would be appreciated
     
  2. jcsd
  3. Jan 23, 2012 #2

    jedishrfu

    Staff: Mentor

    We cant solve the problem but we can give hints

    what is the equation for the surface area of a cone? what values do you need to compute the surface area from this formula?

    draw a picture of x = 9y rotated about the x-axis and identify those values and it should become obvious.
     
  4. Jan 23, 2012 #3

    ElijahRockers

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    Oops. Wrong forum, this should be in Calculus I guess. I am in Cal 3, I put it here because there is really no calculus involved in this question, it is just three dimensional geometry.

    Anyway, I already know the answer, it's

    [itex] x^2 = 81y^2 + 81z^2[/itex]

    I'm just not sure how to arrive at that. I tried to study this on my own last semester, but I still have a difficult time with these 3d objects.

    Anyhow, the equation for a surface area of a cone would be geometrically the same as the surface area of a circle, just with a little piece of the pie missing. I'm not trying to find the area of the object, I'm trying to find the equation that represents the object.

    I drew a little 3D picture, and from looking at it, I know that if I were to take cross-sections with yz planes they would look like bigger circles the further away from x=0 I got.

    Cross sections with xy planes would look like hyperbolas except at z=0, where it would just be the line x=9y.

    Cross sections with xz planes would look almost, if not identical to the xy planes.

    I also know the general form for any surface in R3 I think:

    [itex] Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 [/itex]

    That's if my memory is correct.
     
  5. Jan 23, 2012 #4
    Upon inspection, that "general equation" for a surface in ℝ3 can't be right: it's not even a surface! It's a conic section, a very specific kind of curve, in ℝ2. If you added another variable, say z, to that equation, you'd get the general form of a 2-dimensional quadric (the higher-dimensional analogue of a conic). This still leaves out many possible surfaces, e.g. z=x3+y3.

    As for the original problem, here's my hint: use the Pythagorean theorem.
     
  6. Jan 23, 2012 #5

    ElijahRockers

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    Oh, sure enough. Obviously I've got some serious issues regarding 3D objects, haha...
    Back to the text, I suppose.
     
    Last edited: Jan 23, 2012
  7. Jan 24, 2012 #6

    HallsofIvy

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    It should be clear from the start that rotating a line around an axis gives a cone.

    I would set up parametric equations with r= y and [itex]\theta[/itex] the angle rotated around the x axis. The equation x= 9y becomes [itex]x= 9r[/itex] while y and z being in the plane of rotation, [itex]y= rcos(\theta)[/itex] and [itex]z= rsin(\theta)[/itex]. We can go back to Cartesian coordinates by noting that
    [tex]y^2+ z^2= r^2= x^2/81[/tex]
     
  8. Jan 24, 2012 #7

    ElijahRockers

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    Interesting, I will take a look at it from that angle, thank you so much!
     
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