How to draw the following volume (not using matlab)

Discussion in 'Math' started by u-will-neva-no, Mar 28, 2012.

  1. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Hey again!

    I have two surfaces: z = \sqrt{2}(x^2+y^2)  (1) and  x^2 + y^2 + z^2 = 1  (2)

    now I know that (1) is the shape of a paraboloid and (2) is a sphere but how do the two look together?

    I put x =0 and get z = \sqrt{2} and z = \frac{-1}{\sqrt{2}} but for y=0 I get an imaginary value for the x value (and also 4 solutions which is strange)
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    PM 1Chance...She can help.
     
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  3. u-will-neva-no

    Thread Starter Member

    Mar 22, 2011
    230
    2
    Sorry, I don't understand what you wrote.
     
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    1Chance is the math wiz of AAC.
     
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  5. MrChips

    Moderator

    Oct 2, 2009
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    You know the equation of a circle is
    x^2 + y^2 = r^2

    Thus,
    x^2 + y^2 = 1
    is the equation of a circle with radius = 1


    Similarly,
    x^2 + y^2 + z^2= 1
    is the equation of a sphere with radius = 1

    See if you can figure out the shape of
    x^2 + y^2 = z
     
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  6. panic mode

    Senior Member

    Oct 10, 2011
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    1) z=sqrt(2)(x^2+y^2)
    2) x^2+y^2+z^2=1


    rewrite last one as
    x^2+y^2=1-z^2
    and substitute in first one to get quadratic equation
    z=sqrt(2)(1-z^2)

    with solutions z=1/sqrt(2) and z=-sqrt(2)
    which are z values where surfaces intersect.

    use that z-value in both equations to get curves of intersection (circles).

    then freeze any one of variables to get section view.

    with few points calculated, just connect the dots.
     
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  7. MrChips

    Moderator

    Oct 2, 2009
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    just put square braces around tex
    in front of the equation
    and close with square braces around /tex

    like this

    {tex}z=sqrt2(1-z^2){/tex}

    but replace { } with [ ]

    and you will get

    z=sqrt2(1-z^2)
     
  8. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    test:

     sqrt(1-z^2)

     sqrt x^5

     c=sqrt (a^2+b^2)

    cool ;-)

    thanks
     
  9. Georacer

    Moderator

    Nov 25, 2009
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    panic mode likes this.
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