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February 17th, 2018, 08:45 PM   #1
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Hi! I'm new here.

I wanted help on a simple "Quadric Surface" Question from calculus 3. Our teacher in college didn't explained it in the class.

Convert it into the standard form and sketch it:

$x^2+y^2+z^2=3z$

Thanks!

Last edited by skipjack; February 17th, 2018 at 08:49 PM.
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February 17th, 2018, 09:37 PM   #2
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$x^2 + y^2 + z^2 = 3z$

$x^2 + y^2 + z^2 -3z=0$

$x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 - \left(\dfrac 3 2\right)^2=0$

$x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 =\left(\dfrac 3 2\right)^2$

can you identify what surface this is, it's center, and it's dimensions?
Thanks from greg1313, Country Boy and hunter6
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February 18th, 2018, 02:09 AM   #3
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Quote:
Originally Posted by romsek View Post
$x^2 + y^2 + z^2 = 3z$

$x^2 + y^2 + z^2 -3z=0$

$x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 - \left(\dfrac 3 2\right)^2=0$

$x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 =\left(\dfrac 3 2\right)^2$

can you identify what surface this is, it's center, and it's dimensions?
Thanks man , I got it !
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