Calculus Calculus Math Forum

 February 17th, 2018, 09:45 PM #1 Newbie   Joined: Feb 2018 From: Pakistan Posts: 4 Thanks: 0 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 09:49 PM. February 17th, 2018, 10:37 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,648 Thanks: 1476 $x^2 + y^2 + z^2 = 3z$ $x^2 + y^2 + z^2 -3z=0$ $x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 - \left(\dfrac 3 2\right)^2=0$ $x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 =\left(\dfrac 3 2\right)^2$ can you identify what surface this is, it's center, and it's dimensions? Thanks from greg1313, Country Boy and hunter6 February 18th, 2018, 03:09 AM   #3
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Quote:
 Originally Posted by romsek $x^2 + y^2 + z^2 = 3z$ $x^2 + y^2 + z^2 -3z=0$ $x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 - \left(\dfrac 3 2\right)^2=0$ $x^2 + y^2 + \left(z-\dfrac 3 2\right)^2 =\left(\dfrac 3 2\right)^2$ can you identify what surface this is, it's center, and it's dimensions?
Thanks man , I got it ! Tags calculus, quadric, question, surface Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Smartmanmaths Geometry 1 November 6th, 2017 02:22 PM keishajc Calculus 3 April 23rd, 2017 09:03 AM SirSam Calculus 3 February 23rd, 2017 09:02 AM Jhenrique Calculus 2 December 27th, 2013 08:14 PM aaron-math Calculus 6 February 12th, 2012 08:19 PM

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