April 17th, 2017, 05:13 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 175 Thanks: 2  3D polar coordinates problem
Find the volume below z = r, above the xy plane, and inside . Answer: 4/9 Is there anyone online tool I could use to plot the 3d polar coordinates? I couldn't approach the problem without any visual clues... 
April 17th, 2017, 06:00 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
You can use desmos graphics calculator at https://www.desmos.com/calculator If you click on the wrench icon ("Graph Settings") then "Grid" and click on the polar coordinates icon. 
April 17th, 2017, 06:01 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 
I use the Mathematica function ParametricPlot3D to plot all these things. There is limited access to it at wolframalpha.com You'll have to read up on how to use it. this one is $\displaystyle \int_{\frac{\pi}{2}}^{\frac{\pi}{2}} ~\int_0^{\cos(\theta)} \int_0^r~r~dz~dr~d\theta$ 
April 18th, 2017, 05:04 PM  #4  
Senior Member Joined: Jan 2017 From: Toronto Posts: 175 Thanks: 2  Quote:
How did you come up the boundaries (why pi/2 and pi/2)? It would be great if you could explain a little... Last edited by zollen; April 18th, 2017 at 05:07 PM.  
April 18th, 2017, 05:07 PM  #5 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011  
April 18th, 2017, 05:10 PM  #6 
Senior Member Joined: Jan 2017 From: Toronto Posts: 175 Thanks: 2 
I don't know how to plot z=r polar coordinate in 3D...

April 18th, 2017, 05:12 PM  #7 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011  
April 18th, 2017, 05:29 PM  #8 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011  
April 18th, 2017, 05:38 PM  #9 
Senior Member Joined: Jan 2017 From: Toronto Posts: 175 Thanks: 2 
Thanks. I am going to study your example closely..

April 18th, 2017, 05:54 PM  #10 
Senior Member Joined: Sep 2015 From: USA Posts: 1,944 Thanks: 1011 
this will probably also help $r = \cos(\theta)$ $r^2 = r \cos(\theta)$ wave the magic wand $x^2 + y^2 = x$ $x^2  x + y^2 = 0$ $\left(x\dfrac 1 2\right)^2 +y^2 = \dfrac 1 4$ i.e. circle at $\left( \dfrac 1 2 , 0\right)$ with radius $\dfrac 1 2$ 

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coordinates, polar, problem 
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