April 17th, 2017, 05:13 PM  #1 
Member Joined: Jan 2017 From: Toronto Posts: 82 Thanks: 1  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: 2,524 Thanks: 641 
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: CA Posts: 1,265 Thanks: 650 
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  
Member Joined: Jan 2017 From: Toronto Posts: 82 Thanks: 1  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: CA Posts: 1,265 Thanks: 650  
April 18th, 2017, 05:10 PM  #6 
Member Joined: Jan 2017 From: Toronto Posts: 82 Thanks: 1 
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: CA Posts: 1,265 Thanks: 650  
April 18th, 2017, 05:29 PM  #8 
Senior Member Joined: Sep 2015 From: CA Posts: 1,265 Thanks: 650  
April 18th, 2017, 05:38 PM  #9 
Member Joined: Jan 2017 From: Toronto Posts: 82 Thanks: 1 
Thanks. I am going to study your example closely..

April 18th, 2017, 05:54 PM  #10 
Senior Member Joined: Sep 2015 From: CA Posts: 1,265 Thanks: 650 
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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